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- William Dean Clark
- Sandra K. McCune

*Table of contents : ContentsPreface1 Systems of linear equations and matrices Systems of linear equations General systems of linear equations Matrices Row transformations and equivalence of matrices Row-echelon form Homogeneous systems2 Matrix algebra Matrix arithmetic Inverse of a square matrix Properties of invertible matrices Matrix solutions of systems of linear equations Transpose of a matrix3 Graphing calculators and matrices Matrix menu Inputting and editing a matrix Matrix arithmetic Calculating determinants Transpose of a matrix Solving linear systems using Gauss-Jordan elimination Solving linear systems using X = A[sup(–1)]C4 Special types of square matrices Nonsingular matrices Triangular, diagonal, and scalar matrices Involutory, idempotent, and nilpotent matrices Symmetric and skew-symmetric matrices Orthogonal matrices Hermitian and skew-Hermitian matrices5 Determinants Determinant of a square matrix Cramer’s rule Properties of determinant6 Vectors in R[sup(n)] Vectors in two dimensions Dot product of vectors Vectors in R[sup(n)] Vectors as matrices7 Vector spaces Definitions and terminology of vector spaces Linear independence Basis Dimension Row space, column space, and null space Rank and nullity8 Inner product spaces Definition and terminology for inner product spaces Norm of a vector in an inner product space Cauchy-Schwarz inequality and properties of the norm Orthogonality in inner product spaces Gram-Schmidt procedure9 Linear transformations Definition and terminology for linear transformations Kernel and image of a linear transformation Matrix representations of linear transformations Change of basis Algebra of linear transformations Linear operators on R2 and R310 Eigenvalues and eigenvectors The eigenvalue problem Useful properties of eigenvalues DiagonalizationAnswer key*

PRACTICE MAKES PERFECT ®

Linear Algebra

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PRACTICE MAKES PERFECT ®

Linear Algebra William D. Clark, PhD, and Sandra Luna McCune, PhD

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Copyright © 2013 by The McGraw-Hill Companies, Inc. All rights reserved. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher. ISBN: 978-0-07-177844-2 MHID: 0-07-177844-6 The material in this eBook also appears in the print version of this title: ISBN: 978-0-07-177843-5, MHID: 0-07-177843-8 . All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention of infringement of the trademark. Where such designations appear in this book, they have been printed with initial caps. McGraw-Hill eBooks are available at special quantity discounts to use as premiums and sales promotions, or for use in corporate training programs. To contact a representative please e-mail us at [email protected]. McGraw-Hill, the McGraw-Hill Publishing logo, Practice Makes Perfect, and related trade dress are trademarks or registered trademarks of The McGraw-Hill Companies and/or its affiliates in the United States and other countries and may not be used without written permission. All other trademarks are the property of their respective owners. The McGraw-Hill Companies is not associated with any product or vendor mentioned in this book TERMS OF USE This is a copyrighted work and The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and its licensors reserve all rights in and to the work. Use of this work is subject to these terms. Except as permitted under the Copyright Act of 1976 and the right to store and retrieve one copy of the work, you may not decompile, disassemble, reverse engineer, reproduce, modify, create derivative works based upon, transmit, distribute, disseminate, sell, publish or sublicense the work or any part of it without McGraw-Hill’s prior consent. You may use the work for your own noncommercial and personal use; any other use of the work is strictly prohibited. Your right to use the work may be terminated if you fail to comply with these terms. THE WORK IS PROVIDED “AS IS.” McGRAW-HILL AND ITS LICENSORS MAKE NO GUARANTEES OR WARRANTIES AS TO THE ACCURACY, ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM USING THE WORK, INCLUDING ANY INFORMATION THAT CAN BE ACCESSED THROUGH THE WORK VIA HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WARRANTY, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. McGraw-Hill and its licensors do not warrant or guarantee that the functions contained in the work will meet your requirements or that its operation will be uninterrupted or error free. Neither McGraw-Hill nor its licensors shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any damages resulting therefrom. McGraw-Hill has no responsibility for the content of any information accessed through the work. Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise.

Contents

Preface vii

1

Systems of linear equations and matrices 1 Systems of linear equations 1 General systems of linear equations 4 Matrices 5 Row transformations and equivalence of matrices 6 Row-echelon form 8 Homogeneous systems 12

2

Matrix algebra 14 Matrix arithmetic 14 Inverse of a square matrix 18 Properties of invertible matrices 19 Matrix solutions of systems of linear equations 21 Transpose of a matrix 23

3

Graphing calculators and matrices 25 Matrix menu 25 Inputting and editing a matrix 27 Matrix arithmetic 29 Calculating determinants 32 Transpose of a matrix 33 Solving linear systems using Gauss-Jordan elimination 34 Solving linear systems using X = A–1C 37

4

Special types of square matrices 42 Nonsingular matrices 42 Triangular, diagonal, and scalar matrices 46 Involutory, idempotent, and nilpotent matrices 52 Symmetric and skew-symmetric matrices 53 Orthogonal matrices 56 Hermitian and skew-Hermitian matrices 57

v

5

Determinants 59 Determinant of a square matrix 59 Cramer’s rule 62 Properties of determinant 65

6

Vectors in Rn 68 Vectors in two dimensions 68 Dot product of vectors 72 Vectors in R n 73 Vectors as matrices 75

7

Vector spaces 77 Definitions and terminology of vector spaces 77 Linear independence 81 Basis 83 Dimension 86 Row space, column space, and null space 88 Rank and nullity 92

8

Inner product spaces 97 Definition and terminology for inner product spaces 97 Norm of a vector in an inner product space 100 Cauchy-Schwarz inequality and properties of the norm 103 Orthogonality in inner product spaces 104 Gram-Schmidt procedure 108

9

Linear transformations 110 Definition and terminology for linear transformations 110 Kernel and image of a linear transformation 112 Matrix representations of linear transformations 114 Change of basis 120 Algebra of linear transformations 122 Linear operators on R 2 and R 3 126

10 Eigenvalues and eigenvectors 130 The eigenvalue problem 130 Useful properties of eigenvalues 135 Diagonalization 140 Answer key 144

vi

Contents

Preface

Linear algebra is a subject that has application in a broad spectrum of fields including, for example, the natural sciences, engineering, economics, computer science, cryptography, and other branches of mathematics. Practice Makes Perfect: Linear Algebra is designed to help you to be successful in learning this interesting and practical subject matter. However, the book is not intended to introduce concepts, but rather its primary aim is to reinforce ideas and concepts that you have previously encountered. The topics presented are those that a competent user of linear algebra needs to know. You will find this practice study guide to be a useful supplementary text for your linear algebra course. It can also serve as a refresher text if you are using it to review previously learned linear algebra concepts and techniques. Like most topics worth knowing, learning linear algebra requires diligence and hard work. The foremost purpose of Practice Makes Perfect: Linear Algebra is as a source of solved linear algebra problems. We believe that the best way to develop understanding, while, at the same time, acquiring accuracy and speed in linear algebra skills is to work numerous practice exercises. This book has more than 500 practice exercises from beginning to end. A variety of exercises and levels of difficulty are presented to provide reinforcement of linear algebra knowledge, understanding, and skills. In each chapter, a concept discussion followed by example problems precedes each set of exercises to serve as a concise review for readers already familiar with the topics covered. Concepts are broken into basic components to provide ample practice of fundamental skills. To use Practice Makes Perfect: Linear Algebra in the most effective way, it is important that you work through every exercise. After working an exercise set, use the worked-out solutions to check your understanding of the concepts. We sincerely hope this book will help you acquire greater competence and confidence in using linear algebra in your future mathematical endeavors.

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PRACTICE MAKES PERFECT ®

Linear Algebra

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Systems of linear equations and matrices

·1·

In this chapter, you will learn: ◆ ◆ ◆ ◆ ◆ ◆

Systems of linear equations General systems of linear equations Matrices Row transformations and equivalence of matrices Row-echelon form Homogeneous systems

This chapter provides some basic terminology associated with systems of linear equations and matrices and presents methods for solving linear systems. As you will see, systems of linear equations and matrices go hand-in-hand.

Systems of linear equations A linear equation is an equation of the form a1 x1 + a2 x 2 + $ + an xn = c1 where there are n unknowns (variables), x1 , x 2 , # , xn , and a1 , a2 , # , an and c1 are known constants. An equation is linear if all the variables occur to the first power only and if there are no products or quotients of variables. A solution of a linear equation a1 x1 + a2 x 2 + $ + an xn = c1 in n unknowns is an ordered list of n numbers s1 , s2 , # , sn that satisfy the equation when x1 = s1, x 2 = s2 , . . . , xn = sn. In linear algebra you will need to simultaneously solve linear equations in several variables. You likely have seen two basic algebraic methods of solving simultaneous equations. An example of each method is shown for a quick review. The equation of a linear function in two variables has several forms but for the purposes of this chapter, the form ax + by = c is preferred. You know from elementary algebra that the graph of such a linear equation is a line. When you encounter two linear equations, the basic question underlying a simultaneous solution is “What are the coordinates of the point of intersection, if any, of the two lines that are the graphs of the two equations?” This question has three possible answers. If the two lines do not intersect, there is no solution; if the two lines intersect, there is only one solution, an ordered pair ( x , y ); and if the two lines are equivalent versions of the same line, then there are infinitely many solutions: an infinite set of ordered pairs. This same idea carries over to systems of m equations in n unknowns. That is, Given a system of m equations and n unknowns, only one of the following three possibilities for a solution to the system occurs: 1. There is no solution. 2. There is exactly one solution. 3. There are infinitely many solutions.

1

If there is no solution to the system, the system is inconsistent. If there is at least one solution to the system, the system is consistent. When you are solving two linear equations in two variables (a 2 × 2 system), an example of the standard form of writing them together is 2x − y = 1 x+y=2

Figures 1.1, 1.2, and 1.3 graphically show the three possible outcomes for a 2 × 2 system of equations. 4 3 2

2x – y = 1 x+y=2

1

–2

–1

1

2

3

4

5

–1 –2

Figure 1.1 Graph of a system with exactly one solution (intersecting lines). 4 3 x–y=1 2 x – y = –1 1 –5

–4

–3

–2

–1

1

2

3

4

5

–1 –2 –3

Figure 1.2 Graph of an inconsistent system (parallel lines). 2 2x – 6y = 6 –x + 3y = –3

1

–3

–2

–1

1

2

3

4

–1 –2 –3

Figure 1.3 Graph of a system with infinitely many solutions (same line).

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5

You solve the 2 × 2 system when you answer the question: “What are the coordinates of the point of intersection, if any, of the two lines that are the graphs of the two equations?” Method 1: Substitution. This method requires you to solve one equation for one of the variables in terms of the other variable, and then use substitution to solve the system. PROBLEM SOLUTION

2x − y = 0 x+y=3 Solve the first equation for y in terms of x:

Solve the system: 2x − y = 0 2x = y

Substitute 2x for y into the second equation and solve for x: x + (2 x ) = 3 x =1

Substitute 1 for x into the second equation and solve for y: 1+ y = 3 y=2

The solution is x = 1 and y = 2. That is, the two lines intersect at the point (1, 2).

Method 2: Elimination of a variable. This method involves multiplying the equations by constants to create opposites as coefficients of one variable so that it can be eliminated by adding the two equations. PROBLEM

SOLUTION

Solve the system. 5x + 2 y = 3 2 x + 3 y = −1 To eliminate y, multiply the first equation by 3 and the second equation by −2: 5x + 2 y = 3 2 x + 3 y = −1

Multiply by 3

⎯⎯⎯⎯⎯⎯ → ⎯Multiply ⎯⎯⎯⎯⎯ by −2→

15 x + 6 y = 9 −4 x − 6 y = 2

Add the resulting two equations and solve for x: 15 x + 6 y = 9 −4 x − 6 y = 2 11x = 11 x =1

Substitute 1 for x into one of the original equations and solve for y: 5(1)+ 2 y = 3 y = −1

The solution is x = 1 and y = −1. That is, the two lines intersect at the point (1, − 1).

You can use the two methods exemplified above to solve systems of several linear equations in several variables, but linear algebra is a study of more powerful ways of solving systems of equations. Solving systems of equations is elevated to answering more sophisticated questions Systems of linear equations and matrices

3

than “Where do lines intersect?” Moreover, solving such systems is basic to the study of linear algebra. Chapter 2 presents a discussion of the algebra of matrices and methods of solving simultaneous equations using that algebra. As you might expect, the solution(s) of 2 × 2 systems are ordered pairs; 3 × 3 systems, ordered triples; 4 × 4 systems, ordered quadruples; and, in general, n × n systems, ordered n-tuples. EXERCISE

1·1 For 1 and 2, solve the system by the substitution method.

1.

x+y =3 −2 x + y = −3

2.

5x − 2 y = 3 x + 6y = 1

For 3–5, solve the system using the elimination of variable technique.

3.

x − 2y = 7 3x + 2 y = 5

4.

2x − 3 y = 16 5x − 2y = −4

5.

3x − 2 y = 3 −6 x + 4 y = −8

General systems of linear equations The simple 1 × 2 linear system ax + by = c with nonzero coefficients has infinitely many solutions a c whose graph is a straight line. The solutions can be written as ( x , y ), where y = − x + . The b b variable x is a free variable, meaning its value is unrestricted. The variable y is dependent on x because substituting any value for x will produce a corresponding value for y. By letting x = t , you a c can represent the solutions in parametric form as x = t, y = − t + , where the arbitrary number t b b is called a parameter. Similar parametric substitutions are made if you have several free variables. PROBLEM SOLUTION

Solve the following system: −2 x + 5 y − 4 z = 0 Solve for one variable in terms of the other two: 5 x = y − 2z , where y and z are free variables. In parametric form, the solution is 2 5 x = s − 2t , y = s, and z = t, where s and t are arbitrary. 2

A system of linear equations can have any number of equations and any number of variables. The solution to many such systems involves free variables.

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Linear Algebra

Here are examples of systems of linear equations. 2 x − 3 y = 10 5x + 2 y = 6

3x − y + 2z = 4 2x − 3z = 8 4 x − 5 y − 6z = 3

2 x1 − 3 x 2 + 5 x 3 + x 4 = 1 x1 + 2 x 2 =4 −2 x1 + 4 x 2 − 6 x 3 + 2 x 4 = 0

EXERCISE

1·2 Solve the system of equations.

1. 3 x − 5 y = 8 2.

4. x + 2 y + 3 z + 4w = 0

2 x − 3y = 4 x − 2y = 6

5.

3x − 2 y = 0 6x − 4y = 1

3. 2 x + 3 y − z = 5

Matrices A matrix is a rectangular array of elements. Matrices are used as convenient ways of organizing selected sets of data. A matrix consists of rows and columns of elements within brackets and its size (dimensions) is determined by the number of rows and columns. The notation m × n (read “m by n”) is used to indicate a matrix’s size where m is its number of rows and n is its number of columns. For example the following is a 2 × 3 matrix. The subscript on the brackets is a useful way of indicating a matrix’s size. ⎡3 0 4 ⎤ ⎢ ⎥ ⎣ 6 −2 7 ⎦ 2×3 If a matrix has only one row or one column, it is a row matrix or a column matrix. Row and column matrices are also called row and column vectors, respectively. A matrix is a square matrix if the number of rows is equal to the number of columns. The main diagonal of a square matrix is the diagonal from the upper left element to the lower right element. An identity matrix, denoted by the letter I, is a square matrix all of whose main diagonal elements ⎡1 0 ⎤ are 1s and the remaining elements are 0s. For example, I 2×2 = ⎢ ⎥ is the 2 × 2 identity matrix. ⎣0 1 ⎦ Note: Commonly, a matrix’s main diagonal is called simply its diagonal. A matrix all of whose elements are 0s is called a zero matrix, 0m×n . Two matrices are equal if and only if they are the same size and their corresponding elements are equal. For example, ⎡ 4 ⎡ 2 8⎤ ⎢ 2 ⎢ ⎥=⎢ ⎣ −3 1 ⎦ ⎢ −21 ⎢⎣ 7

⎤ 64 ⎥ ⎥, 1 ⎥⎥ ⎦

⎡1 ⎤ ⎡0 0 0 ⎤ ⎡0 0 ⎤ ⎢ ⎢ ⎥ ⎥ ⎢ ⎥ ≠ ⎢ 0 0 0 ⎥ , and [1 4 6] ≠ ⎢ 4 ⎥ 0 0 ⎣ ⎦ ⎢0 0 0 ⎥ ⎢⎣ 6 ⎥⎦ ⎦ ⎣

Systems of linear equations and matrices

5

EXERCISE

1·3 1. Exhibit the 3 × 1 column vector whose elements are 1, 5, and 7, in this order. 2. Exhibit the I3×3 matrix.

⎡ 1 2⎤ ⎥ ⎢ 3. What is the size of the matrix ⎢ 0 0 ⎥ ? ⎢⎣ 4 5 ⎥⎦ 4. Exhibit the 04× 4 matrix. 5. Exhibit the 1× 3 row vector whose elements are 2, 0, and 9, in this order.

Row transformations and equivalence of matrices Two matrices are equivalent if one is obtained from the other by using one or more of the following row transformations: 1. Multiply each element of a row by the same nonzero constant. 2. Interchange any two rows. 3. Add a nonzero constant multiple of the elements of one row to the corresponding elements

of another row. Here are convenient notations for the three types of row transformations: 1. Multiply row i by the constant c: cRi 2. Interchange rows i and j: Ri ↔ R j 3. Add c times row i to row j: cRi + R j

Note: Equivalence is not to be confused with equality. The symbol ∼ is used to indicate equivalence. Column transformations are similar, but only row transformations will be used for the examples and problems. These concepts are presented at this time because judicious use of these ideas enable you to reduce a matrix that represents a given system of equations to an equivalent matrix in which the solution to its corresponding reduced system of equations is obvious or at least simpler to solve. Systems of equations corresponding to equivalent matrix representations all have the same solution set, so the solution to the reduced system of equations is the solution to the original system as well. For the system of equations

a1 x + b1 y = c1

⎡a b ⎤ , the matrix ⎢ 1 1 ⎥ is the coefficient matrix and a2 x + b2 y = c2 ⎣a2 b2 ⎦

⎡a b c ⎤ the matrix ⎢ 1 1 1 ⎥ is the augmented matrix of the system. Augmented matrices for larger ⎣a2 b2 c2 ⎦ systems of equations are similarly constructed. The following examples illustrate the reduction process.

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Linear Algebra

PROBLEM

SOLUTION

Solve the system:

3x − 2 y = 4 2x − y = 3

⎡3 Form the augmented matrix ⎢ ⎣2 ⎡ 3 −2 4 ⎤ ⎡ −1 0 2 R2 + R1 ⎢ ⎢ ⎥ − 2 − 1 3 ⎣ ⎦ ⎣ 2 −1

−2 4 ⎤ ⎥ and do the indicated row transformations: −1 3 ⎦ ⎡ −1 0 −2 ⎤ −2 ⎤ R1 ⎥ −1 ⎥ 2 R1 + R2 ⎢ 3 ⎦ ⎣ 0 −1 −1 ⎦

⎡1 0 2 ⎤ R2 ⎢ and −1 ⎥ ⎣0 1 1 ⎦

The reduced equivalent system is now

x + 0y = 2 , for which the solution x = 2 0x + y = 1

and y = 1 is obvious. Equivalent systems all have the same solution set, so this solution is the solution to the original system as well. PROBLEM

x − 2 y + 3z = 1 Solve the system: x + 3 y − z = 4 2 x + y − 2z = 13

SOLUTION

Form the augmented matrix and do the indicated row transformations: ⎡1 −2 3 1 ⎤ ⎥ ⎢ 1R1 + R2 ⎢1 3 −1 4 ⎥ − ⎢⎣ 2 1 −2 13 ⎥⎦

⎡1 −2 3 1 ⎤ ⎥ ⎢ ⎢ 0 5 −4 3 ⎥ ⎢⎣ 2 1 −2 13 ⎥⎦

⎡1 −2 3 1 ⎤ ⎥ ⎢ − 2 R1 + R3 ⎢0 5 −4 3 ⎥ ⎢⎣0 5 −8 11⎥⎦ ⎡1 −2 3 1 ⎤ ⎥ ⎢ − 1R2 + R3 ⎢0 5 −4 3 ⎥ ⎢⎣0 0 −4 8 ⎥⎦ ⎡1 −2 3 1 ⎤ ⎥ ⎢ − 1R3 + R2 ⎢0 5 0 −5 ⎥ ⎢⎣0 0 −4 8 ⎥⎦ ⎡1 −2 3 1 ⎤ 1 1 ⎥ ⎢ R2 and − R3 ⎢0 1 0 −1 ⎥ 4 5 ⎢⎣0 0 1 −2 ⎥⎦ ⎡1 0 3 −1 ⎤ ⎥ ⎢ 2 R2 + R1 ⎢0 1 0 −1 ⎥ ⎢⎣0 0 1 −2 ⎥⎦ ⎡1 0 0 5 ⎤ ⎥ ⎢ − 3 + R R 3 1 ⎢0 1 0 −1 ⎥ ⎢⎣0 0 1 −2 ⎥⎦

The solution is x = 5, y = −1, and z = −2. You can check the solution by substituting these values into the original equations. The calculations (5) − 2(−1) + 3(−2) = 1, (5) + 3(−1) − (−2) = 4, and 2(5) + (−1) − 2(−2) = 13 verify the solution. Systems of linear equations and matrices

7

The underlying idea behind using row transformations is to transform the coefficient matrix as closely as possible to an upper triangular matrix whose diagonal elements are all 1s. This result will then make the solution simpler to determine. This idea is formalized in the next section. (See Chapter 4 for a discussion of upper triangular matrices). Note: Graphing calculators have built-in functions that will do row transformations. Intelligent use of these calculators will save valuable time and take the tedium out of calculations. Solving simultaneous equations with the use of a graphing calculator is detailed in Chapter 3. EXERCISE

1·4 Solve using row transformations.

1.

3x + 2 y = 4 x − 2y = 3

2.

5x − y = 3 2x + 2y = 2

3.

3x + 2 y = 4 x + 2y = 3

x − 3y + z = 2 4. 2 x − y − 2 z = 1 3x + 2 y − z = 5 =0 2x − y 2y − z = 0 5. x + 2y − z = 3

Row-echelon form A matrix is in row-echelon form if it satisfies the following three conditions. 1. If there are any rows of all zeros, then they are the lower (bottom) rows of the matrix. 2. If a row does not consist of all zeros, then its first (leftmost) nonzero entry is 1. This 1 is

called a leading 1. 3. In any two successive rows, neither of which consists of all zeros, the leading 1 of the

lower row is to the right of the leading 1 of the row above. A matrix is in reduced row-echelon form if it satisfies items 1–3 and in addition satisfies the following fourth condition. 4. If a column contains a leading 1, then all the other entries of that column are zero.

Reducing the augmented matrix for a system to row-echelon form is called Gaussian elimination. Continuing the reduction to reduced row-echelon form is called Gauss-Jordan elimination.

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Linear Algebra

⎡1 ⎢ 0 The matrix ⎢ ⎢0 ⎢ ⎣0

2 1 0 0

6 4 1 0

0⎤ ⎥ 2⎥ is in row-echelon form and the matrix 2⎥ ⎥ 0⎦

⎡1 0 4 ⎤ ⎢ ⎥ ⎢0 1 3 ⎥ is in reduced ⎢⎣0 0 0 ⎥⎦

row-echelon form. These forms are desired because the solutions of augmented systems are readily obtained if the system has been reduced to one of these forms. PROBLEM

SOLUTION

−2 x + y − z = 4 Use Gauss-Jordan elimination to solve the system: 3x + z = −1 x + 2 y + 3z = 13 ⎡ −2 1 −1 4 ⎤ ⎥ ⎢ Form the augmented matrix ⎢ 3 0 1 −1⎥ and do the indicated row transformations: ⎢⎣ 1 2 3 13 ⎥⎦ ⎡ −2 1 −1 4 ⎤ ⎥ ⎢ ↔ R3 1 ⎢ 3 0 1 −1⎥ R ⎢⎣ 1 2 3 13 ⎥⎦ ⎡1 2 3 13 ⎤ ⎥ ⎢ 2 R1 + R3 ⎢0 −6 −8 −40 ⎥ ⎢⎣0 5 5 30 ⎥⎦ ⎡1 2 3 13 ⎤ ⎥ ⎢ ⎢0 1 1 6 ⎥ ⎢⎣0 −6 −8 −40 ⎥⎦ ⎡1 2 3 13 ⎤ ⎥ ⎢ ⎢0 1 0 4 ⎥ ⎢⎣0 0 1 2 ⎥⎦

⎡ 1 2 3 13 ⎤ ⎥ ⎢ 3R1 + R2 ⎢ 3 0 1 −1⎥ − ⎢⎣ −2 1 −1 4 ⎥⎦ ⎡1 2 3 13 ⎤ 1 ⎥ ⎢ R3 ⎢0 −6 −8 −40 ⎥ 5 ⎢⎣0 1 1 6 ⎥⎦

⎡1 2 3 13 ⎤ ⎥ ⎢ 6 R2 + R3 ⎢0 1 1 6 ⎥ ⎢⎣0 0 −2 −4 ⎥⎦

⎡1 0 3 5 ⎤ ⎥ ⎢ − 2 R2 + R1 ⎢0 1 0 4 ⎥ ⎢⎣0 0 1 2 ⎥⎦

⎡ 1 2 3 13 ⎤ ⎥ ⎢ ⎢ 0 −6 −8 −40 ⎥ ⎢⎣ −2 1 −1 4 ⎥⎦

R ↔ R3 2

⎡1 2 3 13 ⎤ 1 − R3 ⎢⎢0 1 1 6 ⎥⎥ −1R3 + R2 2 ⎢⎣0 0 1 2 ⎥⎦

⎡1 0 0 −1⎤ ⎥ ⎢ − 3R3 + R1 ⎢0 1 0 4 ⎥ ⎢⎣0 0 1 2 ⎥⎦

x = −1 ⇒ y=4 z=2

Note: Read the symbol “ ⇒ ” as implies that. Remark: This transformation process can be shortened at times by doing two things at a time, but doing more than two things at a time is discouraged due to the greater chance of errors. PROBLEM

Solve the given system. x + 2 y + 2z = 1 a. 3x + 5 y =0 2x + 3 y − z = 3

b.

−2 x + 5 y + 4 z = 0 −3x + 4 y − 8z = 0

x + y + 2z = 1 + z =3 c. x 2 x + y + 3z = 5 x + y + 2z = 1 z =3 d. x + 2 x + y + 3z = 4 Systems of linear equations and matrices

9

SOLUTION

x + 2 y + 2z = 1 a. 3x + 5 y =0 2x + 3 y − z = 3

Form the augmented matrix and do the indicated row transformations: ⎡1 2 2 1 ⎤ ⎥ −3R1 + R2 ⎢ ⎢ 3 5 0 0 ⎥ −2 R + R 1 3 ⎢⎣ 2 3 −1 3 ⎥⎦

⎡1 2 2 1 ⎤ ⎡1 2 2 1 ⎤ ⎥ −1R2 ⎢ ⎥ ⎢ 1R2 + R3 ⎢0 −1 −6 −3 ⎥ −1R ⎢0 1 6 3 ⎥ − 3 ⎢⎣0 −1 −5 1 ⎥⎦ ⎢⎣0 1 5 −1⎥⎦

⎡1 2 2 1 ⎤ ⎥ 6 R3 + R2 ⎢ ⎢0 1 6 3 ⎥ −1R 3 ⎢⎣0 0 −1 −4 ⎥⎦

⎡1 2 2 1 ⎤ ⎥ ⎢ 2 R3 + R1 ⎢0 1 0 −21⎥ − ⎢⎣0 0 1 4 ⎥⎦

⎡1 2 0 −7 ⎤ ⎥ ⎢ 2 R2 + R1 ⎢0 1 0 −21⎥ − ⎢⎣0 0 1 4 ⎥⎦

x = 35 ⎡1 0 0 35 ⎤ ⎥ ⇒ y = −21 ⎢ ⎢0 1 0 −21⎥ z=4 ⎢⎣0 0 1 4 ⎥⎦

b.

−2 x + 5 y + 4 z = 0 −3x + 4 y − 8z = 0

Form the augmented matrix and do the indicated row transformations: ⎡ −2 5 4 0 ⎤ ⎡1 1R2 + R1 ⎢ ⎢ ⎥ − ⎣ −3 4 −8 0 ⎦ ⎣ −3 ⎡1 1 ⎡1 1 12 0 ⎤ − 1R2 + R1 ⎢ R2 ⎢ ⎥ 7 ⎣0 1 4 0 ⎦ ⎣0

⎡1 1 12 0 ⎤ 1 12 0 ⎤ R1 +R2 ⎢ ⎥ 3 ⎥ 4 −8 0 ⎦ ⎣0 7 28 0 ⎦ x = −8z 0 8 0⎤ ⎥ ⇒ y = −4 z , where z is a free variable. 1 4 0⎦

There are infinitely many solutions. For instance, one solution is z = −1, x = 8, y = 4 . x + y + 2z = 1 + z =3 c. x 2 x + y + 3z = 5

Form the augmented matrix and do the indicated row transformations: ⎡1 1 2 1 ⎤ ⎥ −1R1 + R2 ⎢ ⎢1 0 1 3 ⎥ −2 R + R 1 3 ⎢⎣ 2 1 3 5 ⎥⎦

⎡1 1 2 1 ⎤ ⎥ ⎢ 1R2 + R3 ⎢0 −1 −1 2 ⎥ − ⎢⎣0 −1 −1 3 ⎥⎦

⎡1 1 2 1 ⎤ ⎥ ⎢ ⎢0 −1 −1 2 ⎥ ⎢⎣0 0 0 1 ⎥⎦

There is no solution because the last row corresponds to the equation 0 x + 0 y + 0z = 1, which of course is not valid.

Tip: Whenever a row of zeros appears in the coefficient matrix and a nonzero element is the constant in that row, the system is inconsistent. x + y + 2z = 1 + z =3 d. x 2 x + y + 3z = 4

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practice makes perfect

Linear Algebra

Form the augmented matrix and do the indicated row transformations: ⎡1 1 2 1 ⎤ ⎥ −1R1 + R2 ⎢ ⎢1 0 1 3 ⎥ −2 R + R 1 3 ⎢⎣ 2 1 3 4 ⎥⎦

⎡1 1 2 1 ⎤ ⎥ ⎢ 1R2 + R3 ⎢0 −1 −1 2 ⎥ − ⎢⎣0 −1 −1 2 ⎥⎦

⎡1 1 2 1 ⎤ ⎥ 1R2 + R1 ⎢ ⎢0 −1 −1 2 ⎥ −1R 2 ⎢⎣0 0 0 0 ⎥⎦

⎡1 0 1 3 ⎤ x = −z + 3 ⎥ ⎢ ⎢0 1 1 −2 ⎥ ⇒ y = − z − 2, where z is a free variable (x and y are called leading ⎢⎣0 0 0 0 ⎥⎦

variables).

Note: See Chapter 3 for calculator methods for solving systems of linear equations using Gauss-Jordan elimination. EXERCISE

1·5 For 1–6, solve the system by Gauss-Jordan elimination.

1.

x+ y=4 2 x + 3 y = 10

4.

5x − 2 y + 6z = 0 −2 x + y + 3 z = 1

2.

2x − 4y = 3 x + 2y = 4

5.

6 a + 6b + 3c = 5 − 2b + 3c = 1 3a + 6b − 3c = −2

x + y + 2z = 9 3. 2 x + 4 y − 3 z = 1 3 x + 6 y − 5z = 0

2x + 2y + 4z = 2 6. −3 x − 3 z = −9 6 x + 3 y + 9 z = 15

For 7–10, suppose the augmented matrix has been reduced to row-echelon form. Solve the system. Note: Use any convenient variable names.

⎡ 1 0 0 −3 ⎤ ⎥ ⎢ 7. ⎢ 0 1 0 0 ⎥ ⎢⎣ 0 0 1 7 ⎥⎦ ⎡ 1 −3 0 0 ⎤ ⎥ ⎢ 8. ⎢ 0 0 1 0 ⎥ ⎢⎣ 0 0 0 1 ⎥⎦

⎡ 1 0 0 −7 8 ⎤ ⎢ ⎥ 9. ⎢ 0 1 0 3 2 ⎥ ⎢ 0 0 1 1 −5 ⎥ ⎦ ⎣ ⎡ 1 −6 ⎢ 0 0 10. ⎢ ⎢0 0 ⎢ ⎢⎣ 0 0

0 1 0 0

0 0 1 0

3 −2 ⎤ ⎥ 4 7⎥ 5 8⎥ ⎥ 0 0⎦

Systems of linear equations and matrices

11

Homogeneous systems A system of equations is said to be homogeneous if the constant terms are all zero. The system has the form a11 x1 + a12 x 2 + + a1n xn = 0 a21 x1 + a22 x 2 + + a2n xn = 0 am1 x1 + am 2 x 2 + + amn xn = 0

Such systems are special because of the solution possibilities. Every homogeneous system always has a solution because x1 = 0, x 2 = 0, # , xn = 0 is a solution, called the trivial solution. Any other solution is a nontrivial solution. In fact, there are only two possibilities for solutions. ◆ ◆

The system has only the trivial solution. The system has infinitely many solutions including the trivial one.

In addition, if a system has more unknowns than equations, the system will have infinitely many solutions. Systems that have only the trivial solution are among the most important systems in the study of linear algebra (see, for example, linear independence in Chapter 7). PROBLEM

Solve the given system. 2 x + y + 3z = 0 =0 a. x + 2 y y+ z =0 b.

SOLUTION

3x + y + z + w = 0 5x − y + z − w = 0

2 x + y + 3z = 0 a. x + 2 y =0 y+ z =0

Form the augmented matrix and do the indicated row transformations: ⎡2 1 3 0⎤ ⎡1 2 0 0 ⎤ ⎥ R2 ↔ R1 ⎢ ⎥ ⎢ 2 R1 + R3 1 2 0 0 ⎥ R ↔ R ⎢0 1 1 0 ⎥ − ⎢ 2 3 ⎢⎣ 0 1 1 0 ⎥⎦ ⎢⎣ 2 1 3 0 ⎥⎦ ⎡1 2 0 0 ⎤ ⎡1 2 0 0 ⎤ ⎥ 1 ⎥ −1R3 + R2 ⎢ ⎢ ⎢0 1 1 0 ⎥ 6 R3 ⎢0 1 1 0 ⎥ −2 R + R 2 1 ⎢⎣0 0 6 0 ⎥⎦ ⎢⎣0 0 1 0 ⎥⎦

b.

⎡1 2 0 0 ⎤ ⎥ ⎢ R2 + R3 ⎢0 1 1 0 ⎥ 3 ⎢⎣0 −3 3 0 ⎥⎦

⎡1 0 0 0 ⎤ x=0 ⎥ ⎢ ⎢0 1 0 0 ⎥ ⇒ y = 0 ⎢⎣0 0 1 0 ⎥⎦ z =0

3x + y + z + w = 0 5x − y + z − w = 0

Form the augmented matrix and do the indicated row transformations: ⎡3 1 1 1 0 ⎤ ⎡3 1 1 1 0⎤ 1R1 + R2 ⎢ ⎢ ⎥− ⎥ ⎣5 −1 1 −1 0 ⎦ ⎣ 2 −2 0 −2 0 ⎦

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practice makes perfect

Linear Algebra

1 ⎡3 1 1 1 0 ⎤ R2 R1 ↔ R2 2 ⎢⎣1 −1 0 −1 0 ⎥⎦

1 R +R ⎡1 −1 0 −1 0 ⎤ ⎡1 −1 0 −1 0 ⎤ 4 2 1 ⎡1 0 1 / 4 0 0 ⎤ 3R1 + R2 ⎢ ⎢ ⎢ ⎥ − ⎥ 1 ⎥ ⎣3 1 1 1 0 ⎦ ⎣0 4 1 4 0 ⎦ R2 ⎣0 1 1 / 4 1 0 ⎦ 4 1 x=− z 4 ⇒ , where z and w are free variables. There are infinitely many solutions. 1 y = − z −w 4

You can see from the examples that for the homogeneous system the coefficient matrix is the only matrix you actually need to work with and not the augmented matrix. This strategy is shown in the following problem. PROBLEM

SOLUTION

x + y − 3z = 0 Determine whether the system 2 x + 2 y + 3z = 0 has a nontrivial solution. 3x − y − z = 0 Reduce the coefficient matrix. ⎡1 1 −3 ⎤ 1 R ⎥ 9 2 ⎢ ⎢0 0 9 ⎥ 1 ⎢⎣0 −4 8 ⎥⎦ − R3 4 x=0 ⇒ y=0 z =0

⎡1 1 −3 ⎤ ⎥ −2 R1 + R2 ⎢ ⎢ 2 2 3 ⎥ −3R + R 1 3 ⎢⎣ 3 −1 −1⎥⎦ ⎡1 0 0 ⎤ ⎥ ⎢ − 1R2 + R1 ⎢0 1 0 ⎥ ⎢⎣0 0 1 ⎥⎦

⎡1 1 −3 ⎤ 2 R2 + R3 ⎥ ⎢ ⎢0 0 1 ⎥ 3R2 + R1 ⎢⎣0 1 −2 ⎥⎦ R ↔ R 2 3

⎡1 1 0 ⎤ ⎥ ⎢ ⎢0 1 0 ⎥ ⎢⎣0 0 1 ⎥⎦

The only solution is the trivial solution. It does not have a nontrivial solution. EXERCISE

1·6 For 1–4, determine whether the system has a nontrivial solution.

x + 2 y − 3z = 0 1. 2 x + 5 y + 2 z = 0 3x − y − 4 z = 0 x − 3y − 4 z + w = 0 2. 2 x − 5 y − 2 z + 3w = 0 x − y − z − 2w = 0 2 x + 4 y − 2z = 0 2 x + 5y + 2z = 0 3. − x − 4 y − 7z = 0 3x + 9 y + 9z = 0 4.

3x + 2 y = 0 5x − 4 y = 0 x + y + 2z

5. Solve the system:

=0

y − 3z + w = 0 3 x + y + z + 2w = 0 x + 3 y − 2 z − 2w = 0

Systems of linear equations and matrices

13

·2·

Matrix algebra In this chapter, you will learn: ◆ ◆ ◆ ◆ ◆

Matrix arithmetic Inverse of a square matrix Properties of invertible matrices Matrix solutions of systems of linear equations Transpose of a matrix

This chapter provides a discussion of the algebra of matrices and methods of solving simultaneous equations using that algebra.

Matrix arithmetic Addition and subtraction of matrices Matrices can be added or subtracted only if they are the same size. The sum of two matrices is the matrix whose elements are the sums of the corresponding elements of the two matrices. Similarly, the difference of two matrices is the matrix whose elements are the differences of the corresponding elements of the two matrices. Here are examples. ◆

◆

◆

⎡1 ⎢⎣4 ⎡1 ⎢⎣− 5

8 ⎤ ⎡− 2 3⎤ ⎡1 + (− 2) 8 + (3) ⎤ ⎡− 1 11⎤ + = = − 3⎥⎦ ⎢⎣ 0 7⎥⎦ ⎢⎣ 4 + (0) − 3 + (7 )⎥⎦ ⎢⎣ 4 4 ⎥⎦ 0 4⎤ ⎡ 0 4 5 ⎤ ⎡ 1 − (0) 0 − (4 ) 4 − (5) ⎤ ⎡ 1 − 4 − 1⎤ −⎢ =⎢ = ⎥ ⎥ 6 2⎦ ⎣− 3 4 11⎦ ⎣− 5 − (− 3) 6 − (4 ) 2 − (11)⎥⎦ ⎢⎣− 2 2 − 9⎥⎦

⎡2 3⎤ ⎡ a b⎤ ⎡2 + a 3 + b⎤ ⎢⎣5 1⎥⎦ + ⎢⎣−c d⎥⎦ = ⎢⎣5 − c 1 + d⎥⎦ PROBLEM

SOLUTION

14

⎡ 4 1⎤ ⎡ 2 −5 ⎤ If A = ⎢ , B=⎢ ⎥ ⎥ , and C = [ 4 2 9 ], then compute as ⎣ −3 6 ⎦ ⎣0 3 ⎦ indicated. a. A + B b. A + C c. B − A ⎡ 4 1 ⎤ ⎡ 2 −5 ⎤ ⎡ 4 + (2) 1 + (−5)⎤ ⎡ 6 −4 ⎤ a. A + B = ⎢ ⎥+ ⎢ ⎥ ⎥=⎢ ⎥ =⎢ ⎣ −3 6 ⎦ ⎣ 0 3 ⎦ ⎣ −3 + (0) 6 + (3) ⎦ ⎣ −3 9 ⎦ b. A + C, because A and C are not the same size, their sum is undefined. ⎡ 2 −5 ⎤ ⎡ 4 1 ⎤ ⎡ 2 − (4 ) −5 − (1)⎤ ⎡ −2 −6 ⎤ c. B − A = ⎢ ⎥− ⎢ ⎥=⎢ ⎥=⎢ ⎥ ⎣ 0 3 ⎦ ⎣ −3 6 ⎦ ⎣0 − (−3) 3 − (6) ⎦ ⎣ 3 −3 ⎦

Multiplication of matrices Two types of multiplication are associated with matrices: scalar multiplication and matrix multiplication.

Scalar multiplication The multiplication of a matrix by a constant is scalar multiplication. The product of a scalar and a matrix is a matrix whose elements are each multiplied by the scalar. Any size matrix can be multiplied by a scalar. For instance, if c is a constant (scalar), ⎡7 3 ⎤ ⎡7c 3c ⎤ c ⎢1 − 2⎥ = ⎢ c − 2c⎥ ⎢ ⎥ ⎢ ⎥ ⎣0 5 ⎦ ⎣ 0 5c ⎦ ⎡4 1 ⎤ Let A = ⎢ ⎥ , find 3A. ⎣ 0 −2 ⎦

PROBLEM SOLUTION

⎡ 4 1 ⎤ ⎡3 ⋅ 4 3 ⋅1 ⎤ ⎡12 3 ⎤ 3⎢ ⎥=⎢ ⎥=⎢ ⎥ ⎣ 0 −2 ⎦ ⎣ 3 ⋅ 0 3 ⋅−2 ⎦ ⎣ 0 −6 ⎦

Clearly, subtraction of matrices can now be defined as A − B = A + (− B) = A + (−1)B.

Matrix multiplication Matrices must have “agreeable” sizes to multiply them. If A and B are matrices and A is size m × k , then for the product AB to be defined, matrix B must have the same number of rows as the number of columns in A and thus must have size k × n. When this inner matching of sizes occurs, the matrices are compatible for multiplication, and the product matrix AB has size m × n . The entries in the product matrix AB are computed in a systematic manner. The element in the ath row and bth column of the product matrix is the sum of the products of the elements of the ath row of A by the corresponding elements in the bth column of B. (Comment: This process probably takes longer to explain than actually to do.) For example, ⎡ 3 2⎤ ⎡ 3 ⋅ 4 + 2 ⋅ 3 ⎤ ⎡ 18 ⎤ ⎡4⎤ If A = ⎢ 1 5⎥ and B = ⎢ ⎥ , then AB = ⎢ 1 ⋅ 4 + 5 ⋅ 3 ⎥ = ⎢ 19 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣3⎦2×1 ⎣− 3 0⎦3×2 ⎣− 3 ⋅ 4 + 0 ⋅ 3⎦ ⎣− 12⎦3×1

It is important to note that the commutative law for multiplication is not valid for matrix multiplication. That is, in general, AB ≠ BA . However, identity matrices have the property that IA = AI = A where A and I are square matrices of the same size. PROBLEM

⎡1 5 ⎤ ⎡ 0 4 2⎤ If A = ⎢ ⎥ and B = ⎢ −5 3 1 ⎥ , then compute as indicated. − 3 5 ⎣ ⎣ ⎦ ⎦ a. AB b. BA

Matrix algebra

15

SOLUTION

⎡1 5 ⎤ ⎡ 0 4 2⎤ ⎡ 1 ⋅ 0 + 5 ⋅ (−5) 1⋅ 4 + 5 ⋅ 3 1 ⋅ 2 + 5 ⋅1 ⎤ a. AB = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣3 −5⎦ ⎣−5 3 1⎦ ⎣3 ⋅ 0 + (−5) ⋅ (−5) 3 ⋅ 4 + (−5) ⋅ 3 3 ⋅ 2 + (−5) ⋅1⎦ ⎡−25 19 7⎤ =⎢ ⎥ ⎣ 25 −3 1⎦

b. BA is undefined because B2×3 and A2×2 are not compatible for multiplication in the order given. PROBLEM

⎡2 0⎤ ⎡1 5 ⎤ If A = ⎢ and B = ⎢ ⎥ ⎥ , then compute as indicated. ⎣1 3 ⎦ ⎣3 2 ⎦

a. AB b. BA SOLUTION

⎡ 2 0 ⎤ ⎡1 5 ⎤ ⎡ 2 10 ⎤ a. AB = ⎢ ⎥⎢ ⎥ ⎥= ⎢ ⎣1 3 ⎦ ⎣3 2 ⎦ ⎣10 11 ⎦ ⎡1 5 ⎤ ⎡ 2 0 ⎤ ⎡7 15 ⎤ b. BA = ⎢ ⎥ ⎢ ⎥ ⎥=⎢ ⎣3 2 ⎦ ⎣1 3 ⎦ ⎣ 8 6 ⎦

Remark: Both of the above problems reinforce the fact that, in general, AB ≠ BA . The remainder of the work in this text requires you to be very familiar with the arithmetic properties of matrices as well as with important algebraic properties. The basic arithmetic properties for matrices under the operations of matrix addition/subtraction, matrix multiplication, and scalar multiplication are given in the following box.

Given matrices A, B, and C, and scalars a and b with the sizes of the matrices such that the indicated operations are defined, then the following properties hold: Matrix addition/subtraction and matrix multiplication

16

A+ B = B+ A A + ( B + C ) = ( A + B) + C A( BC ) = ( AB)C A( B ± C ) = AB ± AC ( B ± C )A = BA ± CA 0+ A= A+0 = A IA = AI = A

commutative law for addition associative law for addition associative law for multiplication left distributive law right distributive law identity for addition identity for multiplication

Scalar multiplication a( B ± C ) = aB ± aC a(bC ) = (ab)C 1A = A

(a ± b)C = aC ± bC a( BC ) = (aB)C = B(aC ) 0A = 0

practice makes perfect

Linear Algebra

You also have the following useful matrix properties: ◆

If A is a square matrix, then A0 = I and An = AA A

(n > 0).

n factors

◆ ◆

◆

◆

◆

n m nm If A is a square matrix and m and n are integers, then Am An = Am+n and ( A ) = A . If A and 0 are matrices such that the operations are defined, then A0 = 0 and 0 A = 0 ; also, 0 − A = − A and A − A = 0. The cancellation law for matrices does not hold. That is, if AB = AC , then it is not true, in general, that B = C . If AB = 0 , it is possible that both A ≠ 0 and B ≠ 0 (Thus, the useful algebraic zero property of real numbers does not carry over to matrices). The commutative law for multiplication does not hold. That is, it is not true, in general, that AB = BA .

PROBLEM

⎡0 1 ⎤ ⎡3 7 ⎤ Suppose A = ⎢ and B = ⎢ ⎥ ⎥ . Show that A and B are matrices such that ⎣0 2 ⎦ ⎣0 0 ⎦ AB = 0 , but A ≠ 0 and B ≠ 0 .

SOLUTION

⎡0 1 ⎤ ⎡ 3 7 ⎤ ⎡0 0 ⎤ AB = ⎢ ⎥⎢ ⎥=⎢ ⎥ , A ≠ 0 and B ≠ 0 . ⎣0 2 ⎦ ⎣0 0 ⎦ ⎣0 0 ⎦

If A is a square matrix, then the trace of A, denoted tr(A), is the sum of the diagonal elements of A. PROBLEM

⎡1 5 ⎤ If A = ⎢ ⎥ , find tr( A). ⎣2 8⎦

SOLUTION

⎡1 5 ⎤ tr( A) = tr ⎢ ⎥ = 1+ 8 = 9 ⎣2 8⎦

EXERCISE

2·1 For 1–5, compute as indicated.

⎡ 2 8⎤ ⎡ 5 4 ⎤ 1. ⎢ ⎥ ⎥ + 3⎢ ⎣ −3 0 ⎦ ⎣ −2 3 ⎦ ⎡ 3 6 0 ⎤ ⎡ −1 6 4 ⎤ 2. ⎢ 4 −2 −5 ⎥ − ⎢ −4 3 −1⎥ ⎥ ⎥ ⎢ ⎢ ⎢⎣ 6 0 −1⎥⎦ ⎣⎢ 5 4 2 ⎥⎦ ⎡ 2 4⎤ ⎡1 6 3⎤ ⎢ ⎥ 3. ⎢ ⎥ ⎢ −3 1 ⎥ 0 5 4 ⎣ ⎦⎢ ⎣ 5 0 ⎥⎦ ⎡1 0⎤ ⎡ 3 6⎤ 4. ⎢ ⎥⎢ ⎥ ⎣ 0 1 ⎦ ⎣ −4 5 ⎦ ⎡0 0 ⎤ ⎡ 5 6 ⎤ 5. ⎢ ⎥⎢ ⎥ ⎣0 0 ⎦ ⎣ 4 9 ⎦ Matrix algebra

17

⎡ 0 1⎤ ⎡3 7 ⎤ ⎡1 5⎤ 6. If A = ⎢ , B=⎢ , and C = ⎢ ⎥ ⎥ ⎥ , show that A( BC ) = ( AB )C . ⎣0 2 ⎦ ⎣1 0⎦ ⎣2 8 ⎦ ⎡1 5⎤ ⎡2 −3 ⎤ and B = ⎢ 7. If A = ⎢ ⎥ ⎥ , show that tr( A + B ) = tr( A) + tr(B ). ⎣2 8 ⎦ ⎣ 5 −4 ⎦ ⎡1 5⎤ 8. For A = ⎢ ⎥ , compute (3x − y ) A. ⎣2 8 ⎦ ⎡ 2 1⎤ ⎡ 1 −2 ⎤ 9. For A = ⎢ and B = ⎢ ⎥ , compute 3 A + 3B . ⎥ ⎣ −2 3 ⎦ ⎣3 4 ⎦ ⎡ 2 1⎤ ⎡ 1 −2 ⎤ 10. If A = ⎢ and B = ⎢ ⎥ , show that A + B = B + A . ⎥ ⎣ −2 3 ⎦ ⎣3 4 ⎦

Inverse of a square matrix The inverse of a square matrix A is a square matrix A−1 of the same size such that AA−1 = A−1 A = I . Finding inverses of matrices is very important in the theory and application of matrices. There are several methods of finding inverses by hand. Two of those methods for 2 × 2 matrices will be demonstrated in the following problem. PROBLEM

SOLUTION

⎡1 1 ⎤ −1 Given A = ⎢ ⎥, find A . ⎣4 2⎦ ⎡ 1 1 ⎤ ⎡a b ⎤ ⎡a b ⎤ Method 1. Let A−1 = ⎢ . It follows that AA−1 = ⎢ ⎥⎢ ⎥= ⎥ ⎣4 2⎦ ⎣ c d ⎦ ⎣c d ⎦ b + d ⎤ ⎡1 0 ⎤ ⎡ a+c ⎢ 4a + 2c 4b + 2d ⎥ = ⎢0 1 ⎥ . From this relationship, you get the two systems of ⎣ ⎦ ⎣ ⎦ equations: b+d = 0 a+c =1 and 4b + 2d = 1 4a + 2c = 0 1 1 Solving these two systems yields, a = −1, b = , c = 2 , and d = − . 2 2 1 1 / 2 − ⎡ ⎤ Thus, A−1 = ⎢ ⎥. ⎣ 2 −1/ 2 ⎦ ⎡ d ⎡a b ⎤ 1 ⎡ d −b⎤ ⎢ ad − bc −1 Method 2. In general, if A = ⎢ ⎥ , then A = ad − bc ⎢−c a ⎥ = ⎢ −c ⎣ ⎦ ⎢ ⎣c d ⎦ ⎢⎣ ad − bc provided ad − bc ≠ 0.

−b ⎤ ad − bc ⎥⎥ , a ⎥ ad − bc ⎥⎦

Note: The quantity ad − bc is the determinant of A (see Chapter 5 for a discussion of determinants). Thus, to determine the A−1 matrix from the original matrix do the following: 1. Interchange the elements on A’s diagonal. 2. Negate the other two elements (but don’t switch them!). 3. Multiply the resulting matrix by the reciprocal of ad − bc .

18

practice makes perfect

Linear Algebra

Warning: This method only works for 2 × 2 matrices. ⎡1 1⎤ So, for A = ⎢4 2⎥ , ⎣ ⎦ 1 ⎡ 2 − 1⎤ 1 ⎡ 2 − 1⎤ ⎡− 1 1/ 2 ⎤ A−1 = (1 ⋅ 2) − (4 ⋅1) ⋅ ⎢− 4 1 ⎥ = − 2 ⋅ ⎢− 4 1 ⎥ = ⎢ 2 − 1/ 2⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Remark: There are manual methods for finding inverses of n × n matrices when n > 2, but those methods can be time-consuming and tedious; and, on the whole, technology devices have supplanted their use. Therefore, such methods are not presented in this text. Instead see Chapter 3 for instructions on finding inverses with the use of graphing calculators. EXERCISE

2·2 1⎤ ⎡4 ⎡ −1 1 ⎤ ⎥ and B = ⎢ −2 4 ⎥ . − 3 − 2 ⎣ ⎦ ⎣ ⎦

Let A = ⎢

For 1–7, compute as indicated.

1. A−1 2. B −1 3. ( AB )−1 4. A−1B −1 5. B −1A−1 6. A( − B −1 ) 7. − A( B −1 ) For 8–10, indicate whether the two expressions are equal or not equal.

8. ( AB )−1 and B −1A−1 9. A( − B −1 ) and − A( B −1 ) 10. ( AB )−1 and A−1B −1

Properties of invertible matrices If A is a square matrix that has an inverse A−1 such that AA−1 = A−1 A = I , then A is invertible (or nonsingular). If A does not have an inverse then A is singular. Some algebraic properties of invertible matrices A and B are the following: ◆ ◆ ◆

( A−1 )−1 = A

If AB is invertible, then both A and B are invertible. The inverse of an invertible matrix is unique. Matrix algebra

19

◆ ◆

1 (kA)−1 = A−1 for any nonzero scalar k. k If AX = C , then X = A−1C (Remark: This equation algebraically solves for X) PROBLEM

Respond as indicated. ⎡ 2 3⎤ −1 a. For A = ⎢ ⎥ , find (2 A) . 1 3 ⎣ ⎦ ⎡3 2 ⎤ ⎡ x ⎤ ⎡3⎤ ⎡x ⎤ b. Solve ⎢ = ⎢ ⎥ for ⎢ ⎥ . ⎥ ⎢ ⎥ ⎣5 4 ⎦ ⎣ y ⎦ ⎣1 ⎦ ⎣y⎦ c. Using matrix properties prove: If BA−1 = C , then B = CA.

SOLUTION

⎡ 2 3⎤ −1 a. For A = ⎢ ⎥ , find (2 A) . 1 3 ⎣ ⎦

⎡ 1 ⎢ 2 3 3 3 3 − − ⎡ ⎤ ⎡ ⎤ 1 1 1 1 ⎛ ⎞ (2 A)−1 = A−1 = ⎜ ⎟ ⎢ =⎢ = ⎢ ⎥ ⎥ 2 2 ⎝ 3 ⎠ ⎣ −1 2 ⎦ 6 ⎣ −1 2 ⎦ ⎢ 1 − ⎢⎣ 6 ⎡3 2 ⎤ ⎡ x ⎤ ⎡3⎤ ⎡x ⎤ b. Solve ⎢ = ⎢ ⎥ for ⎢ ⎥. ⎥ ⎢ ⎥ ⎣5 4 ⎦ ⎣ y ⎦ ⎣1 ⎦ ⎣y⎦

1⎤ − ⎥ 2 ⎥ 1 ⎥ 3 ⎥⎦

⎡3 2 ⎤ −1 1 ⎡ 4 −2 ⎤ For A = ⎢ ⎥ , A = 2 ⎢ −5 3 ⎥ . ⎣ ⎣5 4 ⎦ ⎦ ⎡ x ⎤ 1 ⎡ 4 −2 ⎤ ⎡3 ⎤ 1 ⎡ 10 ⎤ Thus, ⎢ ⎥ = ⎢ ⎥⎢ ⎥ = ⎢ ⎥. ⎣ y ⎦ 2 ⎣ −5 3 ⎦ ⎣1 ⎦ 2 ⎣ −12 ⎦ ⎡x ⎤ ⎡ 5 ⎤ Hence, ⎢ ⎥ = ⎢ ⎥ . ⎣ y ⎦ ⎣ −6 ⎦

c. Using matrix properties prove: If BA−1 = C , then B = CA . Given BA−1 = C

Multiplying both sides on the right by A, ( BA−1 )A = CA

Using the associative law, B( A−1 A) = CA

Therefore, BI = CA

Because I is the identity, B = CA

Note: See Chapter 4 for an additional discussion of inverse matrices.

20

practice makes perfect

Linear Algebra

EXERCISE

2·3 Respond as indicated. −1 ⎡1 5⎤ ⎛1 ⎞ . 1. For A = ⎢ , find A ⎜⎝ ⎟⎠ ⎥ 5 ⎣2 8 ⎦

⎡1 5⎤ 2. For A = ⎢ , compute A−2 . ⎥ ⎣2 8 ⎦ ⎡ 1 5 ⎤ ⎡ x ⎤ ⎡2 ⎤ ⎡x⎤ 3. For ⎢ = ⎢ ⎥ , use the inverse matrix to solve for ⎢ ⎥ . ⎥ ⎢ ⎥ ⎣2 8 ⎦ ⎣ y ⎦ ⎣ 3 ⎦ ⎣y⎦ ⎡ 2 1⎤ ⎡ 1 −2 ⎤ −1 −1 −1 4. For A = ⎢ ⎥ and B = ⎢ −2 3 ⎥ , show that ( AB ) = B A . 3 4 ⎣ ⎦ ⎣ ⎦ ⎡3 2 ⎤ −1 −1 5. For A = ⎢ ⎥ , show that ( A ) = A . 5 4 ⎣ ⎦ ⎡1 5⎤ −2 −1 6. For A = ⎢ ⎥ , show that A A = A (refer to Prob. 2 above). 2 8 ⎣ ⎦ ⎡3 2 ⎤ −1 7. For A = ⎢ ⎥ , show that AA = I . ⎣5 4 ⎦ ⎡2 0⎤ 8. For A = ⎢ , compute A2 − 2 A + I . ⎥ ⎣ 4 1⎦ ⎡ cosθ 9. Find the inverse of ⎢ ⎣ − sinθ

sinθ ⎤ . cosθ ⎥⎦

10. Using matrix properties, prove that if ABA−1 = C , then B = A−1CA.

Matrix solutions of systems of linear equations You can solve linear equations using matrices and their associated algebra. Technological advances have made this approach for the most part the technique of choice because the calculations involved, many of which are wearisome, can be done rapidly using appropriate hardware such as computers or sophisticated calculators. To make effective use of this approach, however, you must completely understand the theory and use of the matrix algebra involved. In this section, the basic concepts are introduced and then exemplified in solving simple equations. By using the definitions of matrix multiplication and matrix equality, it follows that the a1x + b1 y = c1 ⎡a1 b1 ⎤ ⎡x ⎤ ⎡c1 ⎤ linear system is equivalent to the matrix equation ⎢a b ⎥ ⎢y⎥ = ⎢c ⎥ . ⎣ 2 2⎦ ⎣ ⎦ ⎣ 2⎦ a x +b y = c 2

2

2

⎡a1 b1 ⎤ In matrix notation this equation is written AX = C , where A = ⎢a b ⎥ is the coefficient ⎣ 2 2⎦ ⎡c1 ⎤ ⎡x ⎤ matrix, X = ⎢y⎥ is the variable matrix, and C = ⎢c ⎥ is the constant matrix. ⎣ 2⎦ ⎣ ⎦ Matrix algebra

21

−1

If A exists, then A−1 AX = A−1C

Thus, IX = A−1C

yielding, X = A−1C .

The solution to the system is determined by this latter equation. Caution: Keep in mind that X = A−1C does not apply if the coefficient matrix does not have an inverse. PROBLEM

Solve using matrix algebra. 3x − 2 y = 4 a. 2x − y = 3 x+ y+ z =4 b. 2 x − y − 2z = −1 x − 2y − z =1

SOLUTION

a.

3x − 2 y = 4 2x − y = 3

⎡−1 2⎤ ⎡ 3 −2 ⎤ ⎡x ⎤ ⎡4 ⎤ From the linear system, A = ⎢ , X = ⎢ ⎥ and C = ⎢ ⎥ , then A−1 = ⎢ ⎥. ⎥ ⎣−2 3⎦ ⎣ 2 −1 ⎦ ⎣y⎦ ⎣3⎦ ⎡x ⎤ ⎡ −1 2 ⎤ ⎡ 4 ⎤ ⎡ 2 ⎤ The solution is X = A−1C ; that is, ⎢ ⎥ = A−1C = ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ yielding ⎣y⎦ ⎣ −2 3 ⎦ ⎣ 3 ⎦ ⎣1 ⎦

the solution x = 2, y = 1. If you substitute these values into the original

equations, the calculations 3(2) − 2(1) = 4 and 2(2)− 1 = 3 verify the solution. x+ y+ z =4 b. 2 x − y − 2z = −1 x − 2y − z =1

⎡1

⎡x⎤ ⎡4⎤ 1 1⎤ − 1 − 2⎥ , X = ⎢y⎥ , and C = ⎢− 1⎥ . ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎣1 − 2 − 1⎦ ⎣z⎦ ⎣1⎦

From the linear system, A = ⎢2

⎡x ⎤

The solution is X = A−1C ; that is (using a calculator), ⎢ y⎥ = A−1C =

⎢ ⎥ ⎣z ⎦

⎡1 1 1 ⎤ 6 6 ⎥⎡4⎤ ⎡2⎤ ⎢ 2 ⎥ ⎢− 1⎥ = ⎢− 1⎥ yielding the solution x = 2, y = −1, z = 3, which ⎢0 1 2 3 − 3⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢1 1 1 ⎥⎣1⎦ ⎣3⎦ ⎣ 2 − 2 2⎦ can be verified by substitution into the original system.

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Remark: This example illustrates the usefulness and power of technology when working with matrices. Note: See Chapter 3 for an additional discussion of using inverses to solve systems of equations. EXERCISE

2·4 For 1–3, find the inverse using either Method 1 or Method 2 from this section.

⎡ 2 3⎤ 1. ⎢ ⎥ ⎣ −1 4 ⎦ ⎡1 1⎤ 2. ⎢ ⎥ ⎣1 2 ⎦ ⎡2 −2 ⎤ 3. ⎢ ⎥ ⎣ 1 −1⎦ For 4 and 5, solve using inverse matrices.

4. 5.

3x + 2 y = 4 x −2y = 3 5x − y = 3 2x + 2y = 2

Transpose of a matrix The transpose of a matrix A, written AT, is the matrix obtained by changing the rows of A, in order, to respective columns. PROBLEM

As indicated, find the transpose. ⎡3 1 ⎤ a. ⎢ ⎥ ⎣5 −9⎦

T

b. [1 2 5]

T

SOLUTION

⎡3 1 ⎤ ⎡3 5 ⎤ a. ⎢ ⎥ =⎢ ⎥ ⎣5 −9⎦ ⎣1 −9⎦ T

b. [1 2 5]

T

⎡1⎤ ⎢ ⎥ = ⎢2⎥ ⎢⎣5⎥⎦

The following properties of transpose hold: ◆ ◆ ◆

( AT )T = A ( A + B)T = AT + BT (cA)T = cAT Matrix algebra

23

◆ ◆ ◆ ◆

( AB)T = BT AT (Note the order reversal) If A is invertible, then AT is invertible. If A is invertible, then AAT and AT A are invertible. If A is invertible, then ( AT )−1 = ( A−1 )T .

EXERCISE

2·5 Respond as indicated.

⎡3 2 ⎤ ⎡0 4 ⎤ 1. Show that ( AB )T = B T AT for A = ⎢ and B = ⎢ ⎥ ⎥. ⎣0 5⎦ ⎣ 2 −3 ⎦ ⎡3 2⎤ ⎡0 4 ⎤ 2. Show that ( A + B )T = AT + B T for A = ⎢ and B = ⎢ ⎥ ⎥. ⎣0 5⎦ ⎣ 2 −3 ⎦ ⎡3 2 ⎤ 3. Show that (5 A)T = 5 AT for A = ⎢ ⎥. ⎣0 5⎦ ⎡ 6 −2 ⎤ 4. Show that (C T )−1 = (C −1 )T for C = ⎢ ⎥. ⎣4 2 ⎦ ⎡ 6 −2 ⎤ T −1 5. Given C = ⎢ ⎥ , find (CC ) . 4 2 ⎦ ⎣ ⎡ 6 −2 ⎤ 6. Given C = ⎢ , find (C T C )−1 . ⎥ ⎣4 2 ⎦ T

⎡4⎤ ⎢ ⎥ 7. Find ⎢−1⎥ . ⎢⎣ 1 ⎥⎦ T 8. Find [2 −1 5 7] . ⎡ 2 −1 3 7 ⎤ T T 9. Given D = ⎢ ⎥ , show that ( D ) = D . 4 − 5 0 8 ⎣ ⎦ ⎡1 5 2 ⎤ 10. Given E = ⎢⎢ 5 −1 −3 ⎥⎥ , show that E T = E . ⎢⎣2 −3 −4 ⎥⎦

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Linear Algebra

Graphing calculators and matrices

·3·

In this chapter, you will learn: ◆ ◆ ◆ ◆ ◆ ◆ ◆

Matrix menu Inputting and editing a matrix Matrix arithmetic Calculating determinants Transpose of a matrix Solving linear systems using Gauss-Jordan elimination Solving linear systems using X = A −1C

Because you are a serious student of mathematics, we assume you already have a graphing calculator and that you have mastered its elementary operation. In this chapter, you will see that a graphing calculator is an indispensable tool for working with matrices in linear algebra. To demonstrate basic matrix features of graphing calculators, we have elected to use the TI-83 Plus platform. If you have a different calculator, that’s okay. Most graphing calculators will have matrix features like the ones shown here. Consult your user’s guidebook for instructions. After you complete this chapter, you should use your graphing calculator as a handy tool as you work through the remaining material in this book.

Matrix menu You access MATRX , the matrix menu, by pressing 2nd x − 1 . The MATRX menu is quite extensive and is divided into three parts: NAMES, MATH, and EDIT (See below).

You use the NAMES menu to paste the variable name of a matrix onto the home screen. The TI-83 Plus has 10 different variable names, [A] through [J]. To paste a matrix’s variable name to the home screen, select MATRX NAMES, and then press the number of the name you want (or scroll using the arrow keys and then press ENTER ). For instance, to choose matrix [B], select MATRX NAMES, and then press 2 (or scroll to 2:[B] and then press ENTER ). The home screen will show [B] with a blinking cursor to the right. Notice that brackets appear around the letter selected.

25

To access the MATRX MATH menu, select MATRX and then press the right arrow key one time. This menu lists the following commands that you can use with a matrix:

2:T

Calculates the determinant of the matrix, provided the matrix is a square matrix (same number of rows and columns). Returns the transpose of the matrix.

3: dim(

Returns the matrix size (dimensions):{number of rows number of columns}.

4: Fill(

Fills all elements of the matrix with the same scalar value.

5:identity(

Returns the identity matrix of a given order.

6: randM(

Returns a random matrix of integers between -9 and 9 inclusive.

7: augment( 8: Matr䉴list(

Appends two matrices. The two combined matrices must have the same number of rows. Stores a matrix to a list.

9: List䉴matr(

Stores a list to a matrix.

0: cumSum(

Returns cumulative sums of the column elements of the matrix, starting with the first element. Returns the row-echelon form of the matrix. Caution: The number of rows must be less than or equal to the number of columns. Returns the reduced row-echelon form of the matrix. Caution: The number of rows must be less than or equal to the number of columns. Swaps two rows of the matrix.

1: det(

A: ref( B: rref( C: rowSwap(

Adds a row of the matrix to another row and replaces the other row with the result. Multiplies a row of the matrix by a number and stores the result in the same row. Multiplies a row of the matrix by a number and adds the result to another row; result replaces the other row.

D: row+( E: *row( F: *row+(

Certain ones of these commands that are particularly useful in linear algebra are demonstrated in later sections of this chapter. You use the MATRX EDIT menu to select a variable name for a matrix, to specify its size, and to enter its elements (although for the TI-83 Plus calculator, only real number elements are allowed). Depending on available memory, the matrix can have up to 99 rows or columns. The MATRX EDIT feature is presented in the next section. EXERCISE

3·1 For 1–5, fill in the blank with the best response.

1. The matrix menu of a graphing calculator allows users to select a matrix by variable ____________, edit a matrix, and perform math commands on a matrix. 2. The det( command is used only with ____________ matrices. 3. The Fill( command fills all elements of a matrix with the same ____________ value.

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Linear Algebra

4. Two matrices that are combined using the augment( command must have the same number of ____________. 5. The ____________ command returns the reduced row-echelon form of the matrix.

Inputting and editing a matrix ⎡ 2 −3 0 4 ⎤ Suppose you have the 2 × 4 matrix ⎢ ⎥ . To input this matrix, do the following: ⎣1 6 −5 3 ⎦ Instructions

Screen Display

Go to MATRX , press the right arrow key twice to display the MATRX EDIT menu, and then Press 1 to select 1: [A]. Note: If a size appears next to a variable name, a matrix is already stored in that slot. You can overwrite the existing matrix or select a new name.

Note: The default matrix has size 1 × 1.

Press 2 ENTER 4 ENTER to specify a 2 × 4 matrix. Note: The rectangular cursor indicates the current element, whose position and value is displayed at the bottom of the screen. Note: Ellipses at the far right indicate additional elements beyond the viewing screen. Press 2 ENTER to enter 2 as the value in the first row, first column cell. The rectangular cursor moves to the first row, second column.

Press -3 ENTER 0 ENTER 4 ENTER to complete the first row. Caution: Use the negation key (-) , not the minus key - to enter negative values.

Press 1 ENTER 6 ENTER −5 input the second row.

ENTER

3

ENTER

to

To save the matrix, press QUIT ( 2nd MODE ). To display the matrix [A], go to MATRX , Press 1 to select 1: [A], and then press ENTER .

Graphing calculators and matrices

27

You can edit a matrix, such as [A], that is stored in the calculator. For instance, to change the element in the first row, third column of [A] from 0 to 8, do the following: Instructions

Screen Display

Go to MATRX , press the right arrow key twice to display the MATRX EDIT menu, and then Press 1 to select 1: [A]. Use the arrow keys to move to the first row, third column of [A].

Press 8 ENTER to change 0 to 8. The rectangular cursor moves to the first row, fourth column. To save the change, press QUIT ( 2nd MODE ).

Note: Ellipses at the far left indicate additional elements beyond the viewing screen.

PROBLEM

SOLUTION

5x + 2 y = 3 Input the augmented matrix for the system as matrix [B] in the 2 x + 3 y = −1 calculator.

Go to MATRX , press the right arrow key twice to display the MATRX EDIT menu, and then Press 2 to select 2: [B]. Press 2 ENTER 3 ENTER to specify a 2 × 3 matrix. Press 5 ENTER 2 ENTER 3 ENTER to input the first row. Press 2 ENTER 3 ENTER -1 ENTER to input the second row. The screen display shows

To save matrix [B], press

QUIT

(

2nd MODE

).

Note: The TI-83 Plus does not allow the use of complex numbers in matrices. If you enter an imaginary or complex number as a matrix element, ERR: DATA TYPE is displayed. EXERCISE

3·2 Input the indicated matrix into the calculator. (Select any convenient matrix variable name for the matrix.)

⎡3 0 4 ⎤ 1. ⎢ ⎥ ⎣ 6 −2 7 ⎦ 2 8⎤ 2. ⎡ ⎢ −3 1 ⎥ ⎣ ⎦

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Linear Algebra

⎡1 ⎢0 3. ⎢ ⎢0 ⎢ ⎣0

0 1 0 0

0 0 1 0

0⎤ 0⎥ ⎥ 0⎥ ⎥ 1⎦

⎡5 6 ⎤ ⎥ ⎢ 4. ⎢ 0 −2 ⎥ ⎢⎣ 4 9 ⎥⎦ ⎡3⎤ ⎢ −1⎥ ⎢ ⎥ 5. ⎢ 0 ⎥ ⎢ ⎥ ⎣4⎦ 6. [1 0 5 − 2]

⎡5 5 ⎤ 7. ⎢ ⎥ ⎣5 5 ⎦ ⎡ 2 −1 4 3 0 ⎤ ⎥ ⎢ 8. ⎢ 1 6 −2 5 0 ⎥ ⎢⎣ −1 3 8 −2 0 ⎥⎦ 9. The augmented matrix for the system

5x + 2 y = 3 10 x + 6 y = −1

x − 2 y + 3z = 1 10. The coefficient matrix for the system x + 3 y − z = 4 2 x + y − 2 z = 13

Matrix arithmetic You can do arithmetic computations with matrices using the standard math keys, as long as the sizes of the matrices allow the computations to be meaningful. Input the matrices and then use the MATRX NAMES menu to paste the matrix variable names into the computational expression. Tip: If part of the answer is not showing when you perform an allowable computation, use the arrow keys to scroll through the matrix. If a computation is not meaningful, and you try it anyway, the calculator will return an error message. For instance, when matrices are not compatible for addition, subtraction, or multiplication, the calculator shows a message, like the one below, telling you that you have a dimension (size) mismatch.

Graphing calculators and matrices

29

PROBLEM

2 Store the matrices A = ⎡ ⎢⎣1

− 3 0 4⎤ 5 − 2 3 7⎤ ,B=⎡ , and C = ⎢⎣8 6 − 4 2⎥⎦ 6 − 5 3⎥⎦

⎡ 2 8⎤ in the calculator and then perform the indicated computations. If the ⎢⎣− 3 1⎥⎦ computation is not meaningful, write Undefined. a. A + 3B b. A − C c. AB d. CA e. A−1 f. C −1 SOLUTION

a. A + 3B A2 × 4 and B2 × 4 are the same size, so their sum is meaningful. Go to the

MATRX

NAMES menu and on the home screen paste [A], press +, press 3, and then paste [B].

Press

ENTER .

The screen display shows the following answer:

b. A − C A − C is undefined because A2 × 4 and C2 × 2 are not the same size.

c. AB AB is undefined because A2 × 4 and B2 × 4 are not compatible for matrix multiplication. d. CA CA is meaningful because C2 × 2 and A2 × 4 are compatible for multiplication in the order given and the product is a 2 × 4 matrix. Go to the MATRX NAMES menu and on the home screen paste [C] and then paste [A]. Press ENTER . The screen display shows the following answer:

e. A−1 A is not a square matrix, so its inverse is undefined. f. C −1 C is a square matrix, so it might have an inverse. Go to the MATRX NAMES menu and on the home screen paste [C] and then press x −1 ENTER . The screen display shows the following answer with decimal entries:

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Linear Algebra

To change the elements to fractions (if possible), press The screen display now shows the following:

MATH ENTER ENTER .

Tip: You can go directly to the fractional form by entering [C] x −1 Frac , and then pressing Note: See Chapter 2 for an additional discussion of matrix computation.

ENTER .

EXERCISE

3·3

⎡2 −3 0 4 ⎤ ⎡ 5 −2 3 7 ⎤ ,B= ⎢ Store the matrices A = ⎢ ⎥ ⎥, C = ⎣ 1 6 −5 3 ⎦ ⎣ 8 6 −4 2 ⎦

⎡ 2 8⎤ ⎢ −3 1 ⎥ , ⎣ ⎦

⎡3⎤ ⎢ −1⎥ ⎡2 − 3 5 ⎤ ⎡ 5 − 10⎤ ⎡ 4 − 3⎤ ⎢0 1 , G = H = D = ⎢ ⎥ , E = [1 0 5 − 2], F = ⎢ 6 ⎥ , and , ⎢⎣− 1 2 ⎥⎦ ⎢ ⎥ ⎢0⎥ ⎣− 1 2 ⎥⎦ 0 0 − 4⎦ ⎣ ⎢ ⎥ ⎣4⎦ ⎡1 0⎤ in the calculator and then perform the indicated computation. If a computation is I=⎢ ⎣0 1⎥⎦ not meaningful, write Undefined.

1. B − A 2. 2F − 5C + 3G − I 1 3. C 2 4. AC 5. CB 6. D −1 7. F −1 8. G −1 9. DE 10. ED 11. −H 12. 2 A + 3B Graphing calculators and matrices

31

13. F −1F 14. H −1 15. H 5

Calculating determinants ⎡ 3 −2 5 ⎤ Store the matrix A = ⎢⎢ 1 −5 6 ⎥⎥ in the calculator. To calculate the determinant of A, go to the ⎢⎣ −4 2 −1⎥⎦ MATRX MATH menu, and press 1 to paste det( to the home screen. Then go to the MATRX NAMES menu and paste [A]. Press ENTER . The following screen display shows that the determinant of A is −65 :

Note: You can omit the closing parenthesis ( ) ) after [A]. All open parenthetical elements are closed automatically at the end of an expression. Note: See Chapter 5 for an additional discussion of determinants. EXERCISE

3·4

6 2⎤ ⎡− 7 1⎤ ⎡ 5 − 10⎤ , ,B=⎢ Store the matrices A = ⎢⎡ ,C=⎢ ⎥ ⎥ 6 2 − 7 1 ⎣ ⎦ ⎣ ⎦ ⎣− 1 2 ⎥⎦ ⎡2 − 3 5 ⎤ 6 ⎥,E= D = ⎢0 1 ⎢ ⎥ 0 0 − 4⎦ ⎣ ⎡ 2 −1 3 ⎢− 3 4 1 H= ⎢ 5 − 1 − 2 ⎢ ⎢⎣ 7 − 5 1

⎡2 0 0 ⎤ ⎢0 1 0 ⎥ , F = ⎡5 − 2 3 7⎤ , G = ⎢⎣8 6 − 4 2⎥⎦ ⎢ ⎥ ⎣0 0 − 5⎦

5⎤ 0⎥ ⎥, J = 3⎥ 2⎥⎦

⎡ 2 −1 3 ⎢− 3 4 1 ⎢ 5 − 1 − 2 ⎢ 1 ⎢⎣− 3 4

⎡4 − 9 6 ⎤ ⎢2 3 − 1⎥ , ⎢ ⎥ ⎣0 0 0 ⎦

5⎤ ⎡1 0 0⎤ 0⎥ ⎥ , and I = ⎢0 1 0⎥ in the 3⎥ ⎢ ⎥ ⎣0 0 1⎦ 0⎥⎦

calculator and then perform the indicated computation. If a computation is not meaningful, write Undefined. 1. det (A) 2. det (B ) 3. det (A + B ) 4. det (C )

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5. det (2A) 6. 2det (A) 7. det (A−1 ) 8. det (D ) − 2det ( E ) 9. det (F ) 10. det (G ) 11. det (H ) 12. det ( I ) 13. det ( J ) 14. det ( DE ) 15. det ( D ) det ( E )

Transpose of a matrix ⎡2 − 3 0 4⎤ Store the matrix A = ⎢1 6 − 5 3⎥ in the calculator. To find the transpose of A, paste [A] on ⎣ ⎦ the home screen. Next, go to the MATRX MATH menu, press 2 to paste T, and then press ENTER . The following screen display shows the transpose of A:

Note: See Chapter 2 for an additional discussion of transpose. EXERCISE

3·5 5 − 10⎤ ⎡5 − 2 3 7⎤ ⎡− 7 1⎤ Store the matrices A = ⎢ ,B= ⎢ , C = ⎢⎡ , ⎣8 6 − 4 2⎥⎦ ⎣ 6 2⎥⎦ ⎣− 1 2 ⎥⎦ ⎡3⎤ ⎡ 4 −9 6 ⎤ ⎢− 1⎥ ⎡2 − 3 3 6⎤ ⎢− 9 3 − 1⎥ , D = ⎢ ⎥ , E = [1 0 5 − 2], F = ⎢ , G = ⎢ ⎥ ⎣4 7 − 4 5⎥⎦ ⎢0⎥ ⎣ 6 −1 5 ⎦ ⎢⎣ 4 ⎥⎦ ⎡ 2 −1 3 ⎢− 3 4 1 and H = ⎢ 5 − 1 − 2 ⎢ ⎢⎣ 7 − 5 1

5⎤ 0⎥ ⎥ in the calculator and then perform the indicated 3⎥ 2⎥⎦

computation. If a computation is not meaningful, write Undefined.

Graphing calculators and matrices

33

1. AT 2. ( AT )T 3. F T F 4. FF T 5. ( BC )T 6. BT C T 7. C T BT 8. 5( E T ) 9. (5E )T 10. AT + F T 11. ( A + F )T 12. GT − G 13. det( HT ) − det H T

14. det(( H −1 ) ) 15. det(( HT )−1 )

Solving linear systems using Gauss-Jordan elimination As discussed in Chapter 1, Gauss-Jordan elimination is a procedure for solving systems of linear equations such as the following system of three linear equations in three unknowns: a11x + a12 y + a13z = c1 a21x + a22 y + a23z = c2 a31x + a32 y + a33z = c3

The system is consistent if it has a solution; otherwise, the system is inconsistent. The augmented matrix for the system shown is the matrix ⎡a11 a12 ⎢a a ⎢ 21 22 ⎣a31 a32

a13 c1 ⎤ a23 c2 ⎥ ⎥ a33 c3 ⎦

To solve this system, you perform elementary row operations (interchanging two rows, adding two rows, multiplying a row by a number, and multiplying a row by a number and adding the

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Linear Algebra

result to a different row) to transform the augmented matrix into reduced row-echelon form. Once the augmented matrix is in reduced row-echelon form, it is a simple process to obtain the solution in a few easy steps. If the reduced row-echelon matrix contains a row of the form (0 0 0 1), the system is inconsistent; otherwise, the system is consistent. Furthermore, the procedure generalizes to any system of m linear equations in n unknowns. With the graphing calculator, you can use the elementary row operation commands rowSwap(, row+(, *row(, and *row+( under the MATRX MATH menu to complete the transformation of the augmented matrix or you can go directly to the reduced row-echelon form by using the MATRX MATH rref( command with the restriction that the number of rows must be less than or equal to the number of columns (that is, the number of equations must be less than or equal to the number of unknowns.) PROBLEM

Use Gauss-Jordon elimination to solve the given system. a.

4x − 3 y = 3 x + y = −1

b.

x − y −z =0 x+ y+z =0

2x − 4 y − z = 5 c. −3x + y + 2z = 0 2x − 4 y − z = 6 SOLUTION

a.

4x − 3 y = 3 x + y = −1

Letting [A] be the augmented matrix, you have

Using the reduced row-echelon form of [A] yields the unique solution x = 0, y = −1. x − y −z =0 b. x + y + z = 0

Letting [B] be the augmented matrix, you have

Graphing calculators and matrices

35

Using the reduced row-echelon form of [B] yields x = 0 and y + z = 0. Thus, x = 0 and y = −z . Let y = t (an arbitrary real number), then the non-unique solutions have the form x = 0, y = t, z = −t . 2x − 4 y − z = 5 c. −3x + y + 2z = 0 2x − 4 y − z = 6

Letting [C] be the augmented matrix, you have

Because row 3 of the reduced row-echelon form of [C] is (0 0 0 is inconsistent; that is, it has no solution.

Note: See Chapter 1 for an additional discussion of Gauss-Jordan elimination.

EXERCISE

3·6 Use Gauss-Jordon elimination to solve the given system.

1.

2x + 4y = 2 3x − 2 y =1

4 x + 2 y − 6z = 2 2. 3 x − y − 4 z = 7 5x + 2 y − 6z = 5 x − 2y = 1 3. 3 x − 6 y = 3 4 x + 2 y − 6z = 2 4. 3 x − y − 4 z = 7 2 x + y − 3z = 2 2 x + 5y − 8z = 4 5. 2 x + 4 y − 6 z = 2 3 x + 8 y − 13 z = 7 6.

x + 2y = 0 2 x − 3y = 0

4 x + 2 y − 6z = 0 7. 3 x − y − 4 z = 0 5x + 2 y − 6z = 0

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1), the system

4 x + 2 y − 6z = 0 8. −2 x − y + 3 z = 0 5x + 2 y − 6z = 0 4 x + 2 y − 6z = 3 9. −2 x − y + 3 z = 5 5x + 2 y − 6z = 0 10.

4 x + 2 y − 6z = 0 5x + 2 y − 6z = 0

x+ y=4 11. 2 x + 3 y = 10 12.

2x − 4y = 3 x + 2y = 4

x + y + 2z = 9 13. 2 x + 4 y − 3 z = 1 3 x + 6 y − 5z = 0 14.

5x − 2 y + 6z = 0 −2 x + y + 3 z = 1

6a + 6b + 3c = 5 15. − 2b + 3c = 1 3a + 6b − 3c = −2

Solving linear systems using X = A-1C The n × n system of linear equations

a11x1 + a12 x 2 + + a1n xn = c1 a21x1 + a22 x 2 + + a2n xn = c2

can be written as AX = C , where

an1x1 + an 2 x 2 + + ann xn = cn ⎡a11 a12 A is the n × n coefficient matrix ⎢a21 a22 ⎢ ⎢ ⎢⎣an1 an 2

⎡x1 ⎤ … a1n ⎤ ⎥ … a2n , X is the n ×1 column vector of variables ⎢x 2 ⎥ , ⎢% ⎥ ⎥⎥ ⎢ ⎥ ⎢⎣xn ⎥⎦ ann ⎥⎦

⎡c1 ⎤ ⎢c ⎥ and C is the n ×1 column vector of constants ⎢ 2 ⎥ . ⎢% ⎥ ⎣⎢cn ⎦⎥

This system has exactly one solution X = A−1C if A is invertible (nonsingular). Graphing calculators and matrices

37

Caution: If A does not have an inverse, then X = A−1C does not apply. PROBLEM

Use the inverse of the coefficient matrix to solve the given system. When appropriate, give the solution in fractional form. a.

4 x1 − 3 x 2 = 3 x1 + x 2 = −1

4 x1 + x 2 + 3x 3 = −1 b. 3x1 − 2 x 2 + x 3 = 9 − x1 + 5 x 2 + 2 x 3 = 7 2 x1 − 4 x 2 − x 3 = 5 c. −3x1 + x 2 + 2 x 3 = 0 2 x1 − 4 x 2 − x 3 = 6

d.

x1 − x 2 − x 3 = 0 x1 + x 2 + x 3 = 0

4 x1 − x 2 + 2 x 3 = −1 e. x1 − 4 x 2 + x 3 = 6 2 x1 + 2 x 2 − 3x 3 = −20 SOLUTION

4 x1 − 3 x 2 = 3 a. x + x = −1 1 2

Input [A] = 2 × 2 coefficient matrix and [C] = 2 × 1 column matrix of constants as displayed here.

Paste [A] to the home screen. Press x −1 . Paste [C], and then press screen display shows the following answer.

Thus, the solution is x1 = 0, x 2 = −1 4 x1 + x 2 + 3x 3 = −1 b. 3x1 − 2 x 2 + x 3 = 9 − x1 + 5 x 2 + 2 x 3 = 7

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Linear Algebra

ENTER .

The

Input [A] = 3 × 3 coefficient matrix and [C] = 3 × 1 column matrix of constants as displayed here.

Paste [A] to the home screen. Press x −1 . Paste [C], and then press screen display shows the following answer.

Thus, the solution is x1 = −43.25, x 2 = −35.25, x 3 = 69.75 or x1 = − x3 =

279 . 4

ENTER .

The

141 175 , x2 = − , 4 4

2 x1 − 4 x 2 − x 3 = 5 c. −3x1 + x 2 + 2 x 3 = 0 2 x1 − 4 x 2 − x 3 = 6

Input [A] = 3 × 3 coefficient matrix and [C] = 3 × 1 column matrix of constants as shown here.

Paste [A] to the home screen. Press x −1 . Paste [C], and then press screen display shows the following error message:

ENTER .

The

The error message indicates that [A] is singular. Looking back at [A], you see that rows 1 and 3 are the same. When this situation occurs, the matrix is singular (thus, not invertible). So, X = A−1C does not apply. Graphing calculators and matrices

39

d.

x1 − x 2 − x 3 = 0 x1 + x 2 + x 3 = 0

The coefficient matrix is not square, so X = A−1C does not apply. 4 x1 − x 2 + 2 x 3 = −1 e. x1 − 4 x 2 + x 3 = 6 2 x1 + 2 x 2 − 3x 3 = − 20

Input [A] = 3 × 3 coefficient matrix and [C] = 3 × 1 column matrix of constants as displayed here.

Paste [A] to the home screen. Press x −1 . Paste [C], and then press screen display shows the following answer:

Press

MATH ENTER ENTER

Thus, the solution is x1 = −

ENTER .

The

to convert the elements to fractions. The display shows

144 61 46 , x2 = − , x3 = 55 55 11

Note: See Chapter 2 for an additional discussion of solving systems of linear equations using X = A−1C . EXERCISE

3·7 Use the inverse of the coefficient matrix to solve the given system. When appropriate, give the solution in fractional form.

1. 40

2 x1 + 4 x 2 = 2 3 x1 − 2 x 2 = 1

practice makes perfect

Linear Algebra

4 x1 + 2 x 2 − 6 x 3 = 2

2. 3 x1 − x 2 − 4 x 3 = 7 5 x1 + 2 x 2 − 6 x 3 = 5

3.

x1 − 2 x 2 = 1 3 x1 − 6 x 2 = 3

4 x1 + 2 x 2 − 6 x 3 = 2 4. 3 x1 − x 2 − 4 x 3 = 7 2 x1 + x 2 − 3 x 3 = 5 2 x1 + 5 x 2 − 8 x 3 = 4 5. 2 x1 + 4 x 2 − 6 x 3 = 2 3 x1 + 8 x 2 − 13 x 3 = 7

6.

3 x1 + 2 x 2 = 12 3 x1 − 2 x 2 = 0

4 x1 + 2 x 2 − 6 x 3 = 0 7. 3 x1 − x 2 − 4 x 3 = 0 5 x1 + 2 x 2 − 6 x 3 = 0 2 x1 − x 2 + x 3 = 6

8. x1 + x 2 + x 3 = 0 4 x1 + 5 x 2 − x 3 = 2

9.

11x1 + x 2 + 2 x 3 = 2 x1 + 5 x 2 + 2 x 3 = 2 4 x1 + 5 x 2 =0

7 x1 + 3 x 2 − 5 x 3 + 8 x 4 = −3 x + x2 + x3 + 2 x 4 = 3 10. 1 4 x1 + x 2 + x 3 + x 4 = 6 3 x1 + 7 x 2 − x 3 + x 4 = 1

Graphing calculators and matrices

41

Special types of square matrices

·4·

In this chapter, you will learn: ◆ ◆ ◆ ◆ ◆ ◆

Nonsingular matrices Triangular, diagonal, and scalar matrices Involutory, idempotent, and nilpotent matrices Symmetric and skew-symmetric matrices Orthogonal matrices Hermitian and skew-Hermitian matrices

This chapter contains a discussion of different types of square matrices. Square matrices play an integral part in linear algebra.

Nonsingular matrices A square matrix A is nonsingular or invertible if its inverse exists. The inverse is denoted A−1 and is unique (that is, there is at most one). Therefore, AA−1 = A−1 A = I for any nonsingular matrix A. Only square matrices have inverses, and the inverse of a square matrix A exists if and only if its determinant is not zero, denoted det( A) ≠ 0 (see Chapter 5 for a discussion of determinants). The following properties for a nonsingular matrix A hold: ◆ ◆ ◆ ◆ ◆

( A−1 )−1 = A 1 (kA)−1 = A−1 , for a nonzero scalar k k ( A−1 )T = ( AT )−1 Furthermore, for the k n × n nonsingular matrices A1, A2 , # , Ak , ( A1 A2 $ Ak−1 Ak )−1 = Ak −1 Ak−1−1 $ A2−1 A1−1 ; that is, the inverse of the product is the product of the inverses in the reverse order.

PROBLEM

⎡1 2 4 ⎤ Let A = ⎢⎢ 0 −1 1 ⎥⎥ and B = ⎢⎣ 2 3 8 ⎥⎦

⎡ −11 −4 6 ⎤ ⎢ 2 0 −1⎥⎥ . ⎢ ⎢⎣ 2 1 −1⎥⎦

a. Show that A and B are inverses of each other. b. Show that ( A−1 )−1 = A. c. Show that ( B−1 )T = ( BT )−1. 1 d. Show that (4 A)−1 = A−1. 4

42

SOLUTION

First, input A and B into the calculator:

a. Show that A and B are inverses of each other.

⎡1 0 0 ⎤ Thus, AB = BA = ⎢⎢0 1 0 ⎥⎥ = I, so A and B are inverses of each other. ⎢⎣0 0 1 ⎥⎦

b. Show that ( A−1 )−1 = A.

⎡1 2 4 ⎤ Thus, ( A ) = ⎢⎢ 0 −1 1 ⎥⎥ , which is the same as A = ⎢⎣ 2 3 8 ⎥⎦ −1 −1

⎡1 2 4 ⎤ ⎢ 0 −1 1 ⎥ , so ( A−1 )−1 = A. ⎥ ⎢ ⎢⎣ 2 3 8 ⎥⎦

c. Show that ( B−1 )T = ( BT )−1

Special types of square matrices

43

⎡1 0 2 ⎤ Thus, ( B ) = ( B ) = ⎢⎢ 2 −1 3 ⎥⎥ . ⎢⎣ 4 1 8 ⎥⎦ 1 d. Show that (4 A)−1 = A−1 4 −1 T

T −1

⎡ −11 4 1 −1 ⎢⎢ 1 −1 Thus, (4 A) = A = ⎢ 2 4 ⎢ 1 ⎣ 2

⎤ ⎡ −11 −4 6 ⎤ 2 ⎥ 1⎢ 1 ⎥ 0 −1⎥⎥ . 0 − 4 = ⎢ 2 ⎥ 4 ⎢⎣ 2 1 −1⎥⎦ 1 1 ⎥ − 4 4⎦ −1

3

Notice that using MATH •Frac, displays the matrix answers in fractional form. Also, you must use the arrow keys to scroll across the screen to see all the entries. PROBLEM

Use the calculator’s MATRX det( command to determine whether the given matrix is nonsingular or singular. ⎡ 1 −1 2 ⎤ a. A = ⎢⎢ 2 −1 7 ⎥⎥ ⎢⎣ −4 5 −3 ⎥⎦ ⎡ 5 −1 −3 ⎤ b. B = ⎢⎢ 4 −2 6 ⎥⎥ ⎢⎣ −2 1 −3 ⎥⎦ ⎡1 5 −2 8 ⎤ ⎢ 2 3 −7 −1⎥ ⎥ c. C = ⎢ ⎢0 0 0 0 ⎥ ⎢ ⎥ ⎣1 −5 4 3 ⎦

SOLUTION

⎡ 1 −1 2 ⎤ a. A = ⎢⎢ 2 −1 7 ⎥⎥ ⎢⎣ −4 5 −3 ⎥⎦

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det( A) = 2 ≠ 0 , so A is nonsingular. ⎡ 5 −1 −3 ⎤ b. B = ⎢⎢ 4 −2 6 ⎥⎥ ⎢⎣ −2 1 −3 ⎥⎦

det( B) = 0 , so B is singular. ⎡1 5 −2 8 ⎤ ⎢ 2 3 −7 −1⎥ ⎥ c. C = ⎢ ⎢0 0 0 0 ⎥ ⎥ ⎢ ⎣1 −5 4 3 ⎦

det(C ) = 0 , so C is singular. EXERCISE

4·1

⎡ 3 −1⎤ ⎡ 2 1⎤ For 1–4, let A = ⎢ ⎥ and B = ⎢ 5 3 ⎥ . Give answers in fractional form as needed. − 5 2 ⎦ ⎦ ⎣ ⎣

1. Show that A and B are inverses of each other. 2. Show that ( A−1 )−1 = A. 3. Show that ( B −1 )T = ( BT )−1. 1 4. Show that (10 A)−1 = A−1. 10 1 5. Show that ( A + B )−1 = I . 5 6. Show that A−1 + B −1 = 5I. 7. Show that ( A2 )−1 = ( A−1 )2 . Special types of square matrices

45

⎡1 4 2⎤ ⎡ −1 0 0 ⎤ ⎡ −1 9 1 ⎤ For 8–12, let C = ⎢⎢ −2 5 0 ⎥⎥ , D = ⎢⎢ 0 5 0 ⎥⎥ , and E = ⎢⎢ 0 5 3 ⎥⎥ . Give answers in fractional form as needed. ⎢⎣ 0 0 −4 ⎥⎦ ⎢⎣ 0 0 −4 ⎥⎦ ⎣⎢ 3 −6 −1⎥⎦

8. Find C −1. 9. Find D −1. 10. Find E −1 . 11. Show that (C + D )−1 ≠ C −1 + D −1. 12. Show that (CDE )−1 = E −1D −1C −1 . For 13–15, use the calculator’s or singular.

MATRX det(

command to determine whether the given matrix is nonsingular

⎡3 2 ⎤ 13. F = ⎢ ⎥ ⎣5 3 ⎦ ⎡1 2 4⎤ 14. G = ⎢⎢ 0 −1 1 ⎥⎥ ⎢⎣ 2 4 8 ⎥⎦ ⎡ 1 5 −2 8 ⎤ ⎢ 2 3 −7 −1⎥ ⎥ 15. H = ⎢ ⎢0 1 0 2 ⎥ ⎥ ⎢ ⎣ 1 −5 4 3 ⎦

Triangular, diagonal, and scalar matrices A square matrix is upper triangular if all its elements below the main diagonal are 0s. For example, ⎡− 1 9 1 ⎤ the matrix ⎢ 0 5 3 ⎥ is upper triangular. Note: Hereafter a matrix’s main diagonal will be ⎢ ⎥ ⎣ 0 0 − 4⎦ called simply its diagonal. A square matrix is lower triangular if all its elements above the diagonal are 0s. For example, ⎡− 3 0 0 ⎤ the matrix ⎢ 0 1 0 ⎥ is lower triangular. ⎢ ⎥ ⎣ 2 7 − 4⎦ Triangular matrices must have only 0s either below the diagonal (upper triangular) or above the diagonal (lower triangular). However, 0s can occur as other elements, including on the diago⎡0 ⎢0 nal, as well. For instance, the matrix ⎢ ⎢ ⎢⎣0

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Linear Algebra

0 … 0⎤ 0 … 0⎥ is both upper and lower triangular. ⎥⎥ 0 … 0⎥⎦ n × n

Triangular matrices have the following properties: ◆

◆

◆ ◆ ◆ ◆ ◆ ◆

If A and B are n × n upper triangular matrices, then A + B , AB, and kA are upper triangular matrices. If A and B are lower triangular matrices, then A + B, AB, and kA are lower triangular matrices. If A is upper triangular, then AT is lower triangular. If A is lower triangular, then AT is upper triangular. If A is a triangular matrix, then det( A) is the product of A’s diagonal elements. A triangular matrix A is nonsingular if and only if each diagonal element aii ≠ 0. If A is a nonsingular upper triangular matrix, then A−1 is upper triangular. If A is a nonsingular lower triangular matrix, then A−1 is lower triangular. PROBLEM

SOLUTION

⎡ −1 9 1 ⎤ Given upper triangular matrices A = ⎢⎢ 0 5 3 ⎥⎥ and B = ⎢⎣ 0 0 −4 ⎥⎦ a. Show that A + B is upper triangular. b. Show that AB is upper triangular. c. Show that 5B is upper triangular. d. Show that AT is lower triangular. e. Show that A−1 is upper triangular. f. Find det( A).

⎡ 2 1 −1⎤ ⎢ 0 −3 5 ⎥ . ⎥ ⎢ ⎢⎣ 0 0 4 ⎥⎦

First, input A and B into the calculator:

a. Show that A + B is upper triangular.

A + B is upper triangular because it is a square matrix whose elements below the diagonal are 0s. b. Show that AB is upper triangular.

AB is upper triangular because it is a square matrix whose elements below the diagonal are 0s. Special types of square matrices

47

c. Show that 5B is upper triangular.

5B is upper triangular because it is a square matrix whose elements below the diagonal are 0s. d. Show that AT is lower triangular.

AT is lower triangular because it is a square matrix whose elements above the diagonal are 0s. e. Show that A−1 is upper triangular.

A−1 is upper triangular because it is a square matrix whose elements below the diagonal are 0s. f. Find det( A). ⎡ −1 9 1 ⎤ det( A) = ⎢⎢ 0 5 3 ⎥⎥ = −1 ⋅ 5 ⋅−4 = 20 ⎢⎣ 0 0 −4 ⎥⎦

A diagonal matrix is a square matrix in which all the non-diagonal elements are 0s. For ⎡− 1 ⎡2 0⎤ ⎢ instance, the matrices ⎢ , 0 ⎣0 3⎥⎦ ⎢ 0 ⎣ onal matrices are both upper and diagonal.

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Linear Algebra

0 0⎤ ⎡3 ⎥ ⎢0 , and 5 0 ⎥ ⎢ 0 − 4⎦ ⎣0 lower triangular

0 0⎤ 0 0 ⎥ are diagonal matrices. Clearly, diag⎥ 0 − 5⎦ because they have 0s below and above the

Diagonal matrices have the following properties: ⎡a11 0 … 0 ⎤ ⎢ 0 a22 0 ⎥ Given n × n diagonal matrices A = ⎢ ⎥ and B = ⎢ ⎥ 0 … ann ⎥⎦ ⎢⎣ 0

⎡b11 0 … 0 ⎤ ⎢ 0 b22 0 ⎥ ⎢ ⎥ , then ⎢ ⎥ 0 … bnn ⎥⎦ ⎢⎣ 0

◆

0 0 ⎤ … ⎡a11 + b11 ⎢ 0 a22 + b22 0 ⎥ A+ B = ⎢ ⎥⎥ ⎢ 0 … ann + bnn ⎥⎦ ⎢⎣ 0

◆

0 … 0 ⎤ ⎡a11b11 ⎢ 0 a22b22 0 ⎥ AB = ⎢ ⎥ = BA ⎢ ⎥ … annbnn ⎥⎦ 0 ⎣⎢ 0

◆

⎡ka11 ⎢ 0 kA = ⎢ ⎢ ⎢⎣ 0

◆

AT = A

◆

a11 0 … 0 0 a22 0 det( A) = = a11a22 ann 0 0 … ann

◆

A diagonal matrix A is nonsingular if and only if each diagonal element aii ≠ 0.

◆

⎡1 ⎢ a11 ⎢ 0 A−1 = ⎢ ⎢ ⎢ ⎢⎣ 0

◆

⎡(a11 )p ⎢ 0 Ap = ⎢ ⎢ ⎢⎣ 0

0 … 0 ⎤ ka22 0 ⎥ , where k is a scalar. ⎥⎥ 0 … kann ⎥⎦

…

0 1

a22 …

0 0 (a22 ) 0

p

0 ⎤ ⎥ 0 ⎥ ⎥ , provided no aii is 0. ⎥ 1 ⎥ ann ⎥⎦

… 0 ⎤ 0 ⎥ ⎥ ⎥ … (ann )p ⎥⎦

A diagonal matrix all of whose diagonal elements are the same scalar is a scalar matrix. For ⎡2 0⎤ instance, ⎢ , ⎣0 2⎥⎦

⎡− 3 0 0 0 ⎤ ⎢ 0 −3 0 0 ⎥ ⎢ 0 0 − 3 0 ⎥ , and In are scalar matrices. ⎥ ⎢ ⎢⎣ 0 0 0 − 3⎥⎦ Special types of square matrices

49

PROBLEM

⎡ −1 0 0 ⎤ Given diagonal matrices A = ⎢⎢ 0 5 0 ⎥⎥ and B = ⎢⎣ 0 0 7 ⎥⎦ a. Show that A + B is the scalar matrix 3I. b. Compute AB. c. Compute 5B. d. Find AT . e. Find A−1. f. Find A3 . g. Find det( A).

⎡4 0 0 ⎤ ⎢ 0 −2 0 ⎥ . ⎥ ⎢ ⎢⎣ 0 0 −4 ⎥⎦

(Try this problem without using graphing calculator matrix features.) SOLUTION

a. Show that A + B is the scalar matrix 3I. ⎡ −1 0 0 ⎤ A + B = ⎢⎢ 0 5 0 ⎥⎥ + ⎢⎣ 0 0 7 ⎥⎦

⎡4 0 0 ⎤ ⎢ 0 −2 0 ⎥ = ⎥ ⎢ ⎢⎣ 0 0 −4 ⎥⎦

⎡3 0 0 ⎤ ⎡1 0 0 ⎤ ⎢0 3 0 ⎥ = 3 ⎢0 1 0 ⎥ ⎥ ⎥ ⎢ ⎢ ⎢⎣0 0 3 ⎥⎦ ⎢⎣0 0 1 ⎥⎦

b. Compute AB. 0 ⎤ ⎡ −1 0 0 ⎤ ⎡ 4 0 0 ⎤ ⎡ −4 0 AB = ⎢⎢ 0 5 0 ⎥⎥ ⎢⎢ 0 −2 0 ⎥⎥ = ⎢⎢ 0 −10 0 ⎥⎥ ⎢⎣ 0 0 7 ⎥⎦ ⎢⎣ 0 0 −4 ⎥⎦ ⎢⎣ 0 0 −28 ⎥⎦

c. Compute 5B. 0 ⎤ ⎡ 4 0 0 ⎤ ⎡ 20 0 ⎢ ⎥ ⎢ 5 ⎢ 0 −2 0 ⎥ = ⎢ 0 −10 0 ⎥⎥ ⎢⎣ 0 0 −4 ⎥⎦ ⎢⎣ 0 0 −20 ⎥⎦

d. Find AT. T

⎡− 1 0 0⎤ ⎡− 1 0 0⎤ ⎢ 0 5 0⎥ = ⎢ 0 5 0⎥ ⎥ ⎢ ⎥ ⎢ ⎣ 0 0 7⎦ ⎣ 0 0 7⎦ e. Find A−1. ⎡ −1 0 0 ⎤ ⎢ 0 5 0⎥ ⎥ ⎢ ⎢⎣ 0 0 7 ⎥⎦

−1

f. Find A3 .

⎡ −1 0 ⎢ = ⎢ 0 15 ⎢ ⎢0 0 ⎣

0 ⎤ ⎥ 0 ⎥ ⎥ 1 ⎥ 7⎦

3

0 ⎤ ⎡ −1 0 0 ⎤ ⎡ −1 0 ⎢ 0 5 0 ⎥ = ⎢ 0 125 0 ⎥ ⎥ ⎢ ⎥ ⎢ ⎢⎣ 0 0 7 ⎥⎦ ⎢⎣ 0 0 343 ⎦⎥ g. Find det( A). −1 0 0 det( A) = 0 5 0 = −1 ⋅ 5 ⋅ 7 = −35 0 0 7

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Linear Algebra

EXERCISE

4·2 For 1–5, use U = upper triangular matrix only, L = lower triangular matrix only, and D = diagonal matrix to identify each of the following.

⎡ 1 0⎤ 1. ⎢ ⎥ ⎣ −2 4 ⎦ ⎡1 0 0⎤ 2. ⎢⎢ 0 5 0 ⎥⎥ ⎢⎣ 0 0 6 ⎥⎦ ⎡1 ⎢0 3. ⎢ ⎢0 ⎢ ⎣0

1 4 0 0

0 2⎤ 3 −5 ⎥ ⎥ 3 0⎥ ⎥ 0 5⎦

⎡3 0 0 ⎤ 4. ⎢⎢ 0 3 0 ⎥⎥ ⎢⎣ 0 0 3 ⎥⎦ ⎡2 0 0 ⎤ 5. ⎢⎢9 −2 0 ⎥⎥ ⎢⎣ 1 −4 3 ⎥⎦ ⎡ −4 0 0 ⎤ ⎡ −2 0 0 ⎤ ⎢ ⎥ For 6–10, let A = ⎢ 5 −3 0 ⎥ and B = ⎢⎢ 6 −1 0 ⎥⎥ be lower triangular matrices. ⎢⎣ 7 ⎢⎣ 1 −2 5 ⎥⎦ 4 1 ⎥⎦

6. Show that A + B is lower triangular. 7. Show that AB is lower triangular. 8. Show that 5B is lower triangular. 9. Show that AT is upper triangular. 10. Show that A−1 is lower triangular. ⎡2 0 0 ⎤ For 11–15, let A = ⎢⎢ 0 −4 0 ⎥⎥ and B = ⎢⎣ 0 0 3 ⎥⎦

⎡3 0 0 ⎤ ⎢ 0 9 0 ⎥ be diagonal matrices. ⎢ ⎥ ⎢⎣ 0 0 −5 ⎥⎦

11. Compute A + B. 12. Compute AB. 13. Compute 5B . 14. Find A−1. 15. Find A4.

Special types of square matrices

51

Involutory, idempotent, and nilpotent matrices A square matrix A is involutory if it is its own inverse; that is, if A = A−1 or equivalently if A2 = I . Clearly, because ( In )2 = In , the identity matrix is involutory. PROBLEM

⎡2 3⎤ Show that A = ⎢ ⎥ is involutory. ⎣ −1 −2 ⎦

SOLUTION

⎡2 3⎤ A2 = ⎢ ⎥ ⎣ −1 −2 ⎦

⎡2 3⎤ ⎢ −1 −2 ⎥ = ⎣ ⎦

⎡ 4 − 3 6 − 6 ⎤ ⎡1 0 ⎤ ⎢ −2 + 2 −3 + 4 ⎥ = ⎢0 1 ⎥ = I ⎣ ⎦ ⎣ ⎦

A square matrix A is idempotent if A2 = A. Given that ( In )2 = In , the identity matrix is idempotent, and is the only nonsingular idempotent matrix. PROBLEM

⎡1 −1⎤ Show that A = ⎢ ⎥ is idempotent. ⎣0 0 ⎦

SOLUTION

⎡1 −1⎤ ⎡1 −1⎤ ⎡1 + 0 −1 + 0 ⎤ ⎡1 −1⎤ A2 = ⎢ ⎥=⎢ ⎥⎢ ⎥=⎢ ⎥=A ⎣0 0 ⎦ ⎣0 0 ⎦ ⎣0 + 0 0 + 0 ⎦ ⎣0 0 ⎦

A square matrix A is nilpotent if there is a positive integer p such that A P = 0 (the least such integer p is the index of nilpotence of A). All nilpotent matrices are singular (that is, not invertible). PROBLEM

⎡ 0 0 0⎤ Show that A = ⎢⎢ 5 0 0 ⎥⎥ is nilpotent of index 3. ⎢⎣ −2 3 0 ⎥⎦

SOLUTION

Because A3 = 0 and A2 ≠ 0 , A is nilpotent of index 3. EXERCISE

4·3 For 1–5, fill in the blank with the best response.

1. A square matrix A is ____________ if A2 = I . 2. A square matrix A is idempotent if ____________ equals A. 3. The only nonsingular n × n idempotent matrix is the n × n ____________ matrix. 4. A square matrix A is nilpotent if there is an integer p such that AP equals the ____________ matrix. 5. All nilpotent matrices are ____________ (nonsingular, singular).

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Linear Algebra

For 6–10, respond as indicated.

2⎤ ⎡ 5 6. Show that A = ⎢ ⎥ is involutory. ⎣ −12 −5 ⎦ ⎡ 4 −1 −4 ⎤ ⎢ ⎥ 7. Show that B = ⎢ 3 0 −4 ⎥ is involutory. ⎢⎣ 3 −1 −3 ⎥⎦ 1⎤ ⎡3 8. Show that C = ⎢ is idempotent. − 6 − 2 ⎥⎦ ⎣ ⎡ −1.5 0.5 2 ⎤ ⎢ ⎥ 9. Show that D = ⎢ −1.5 0.5 2 ⎥ is idempotent. ⎢⎣ −1.5 0.5 2 ⎥⎦ ⎡ 1 5 −2 ⎤ ⎢ ⎥ 10. Show that A = ⎢ 1 2 −1⎥ is nilpotent of index 3. ⎢⎣3 6 −3 ⎥⎦

Symmetric and skew-symmetric matrices A square matrix A is symmetric if AT = A . That is, if A = ⎡⎣aij ⎤⎦

n×n

, where aij is the element in the

ith row and jth column of A, then A is symmetric if each aij = a ji . For instance, the matrices 3⎤ ⎡− 1 0 ⎡2 5 ⎤ ⎢ ⎥ ⎢⎣5 − 1⎥⎦ , ⎢ 0 4 − 2⎥ , and In are symmetric. ⎣ 3 − 2 − 5⎦ PROBLEM

Show that the given matrix is symmetric. ⎡2 5 ⎤ a. A = ⎢ ⎥ ⎣5 −1⎦ ⎡ −1 0 3 ⎤ ⎥ ⎢ b. B = ⎢ 0 4 −2 ⎥ ⎢⎣ 3 −2 −5 ⎥⎦

SOLUTION

⎡2 5 ⎤ a. A = ⎢ ⎥ ⎣5 −1⎦ T

⎡2 5 ⎤ ⎡2 5 ⎤ ⎢5 −1⎥ = ⎢5 −1⎥ ⎣ ⎦ ⎣ ⎦

Thus, because AT = A , A is symmetric. ⎡ −1 0 3 ⎤ b. B = ⎢⎢ 0 4 −2 ⎥⎥ ⎢⎣ 3 −2 −5 ⎥⎦ Special types of square matrices

53

T

⎡ −1 0 3 ⎤ ⎡ −1 0 3 ⎤ ⎢ 0 4 −2 ⎥ = ⎢ 0 4 −2 ⎥ ⎢ ⎥ ⎥ ⎢ ⎢⎣ 3 −2 −5 ⎥⎦ ⎢⎣ 3 −2 −5 ⎥⎦

Thus, because BT = B , B is symmetric.

If A and B are n × n symmetric matrices, the following properties hold: ◆ ◆ ◆ ◆ ◆ ◆

A + B is symmetric kA is symmetric for a scalar k. A is symmetric if and only if AT is symmetric. AB is symmetric if and only if AB = BA. If A is nonsingular, A−1 is symmetric. A p is symmetric for p a nonnegative integer.

A square matrix A is skew-symmetric if AT = − A. That is, if A = ⎡⎣aij ⎤⎦

n×n

, where aij is the ele-

ment in the ith row and jth column of A, then A is skew-symmetric if each aij = −a ji . It’s plain to see that a skew-symmetric matrix must have all 0s on its diagonal. For instance, the matrices ⎡ 0 1 − 3⎤ ⎡0 − 3⎤ ⎢ , 2 ⎥ , and 0n × n are skew-symmetric. ⎢⎣3 0 ⎥⎦ ⎢− 1 0 ⎥ ⎣ 3 −2 0 ⎦ PROBLEM

Show that the given matrix is skew-symmetric. ⎡0 −3 ⎤ a. C = ⎢ ⎥ ⎣3 0 ⎦ ⎡ 0 1 −3 ⎤ b. D = ⎢⎢ −1 0 2 ⎥⎥ ⎢⎣ 3 −2 0 ⎥⎦

SOLUTION

⎡0 −3 ⎤ a. C = ⎢ ⎥ ⎣3 0 ⎦ T

⎡0 −3 ⎤ ⎡ 0 3⎤ ⎡0 −3 ⎤ C =⎢ =⎢ = −⎢ ⎥ ⎥ ⎥ = −C ⎣3 0 ⎦ ⎣ −3 0 ⎦ ⎣3 0 ⎦ T

Thus, because C T = −C , C is skew-symmetric. b.

⎡ 0 1 −3 ⎤ D = ⎢⎢ −1 0 2 ⎥⎥ ⎢⎣ 3 −2 0 ⎥⎦ T

⎡ 0 1 −3 ⎤ ⎡ 0 1 −3 ⎤ ⎡ 0 −1 3 ⎤ ⎥ ⎢ ⎥ ⎢ T D = ⎢ −1 0 2 ⎥ = ⎢ 1 0 −2 ⎥ = − ⎢⎢ −1 0 2 ⎥⎥ = − D ⎢⎣ −3 2 0 ⎥⎦ ⎢⎣ 3 −2 0 ⎥⎦ ⎢⎣ 3 −2 0 ⎥⎦

Thus, because D T = − D , D is skew-symmetric.

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If A and B are n × n skew-symmetric matrices, the following properties hold: ◆ ◆ ◆ ◆ ◆

A + B is skew-symmetric. kA is skew-symmetric for a scalar k. A is skew-symmetric if and only if AT is skew-symmetric. If A is skew-symmetric and nonsingular, A−1 is skew-symmetric. AB is skew-symmetric if and only if AB = − BA.

EXERCISE

4·4 For 1–5, fill in the elements to produce the type of matrix indicated.

⎡ 0 4 −3 ⎤ 1. ⎢ ? ? ? ⎥⎥ , skew-symmetric ⎢ ⎢⎣ 3 −8 0 ⎥⎦ ⎡2 4 ? ⎤ 2. ⎢ ? 6 ? ⎥ , symmetric ⎥ ⎢ ⎢⎣ 3 −8 0 ⎥⎦ ⎡ −7 4 ? ⎤ 3. ⎢ ? 5 −2 ⎥ , symmetric ⎥ ⎢ ⎢⎣ 3 ? 9 ⎥⎦ ⎡ 0 10 7 −9 ⎤ ⎢? 0 ? 2 ⎥ ⎥ , skew-symmetric 4. ⎢ ⎢? 5 0 ? ⎥ ⎢ ⎥ ⎣ ? ? 15 ? ⎦ ⎡0 5. ⎢ ⎣?

2⎤ ⎥ , skew-symmetric ? ⎦

⎡ 0 4⎤ For 6–15, let A = ⎢ ⎥,B= ⎣ −4 0 ⎦ ⎡1 0⎤ I2 = ⎢ ⎥. ⎣ 0 1⎦

⎡0 1 8 ⎤ ⎡ 0 −2 ⎤ 3⎥ ⎢ 1 0 −2 ⎥ , C = ⎢ ,D= ⎢ ⎥ ⎢2 ⎥ 0 ⎢⎣ 8 −2 0 ⎥⎦ ⎣ 3 ⎦

⎡0 0 0 ⎤ ⎢ 0 1 0 ⎥ , E = ⎡2 3 −4 −7 ⎤ , and ⎢ 3 −4 7 ⎢ ⎥ 0 ⎥⎦ ⎣ ⎢⎣ 0 0 0 ⎥⎦

6. Is A symmetric or skew-symmetric? 7. Is B symmetric or skew-symmetric? 8. Is C symmetric or skew-symmetric? 9. Is D symmetric or skew-symmetric? 10. Is E symmetric or skew-symmetric? 11. Is I2 symmetric or skew-symmetric?

⎡ 0 2 −1⎤ ⎡ −1 2 ⎤ ⎡ −4 5 ⎤ ⎢ ⎥ For 12–15, let A = ⎢ ⎥ , B = ⎢ 5 −6 ⎥ , and C = ⎢ −2 0 −5 ⎥ . ⎦ ⎣ 2 −3 ⎦ ⎣ ⎢⎣ 1 5 0 ⎥⎦

12. Show that A + B is symmetric.

Special types of square matrices

55

13. Show that A−1 is symmetric. 14. Show that B 5 is symmetric. 15. Show that C T is skew-symmetric.

Orthogonal matrices A nonsingular matrix A is orthogonal if AT = A−1. It follows that A is orthogonal if and only if AT A = AAT = I .

PROBLEM

⎡0 0 1 ⎤ Show that A = ⎢⎢1 0 0 ⎥⎥ is orthogonal. ⎢⎣0 1 0 ⎥⎦

SOLUTION

⎡ 0 1 0 ⎤ ⎡ 0 0 1 ⎤ ⎡1 0 0 ⎤ A A = ⎢⎢0 0 1 ⎥⎥ ⎢⎢1 0 0 ⎥⎥ = ⎢⎢0 1 0 ⎥⎥ = I and ⎢⎣1 0 0 ⎥⎦ ⎢⎣0 1 0 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ T

⎡0 0 1 ⎤ ⎡0 1 0 ⎤ AA = ⎢⎢1 0 0 ⎥⎥ ⎢⎢0 0 1 ⎥⎥ = ⎢⎣0 1 0 ⎥⎦ ⎢⎣1 0 0 ⎥⎦ T

⎡1 0 0 ⎤ ⎢0 1 0 ⎥ = I ⎥ ⎢ ⎢⎣0 0 1 ⎥⎦

Because AT A = AAT = I , A is orthogonal. Note: It is sufficient to show either AT A = I or AAT = I . You do not have to show both.

If A and B are n × n orthogonal matrices, the following properties hold: ◆ ◆ ◆

AB is orthogonal. A−1 is orthogonal. The determinant of A is either 1 or −1.

Note: See Chapter 5 for a discussion of determinants. Also, see Chapter 3 for instructions on calculating determinants using a graphing calculator. EXERCISE

4·5 For 1–2, fill in the blank with the best response.

1. A nonsingular matrix A is orthogonal if AT = ____________. 2. A matrix A is orthogonal if and only if AT A = AAT = ____________. For 3–5, respond as indicated.

⎡ 0 0 1⎤ 3. Given A = ⎢ 1 0 0 ⎥ , find det(A). ⎥ ⎢ ⎢⎣ 0 1 0 ⎥⎦

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⎡ 0 −1 0 ⎤ 4. Given B = ⎢ 1 0 0 ⎥ , show that B is orthogonal. ⎥ ⎢ ⎢⎣ 0 0 1 ⎥⎦ ⎡ 1 2 ⎢ 5. Given C = ⎢ 1 ⎢⎣ 2

⎤ 2⎥ ⎥ , show that C is orthogonal. 1 2 ⎥⎦

− 1

Hermitian and skew-Hermitian matrices A complex matrix A is a matrix with complex numbers as elements. When you replace the elements of A with their complex conjugates, the resulting matrix is referred to as the conjugate of A, denoted A. Clearly, a matrix A is a real matrix, meaning it has only real elements, if and only if A = A. The transpose of the conjugate of a complex matrix A is the conjugate transpose (also, called the Hermitian transpose) of A, denoted A∗. A complex matrix A is Hermitian if A∗ = A; ⎡ 1 1 − i⎤ is Hermitian if each aij = a ji . For instance, the complex matrix ⎢ is that is, A = ⎡⎣aij ⎤⎦ n×n ⎣1 + i − 2 ⎥⎦ Hermitian. Note that all the diagonal elements of a Hermitian matrix must be real numbers. PROBLEM

⎡ 1 1− i⎤ Show that the complex matrix A = ⎢ ⎥ is Hermitian. ⎣1 + i −2 ⎦

SOLUTION

⎡ 1 1− i⎤ ⎡ 1 1+ i⎤ Given A = ⎢ ⎥ , then A = ⎢1 − i −2 ⎥ ; and 1 + − 2 i ⎣ ⎦ ⎣ ⎦ ⎡ 1 1− i⎤ A∗ = ( A)T = ⎢ ⎥= A ⎣1 + i −2 ⎦

Because A∗ = A, A is Hermitian.

If A and B are n × n Hermitian matrices, the following properties hold: ◆ ◆ ◆

( A∗ )∗ = A ( A + B)∗ = A∗ + B∗ ( AB)∗ = B∗ A∗

A complex matrix A is skew-Hermitian if A∗ = − A ; that is, if A = ⎡⎣aij ⎤⎦ n × n is skew-Hermitian

⎡ 2i − 3 + i⎤ is skew-Hermitian. Note that if each aij = −(a ji ) . For instance, the complex matrix ⎢ −i ⎥⎦ ⎣3 + i all the diagonal elements of a skew-Hermitian matrix must be pure imaginary numbers.

Special types of square matrices

57

PROBLEM

⎡ 2i −3 + i ⎤ Show that the complex matrix A = ⎢ is skew-Hermitian. −i ⎥⎦ ⎣3 + i

SOLUTION

⎡ −2i −3 − i ⎤ ⎡ 2i −3 + i ⎤ Given A = ⎢ , then A = ⎢ ; and ⎥ i ⎥⎦ −i ⎦ ⎣3 − i ⎣3 + i ⎡ −2i 3 − i ⎤ A∗ = ( A)T = ⎢ = −A i ⎥⎦ ⎣ −3 − i

Because A∗ = − A, A is skew-Hermitian. EXERCISE

4·6 For 1–8, fill in the blank with the best response.

1. A matrix A is a real matrix if and only if A = ____________. 2. A complex matrix A is Hermitian if A = ____________. 3. All the diagonal elements of a Hermitian matrix are ____________ numbers. 4. A complex matrix A is skew-Hermitian if A* = ____________. 5. All the diagonal elements of a skew-Hermitian matrix are ____________ (two words) numbers. ⎡ 0 For 6–10, let A = ⎢ ⎣− 4 + i

⎡ −1 8 −7 ⎤ 2+i⎤ 5 + 2i ⎤ −4 − i ⎤ ⎡ 1 ⎡ 2i ⎢ 8 4 0 ⎥ , and E = ⎡ i 0 ⎤ . , D = , C = , B = ⎢ 0 −i ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 1 ⎦ 0 ⎦ ⎦ ⎣ ⎣ 3 − 2i ⎣ −5 + 2i −6i ⎦ ⎣⎢ −7 0 1 ⎥⎦

6. Is A Hermitian or skew-Hermitian? 7. Is B Hermitian or skew-Hermitian? 8. Is C Hermitian or skew-Hermitian? 9. Is D Hermitian or skew-Hermitian? 10. Is E Hermitian or skew-Hermitian?

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Determinants

·5·

In this chapter, you will learn: ◆ Determinant of a square matrix ◆ Cramer’s rule ◆ Properties of determinant

Determinants have important applications in the study of linear systems and in the study of invertible matrices. This chapter presents a discussion of determinants and their properties.

Determinant of a square matrix The determinant of a matrix is a number associated with the matrix and its elements and is only meaningful if the matrix is square. For 2 × 2 matrices the determinant is simple to calculate and is defined by ⎛ ⎡a b⎤ ⎞ a b det ⎜ ⎢ = = ad − bc. ⎝ ⎣c d⎥⎦ ⎟⎠ c d

Note: Vertical bars are used to denote determinants. PROBLEM

⎡ 2 −3 ⎤ Find the determinant for A = ⎢ ⎥. ⎣4 5 ⎦

SOLUTION

A=

2 −3 = (2 ⋅ 5) − (−3 ⋅ 4 ) = 22 4 5

A third order (3 × 3) determinant is a little more tedious to calculate, and there are several different methods to perform the calculation. One of these methods uses three 2 × 2 determinants called minors and will be illustrated first. This particular method is referred to as cofactor expansion along the first row (actually, you can use any row or column) as shown here. ⎛ ⎡a11 a12 det ⎜ ⎢a21 a22 ⎜ ⎢a ⎝ ⎣ 31 a32

a11 a12 a13 ⎤⎞ ⎥ ⎟ a23 = a21 a22 ⎥ a31 a32 a33 ⎦⎟⎠

a13 a a23 = a11 22 a32 a33

a a a23 a a − a12 21 23 + a13 21 22 a31 a32 a31 a33 a33

Tip: To find the minor associated with a11, mentally cross out the row and column containing a11 and the remaining 2 × 2 determinant is the minor needed. A similar technique determines the other two minors. In applying this method, an error

59

that is frequently made is to forget the negative sign on the second term of the expansion. The sign is easily determined by remembering and using the following sign array: ⎡+ ⎢− ⎢+ ⎢ ⎢− ⎢⎣

− + − +

+ − + −

− + − +

⎤ ⎥ ⎥⎥ ⎥ ⎥⎦

+ − + −

In general, the minor, Mij , of a matrix element aij is the determinant of the submatrix that remains after crossing out the row and column that contains aij . The cofactor, Cij , of aij is defined as Cij = (− 1) i+ j Mij . With this notation, the determinant equation above takes on the form det( A) = a11C11 + a12C12 + a13C13.

PROBLEM

⎡ 3 0 4⎤ Find the determinant for B = ⎢⎢ −1 6 2 ⎥⎥ using cofactor expansion along the first row. ⎢⎣ 5 −3 6 ⎥⎦

SOLUTION

3 0 4 −1 2 −1 6 6 2 B = −1 6 2 = 3 −0 +4 = 3(36 + 6) − 0(−6 − 10) + 4(3 − 30) −3 6 5 6 5 −3 5 −3 6

= 126 − 108 = 18

Determinants of larger matrices can be similarly calculated, but only 2 × 2 and 3 × 3 determinants will be used for the examples and problems in this section. Another method for finding the determinant of a 3 × 3 matrix is to write a copy of the first two columns adjacent to the last two columns, and then compute the determinant by multiplying the six diagonal elements as illustrated in Fig. 5.1. This method is the added columns method. –

–

–

a11

a12

a13

a11

a12

a21

a22

a23

a21

a22

a31

a32

a33

a31

a32

+

+

+

Figure 5.1 3 × 3 det (A) = a11a22a33 + a12a23a31 + a13a21a32 − a31a22a13 − a32a23a11 − a33a21a12.

The determinant is the result obtained by adding the downward diagonal products and subtracting the upward diagonal products. By reducing calculation time and tedium, graphing calculators are useful tools for calculating determinants of matrices. (See Chapter 3 for instructions on computing numerical determinants using a graphing calculator.) Given this circumstance, you might wonder why you should learn how to calculate determinants by hand. One reason is so that you will understand the process. Another reason is that sometimes, you might need to find determinants of matrices that contain variable elements (see Cramer’s rule in the next section and Chapter 10 for examples).

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EXERCISE

5·1 For 1–5, find the determinant of the given matrix.

⎡2 5 ⎤ 1. A = ⎢ ⎥ ⎣ 1 −4 ⎦ ⎡ 3 −6 ⎤ 2. A = ⎢ ⎥ ⎣ −1 2 ⎦ ⎡3 4 2⎤ 3. A = ⎢⎢ 0 7 −3 ⎥⎥ ⎢⎣ −2 1 5 ⎥⎦ ⎡1 0 0⎤ 4. A = ⎢⎢ 0 1 0 ⎥⎥ ⎢⎣ 0 0 1 ⎥⎦ ⎡ 2 4⎤ 5. A = ⎢⎢ −3 1 ⎥⎥ ⎢⎣ 5 0 ⎥⎦

⎡3 4 2⎤ For 6–8, find the cofactor of the designated element for the matrix A = ⎢⎢ 0 7 −3 ⎥⎥ . ⎢⎣ −2 1 5 ⎥⎦

6. The element 7 7. The element 0 8. The element 1

⎡ 1 2 3⎤ 9. If A = ⎢⎢ 0 0 5 ⎥⎥ , find det( A) by expanding by cofactors along the “smart” row. Why do ⎢⎣ 6 −2 1⎥⎦ you think the word “smart” was used? For 10–11, compute the determinant of the given matrix using the added columns method.

⎡ 2 6 −3 ⎤ 10. A = ⎢⎢ 4 −1 2 ⎥⎥ ⎢⎣ 4 0 3 ⎥⎦ ⎡3 4 2⎤ 11. A = ⎢⎢ 0 7 −3 ⎥⎥ ⎢⎣ −2 1 5 ⎥⎦ For 12–15, compute the determinant of the given matrix using a calculator (refer to Chapter 3, as needed).

⎡ 1 2 3⎤ 12. A = ⎢⎢ 0 0 5 ⎥⎥ ⎢⎣ 6 −2 1⎥⎦ Determinants

61

⎡ 4 0 3⎤ 13. A = ⎢⎢ 4 −1 2 ⎥⎥ ⎢⎣ 2 6 3 ⎥⎦ ⎡ 3 0 −2 ⎤ 1 ⎥⎥ 14. A = ⎢⎢ 4 7 ⎢⎣ 2 −3 5 ⎥⎦ ⎡1 ⎢3 15. A = ⎢ ⎢2 ⎢ ⎣1

0 1 1 0

6 0 2 4

7⎤ 3⎥ ⎥ 3⎥ ⎥ 4⎦

Cramer’s rule Determinants can be used to solve systems of equations by a clever method known as Cramer’s Rule. Consider the 2 × 2 system of linear equations a1x + b1 y = k1 a2 x + b2 y = k2

For this system, by Cramer’s rule the corresponding solution is found using a1 b1 k b a b a x = 1 1 and 1 1 y = 1 a2 b2 k2 b2 a2 b2 a2

k1 . k2

These two equations can be written as Dx = X and Dy = Y , where the determinant D is the coefficient determinant, and the determinants X and Y are defined by replacing the corresponding variables column in the coefficient determinant by the constants column. Similarly, for the 3 × 3 system of linear equations a1x + b1 y + c1z = k1 a2 x + b2 y + c2 z = k2 , a3 x + b3 y + c3z = k3

Cramer’s rule yields a1 b1 a2 b2 a3 b3

c1 k1 b1 c2 x = k2 b2 c3 k3 b3

c1 a1 b1 c2 , a2 b2 c3 a3 b3

c1 a1 c2 y = a2 c3 a3

k1 k2 k3

c1 a1 b1 c2 , and a2 b2 c3 a3 b3

c1 a1 b1 c2 z = a2 b2 c3 a3 b3

k1 k2 , k3

which you can write as Dx = X , Dy = Y , and Dz = Z with D, X, Y, and Z determined in a manner analogous to that for 2 × 2 systems of equations.

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In either case, if D ≠ 0 there is a unique solution to the system. You can write the solution for X Y X Y Z the 2 × 2 system as x = and y = ; and for the 3 × 3 system, as x = , y = , and z = . D D D D D If D = 0 and either X , Y , or Z ≠ 0 , then there is no solution to the system. If D = 0 and X = Y = Z = 0 , then there are infinitely many solutions to the system. PROBLEM

Solve the system of equations using Cramer’s rule. 4x + 2 y = 3 −2 x + y = 5

SOLUTION

3 5 x= 4 −2

2 1 3 ⋅1 − 5 ⋅ 2 7 = =− , 2 4 ⋅1 + 2 ⋅ 2 8 1

y=

4 3 −2 5 8

=

4 ⋅ 5 + 2 ⋅ 3 13 = 8 4

⎛ 7⎞ ⎛ 7 ⎞ 13 ⎛ 13 ⎞ The calculations 4 ⎜ − ⎟ + 2 ⎜ ⎟ = 3 and −2 ⎜ − ⎟ + = 5 verify the solution. ⎝ 8⎠ ⎝ 8⎠ 4 ⎝ 4⎠ PROBLEM

Solve the system of equations using Cramer’s rule. x+ y+ z = 4 2 x − y − 2z = −1 x − 2y − z = 1

SOLUTION

1 1 1 −1 −2 2 −2 2 −1 D = 2 −1 −2 = 1 −1 +1 = −3 + 0 − 3 = −6 −2 −1 1 −1 1 −2 1 −2 −1 4 1 1 1 4 1 1 1 4 X = −1 −1 −2 = − 12 , Y = 2 −1 −2 = 6 , Z = 2 −1 −1 = −18 1 −2 −1 1 1 −1 1 −2 1

Thus, x=

6 −12 −18 = 2, y = = −1 , and z = =3 −6 −6 −6

The calculations 2 + (−1) + 3 = 4 , 2(2) − (−1) − 2(3) = −1 , and 2 − 2(−1) − 3 = 1 verify the solution. PROBLEM

Solve the system of equations using Cramer’s rule. 2 x − 4 y + 2z = 3 x+ y− z =2 3x − 6 y + 3z = 2

Determinants

63

SOLUTION

2 −4 2 1 −1 1 −1 1 1 D = 1 1 −1 = 2 +4 +2 = −6 + 24 − 18 = 0 −6 3 3 3 3 −6 3 −6 3 3 −4 2 X = 2 1 −1 = −5 ≠ 0 . Consequently there is no solution. 2 −6 3 PROBLEM

Solve the system of equations using Cramer’s rule. 2 x + 2 y − 2z = 4 x+ y− z =2 3x − 6 y + 3z = 2

SOLUTION

2 2 −2 D = 1 1 −1 = 2(−3) − 2(6) − 2(−9) = 0 3 −6 3 4 2 −2 X = 2 1 −1 = 4(−3) − 2(8) − 2(−14 ) = 0 2 −6 3 2 4 −2 Y = 1 2 −1 = 2(8) − 4(6) − 2(−4 ) = 0 3 2 3 2 2 4 Z = 1 1 2 = 2(14 ) − 2(−4 ) + 4(−9) = 0 3 −6 2

Now, try to solve two equations such that the coefficient determinant involving just two of the variables is not 0. In this case, use the second and third equations and write them as x + y = 2+ z . Now, manipulating z as a constant, the solution to this 2 × 2 3x − 6 y = 2 − 3z system in x and y is 2+z 1 2 − 3z −6 (2 + z )(−6) − (2 − 3z ) −14 − 3z 14 + 3z = = = x= 1 1 −9 −9 9 3 −6 1 2+z 3 2 − 3z

(2 − 3z ) − 3(2 + z ) −4 − 6z 4 + 6z , = = −9 −9 −9 9 where z is a free variable that can be any real number. Hence, there are infinitely 14 4 many solutions. One solution is z = 0 , x = , and y = . 9 9 y=

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=

EXERCISE

5·2 For 1–5, solve using Cramer’s rule.

1.

2x − 3 y = 16 5x − 2y = −4

2.

x − 2y = 7 3x + 2 y = 5

x − 2 y − 3 z = −20 3. 2 x + 4 y − 5 z = 11 3 x + 7 y − 4 z = 33 x + 2y − z = 7 4. 4 x + 3 y + 2 z = 1 9 x + 8 y + 3z = 4 2 x − 4 y + 7z = 5 5. 3 x + 2 y − z = 2 x − 10 y + 15 z = 8 For 6–10, with the aid of a calculator, solve using X = A−1C when applicable; otherwise, use other methods if det( A) = 0.

6.

3 x − 2 y = −16 2 x − 5 y = −4 x−

z =1

7. 9 x − y + 4 z = 0

8x + 9y − z = 2 x − y + 2z = 4 8. 3 x + 2 y + z = 5 3 x − 3 y + 6 z = 12 9.

x − 5y = 3 2 x − 10 y = 6

0.7 x − 2.1y + z = 3.7 10. 4 x − 0.4 y + 0.23 z = 8.2 2 x − 3 y + 3.2 z = 4

Properties of determinant For a given n × n matrix A, you have the following properties: ◆ ◆ ◆

det( A) = det( AT ) If A is a triangular matrix, then det( A) = a11a22 $ann . If A has a row or column of zeros, then det( A) = 0 . Determinants

65

◆

◆

◆

◆

◆ ◆ ◆ ◆ ◆

If B is a matrix obtained by multiplying a single row or column of A by a scalar k, then det( B) = k det( A). If B is a matrix obtained by interchanging two rows or columns of A, then det( B) = − det( A). If B is a matrix obtained by multiplying one row/column of A by a scalar k, and adding the result to another row/column, then det( B) = det( A). (Notice that this row/column transformation does not affect the determinant.) If two rows/columns of a matrix are proportional (that is, one is a scalar multiple of the other), then det( A) = 0. det(kA) = kn det( A) If A and B are matrices of the same size, then det( AB) = det( A)det( B) . 1 If det( A) ≠ 0, then det( A−1 ) = . det( A) A is invertible if and only if det( A) ≠ 0 . det( A + B) ≠ det( A) + det( B)

EXERCISE

5·3 For 1–7, respond as indicated.

⎡ 8 0 0.03 ⎤ 1. If A = ⎢⎢ 0 5 5.71⎥⎥ , then det( A) = ____________. ⎢⎣ 0 0 3 ⎥⎦ ⎛ ⎡ 3 1 −2 ⎤⎞ 2. det ⎜ ⎢⎢ 0 1 3 ⎥⎥⎟ = ____________. ⎜ ⎟ ⎜⎝ ⎢ 0 1 4 ⎥⎟⎠ ⎦ ⎣ ⎛ ⎡ 8 0 0.03 ⎤⎞ 3. det ⎜ 4 ⎢⎢ 0 5 5.71⎥⎥⎟ = ____________. ⎜ ⎟ ⎜⎝ ⎢ 0 0 3 ⎥⎦⎟⎠ ⎣ ⎡ 2 3⎤ −1 4. If A = ⎢ ⎥ , then det( A ) = ____________. 4 1 ⎣ ⎦ ⎛ ⎡ a b c ⎤⎞ ⎛ ⎡ a b c ⎤⎞ ⎥ ⎢ ⎜ ⎟ 5. If det ⎢ 3 11 0 ⎥ = 12 , then det ⎜ ⎢⎢ 2 5 −1⎥⎥⎟ = ____________. ⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎢ 2 5 −1⎥⎟⎠ ⎜⎝ ⎢ 3 11 0 ⎥⎟⎠ ⎦ ⎦ ⎣ ⎣ ⎡1 7⎤ ⎡ 3 −1⎤ 6. If A = ⎢ and B = ⎢ ⎥ ⎥ , show that det( AB ) = det( A)det( B ). ⎣0 2 ⎦ ⎣2 −1⎦ ⎡1 7⎤ ⎡ 3 −1⎤ and B = ⎢ 7. If A = ⎢ ⎥ , show that det( A + B ) ≠ det( A) + det( B ). ⎥ ⎣2 −1⎦ ⎣0 2 ⎦

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For 8 and 9, use a calculator to find the determinant of the given matrix.

⎡2 3 2 ⎤ 8. A = ⎢⎢ 0 7 6 ⎥⎥ ⎢⎣ 5 5 4 ⎥⎦ ⎡7 8 9 ⎢6 5 4 9. A = ⎢ ⎢ 1 −9 −3 ⎢ 1 ⎣5 0

0⎤ 8⎥ ⎥ 7⎥ ⎥ 2⎦

⎡2 3 2 ⎤ 10. Is A = ⎢⎢ 0 7 6 ⎥⎥ an invertible matrix? Explain. ⎢⎣ 5 5 4 ⎥⎦

Determinants

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Vectors in Rn

·6·

In this chapter, you will learn: ◆ ◆ ◆ ◆

Vectors in two dimensions Dot product of vectors Vectors in Rn Vectors as matrices

This chapter provides a geometric introduction to the concept of vector in the context of Euclidean two-space, followed by a generalization of the concept to Euclidean n-space. It includes a discussion of properties of vectors and their associated arithmetic.

Vectors in two dimensions Many physical quantities such as area, temperature, and volume are adequately described by using a single attribute, magnitude. Quantities such as these are scalar quantities. Other concepts such as velocity and force require two attributes, magnitude and direction, to describe their behavior. Quantities such as these are vector quantities. A description of a vector quantity is complete when you give both its magnitude and direction. Vector quantities are represented in various ways. Two of the most common ways of representing vector quantities in two-space are by ordered pairs and by directed line segments or arrows. In this section, the algebraic representation of vectors is by ordered pairs and the geometric representation of vectors is directed line segments (Fig. 6.1). The length of the segment represents the magnitude, and the arrow head signifies the direction. The notation AB indicates the vector with initial point A and terminal point B. Symbolically, lowercase bold letters denote vectors. B u A

Figure 6.1 The vector u = AB.

Vectors are equal if they have the same magnitude and direction. The multiplication of a vector by a constant is scalar multiplication. Scalar multiplication may alter the magnitude or direction or both. Multiplying by a scalar with absolute value less than 1 will compress the vector, while multiplying by a scalar with absolute value greater than 1 will stretch the vector. Multiplying by a negative scalar

68

reverses the vector’s direction. Hence, w = 2v is a vector in the same direction as v with twice its magnitude, but w = −2 v is a vector in the opposite direction as v with twice its magnitude. Geometric vector representations are convenient means of representing the actions of several vectors, such as forces of different magnitudes acting on a single point from different directions. However, graphical representations of vectors in a coordinate plane use some conventions of which you should be cognizant. Figure 6.2 shows vectors with different placements in the plane. The vectors u, v, and w are equal even though they have different initial and terminal points. Thus, the placement in the plane is irrelevant. (3, 3)

3

(–3, 2)

2 u –6

v

1

–5 –4

–3

–2

–1

(0, 0)

2

(3, 0) 4

5

w

(–6, –1)

–2 (0, –3) –4

Figure 6.2 EquaI vectors.

If a vector v is represented algebraically by an ordered pair (3, 3), then the geometric representation of the vector is the vector v in Fig. 6.2 with initial point at the origin. A vector whose initial point is at the origin is in the position vector location. The vectors u and w shown in Fig. 6.1 are equal to the vector v = (3, 3), but neither is in the position vector location. Geometrically, the sum or resultant of two vectors u and v is a vector w formed by placing the initial point of v on the terminal point of u and then joining the initial point of u to the terminal point of v (Fig. 6.3). The sum is written u + v. 6

v

5 4 w=u+v

3 2

u

1 –6

–5 –4

–3

–2

–1

1

2

3

4

5

6

7

–1 v –3 –4

Figure 6.3 Addition of vectors.

The difference (subtraction) of two vectors is defined as u − v = u + (− v ) where –v is the vector with the same magnitude as v, but in the opposite direction. It is convenient to see the resultant and difference vectors displayed on the same figure (Fig. 6.4). Vectors in R n

69

3 2 1 –3

–2

–1

u–v

v

v 1

2

u

–1

–v

–2

u+v

3

4

5

6

–v

u–v

–3 –4

Figure 6.4 Addition and subtraction of vectors.

The dashed vectors are the vectors that are not in the position vector location and the solid vectors are in the position vector location. A mnemonic helpful in sketching the vector u − v is to read it as “the vector from v to u.” PROBLEM

If u = (2, −1) and v = (2, 4 ), graphically represent a. −3u b. u – v c. u + v 1 d. − 3u + v 2

SOLUTION 5

(2, 4)

4 3 v

2 4

u–v

1

(–6, 3) 3 –3u

–2

–1

2

–1

1

–2 u–v –3

u

1

3

4

(2, –1)

–7

–6

–5

–4

–3

–2

–1

2

u

–1

–4 (2, –1)

–2

–5

(0, –5)

(a)

5

(b)

6

(2, 4)

(–5, 5)

4 3

u

5

1 v 2

(4, 3)

v

–3u + 1 v

2

3

u+v

(–6, 3)

1

2 –3u

–1

1 –1 –2

u

2

3

4

(2, –1) (c)

Figure 6.5

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2

Linear Algebra

1

5 –7

–6

–5

–4

–3

–2

–1

1 v 2

1 –1

(d)

(1, 2)

2

The following properties of vectors hold: If u and v are expressed in component form, u = (a, b) and v = (c , d ) , and k is a scalar, then u + v = (a + c , b + d ) u − v = (a − c , b − d ) ku = (ka, kb)

u = v if and only if a = c and b = d . 0 = (0, 0) is the zero vector.

The norm (or magnitude) of a vector v = (a, b) is denoted v and in two dimensions is defined v is a unit vector in the as v = a2 + b2 . A unit vector is a vector having norm 1. If v ≠ 0 , then v same direction as v. A form of vectors used extensively in linear algebra and in applications that is well suited for algebraic manipulations uses special unit vectors associated with the rectangular coordinate system. In two dimensions, these special vectors are i = (1, 0) and j = (0,1). Every two dimensional vector can be expressed as a linear combination of the two unit vectors, i and j. That is, if v = (a, b) , then v = a i + b j . The vectors i = (1, 0) and j = (0,1) are said to be a basis that spans (generates) all vectors in two-space. This notion underlies all of linear algebra. A vector also can be viewed as a row matrix (vector), u = [a b] or a column matrix (vector), ⎡a⎤ u = ⎢ ⎥ . These forms are used extensively in linear algebra. ⎣b⎦ PROBLEM SOLUTION

Show that (11, 3) can be written as a linear combination of (1, 3) and (3, −1).

Find a and b such that a(1, 3) + b(3, −1) = (11, 3). This leads to solving the two a + 3b = 11 simultaneous equations . 3a − b = 3

The solution to this system is a = 2 and b = 3 . Thus, (11, 3) = 2(1, 3) + 3(3, − 1). EXERCISE

6·1 Respond as indicated.

1. Sketch the vectors u = (5, 2), v = ( −3, 2), and u + v in the coordinate plane. 2. If u = (9, 0) and v = (5, −4 ), compute u − v , 4v , v − u, and 3u − 4 v . 3. Write the vector (5, 8) in ai + bj form. 4. Find the norm of v = 5i − 3 j . 5. If u = 3i + 6 j and v = −4i + 2 j , compute u + v and write the answer in ai + bj form. 6. Write the vector (3, 5) as a column vector.

⎡2 ⎤ 7. Multiply [ 2 3] and ⎢ ⎥ as matrices. ⎣3 ⎦ 2 8. Calculate (2, 3) . 9. Compare the results of problems 7 and 8. What do you notice? 10. If ai + bj = 0 , then what must be true of a and b?

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Dot product of vectors The dot product (also called the inner product) of two vectors is a special type of vector product defined as follows: If u = (a, b) and v = (c , d ), then the dot product of u and v is u u v = ac + bd. Notice that the result of a dot product is a scalar, not a vector! The basic arithmetic properties for vectors in two-space under the operation of dot product are given below. Given vectors u, v, and w and scalar k, then the following properties hold: uuv = v uu 2 uuu = u w u (u + v ) = w u u + w u v k( u u v ) = ( k u ) u v = u u ( k v ) 0 u u = u u 0 = 0 , where 0 is the zero vector u v uuv u = u v u ⋅ v

The dot product is useful in many applications, one of which is finding the angle between two vectors. If u and v are nonzero vectors, then you can determine the angle θ between u and v using the equation ⎛ uuv θ = cos−1 ⎜ ⎝ u ⋅ v

⎞ ⎟ , where 0 ≤ θ ≤ π ⎠ uuv Furthermore, you can express the equation as cos θ = or u u v = u ⋅ v cosθ . u ⋅ v PROBLEM

Find the angle between the vectors u = 2i + 3 j and v = i − 2 j .

SOLUTION

⎛ uu v ⎞ 2−6 −4 cosθ = ⎜ = ≈ −0.496139 ⎟= ⋅ u v 13 ( 5 ) ⎝ ⎠ ( 4 + 9 )( 1 + 4 )

Thus, θ = cos−1 (− 0 .496139) ≈ 120 °. The form u u v = u ⋅ v cosθ is an alternate form of the dot product. Obviously, if the angle between two vectors is 90°, then you have u u v = u ⋅ v cos90 ° = u ⋅ v ⋅ 0 = 0 . This result leads naturally to the following: If u and v are vectors, then u u v = 0 if and only if u and v are perpendicular (orthogonal). EXERCISE

6·2 For, 1–3, find the angle between the two given vectors.

1. u = 5i + 2 j and v = −3i − 3 j 2. u = ( −3, 6) and v = (2, −5) 3. u = −4i + 3 j and v = −6i − 8j

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4. Prove that i and j are orthogonal. 5. Find all vectors, u, such that u = 3 and u is perpendicular to v = 2i + 2 j. 6. For u = −5i + 3 j and v = −6i − 2j , find u u v . 7. Find the unit vector associated with u = 5i + 2 j. 8. Express (5, 7) as a linear combination of i and j . 9. Show that (1, 2) and (2, −1) are orthogonal. 10. Show that (3,1) can be expressed as a linear combination of (1, 2) and (2, −1).

Vectors in Rn Now that you have been introduced to vector concepts, the focus of this chapter turns to vectors of n-dimensions, which, as you might surmise, have the form u = (u1 , u2 ,. . . , un ) . For n > 3 , this concept is an abstract generalization for which there is no geometry available to help with visualization of the ideas, as you have for two- and three-dimensional vectors. However, most of the arithmetic properties of vectors in two-dimensions with which you are familiar carry over to n-dimensions, and the goal is to study these generalizations. If n is a positive integer, then an ordered n-tuple of real numbers has the form (a1 , a2 ,# , an ). The set of all ordered n-tuples of real numbers is called Euclidean n-space (or simply n-space), and is denoted Rn . The n-tuples also are referred to as n-dimensional vectors. Analogous to the two-dimensional case, two vectors, u = (u1 , u2 ,. . ., un ) and v = (v1 , v2 ,. . ., vn ), in Rn , are equal if and only if u1 = v1 , u2 = v2 ,. . ., un = vn . Likewise, if u = (u1 , u2 ,. . ., un ) and v = (v1 , v2 ,. . ., vn ) are vectors in Rn and k is a scalar, then the standard operations on Rn are addition and scalar multiplication defined as follows: Addition: u + v = (u1 + v1 , u2 + v2 ,. . ., un + vn ) Scalar multiplication: ku = (ku1 , ku2 ,. . ., kun ) Additionally, in Rn , 0 = (0, 0, # , 0) is the zero vector; and the difference of two vectors is defined by u − v = u + ( − v ), where − v = (−v1 , −v2 ,. . . , −vn ) . The basic arithmetic properties for vectors in Rn under the operations of addition and scalar multiplication are given below. If u, v, and w are vectors in Rn and a and b are scalars, then the following properties hold: u+v = v+u

u + (v + w ) = (u + v ) + w

u+0=0+u=u a(b u ) = (ab)u

u + (− u )=(− u ) + u = 0 a(u + v ) = a u + a v

(a + b)u = au + b u

1 u = u and 0 u = 0

PROBLEM

For u = (−3, 2,1, 0) and v = (3,1, 4 , −5), find a. u − v b. 2u − 3 v

SOLUTION

a. u − v = u + ( − v ) = (−3, 2,1, 0) + (−3, −1, −4 , 5) = (−3 − 3, 2 −1,1 − 4 , 0 + 5) = (−6,1, −3, 5) b. 2u − 3 v = 2(−3, 2,1, 0) − 3(3,1, 4 , −5) = (−6 − 9, 4 − 3, 2 − 12, 0 + 15) = (−15,1, −10,15) Vectors in R n

73

If u = (u1 , u2 ,. . . , un ) and v = (v1 , v2 ,. . . , vn ), then the Euclidean inner product (or dot product) is defined as u u v = u1v1 + u2v2 + $ + unvn . Properties of the Euclidean inner product are given below.

If u, v, and w are vectors in Rn and a is a scalar, the following properties hold: uuv = v uu

u u u ≥ 0 , and the equality holds if and only if u = 0 .

(u + v) u w = u u w + v u w

u and v are orthogonal if and only if u u v = 0.

(a u ) u v = a(u u v ) PROBLEM

For u = ( − 3,2,1,0 ) and v = (3,1,4,−5) compute u u v .

SOLUTION

u u v = −3 ⋅ 3 + 2 ⋅1 + 1 ⋅ 4 + 0 ⋅−5 = −3

The Euclidean norm of a vector u = (u1 , u2 ,. . . , un ) in Rn is defined by u = u u u = u12 + u22 + $ + un2 and has the following properties. Properties of the Euclidean norm If u and w are vectors in Rn and a is a scalar, then the following properties hold: u ≥0 au = a

u = 0 if and only if u = 0 u + v ≤ u + v (Triangle inequality)

u

If v ≠ 0 , then

v is a unit vector and is the normalization of v. (Note: In two- and threev

dimensions, this vector is a unit vector in the same direction as v.) The distance between vectors u and v is defined by u − v = (u1 − v1 )2 + (u2 − v 2 )2 + $ + (un − vn )2 . PROBLEM

Let u = (−3, 2,1, 0) and v = (3,1, 4 , −5), find a. u b.

1 u u

c. u − v SOLUTION

a. u = (−3)2 + 22 + 12 + 02 = 14 b.

1 1 u= (−3, 2,1, 0) u 14

c. u − v = (−3 − 3)2 + (2 − 1)2 + (1 − 4 )2 + (0 − (−5))2 = 36 + 1 + 9 + 25 = 71 ≈ 8.426

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EXERCISE

6·3 For 1–9, let u = ( 4 , 7, −3, 2) and v = (5, −2, 8,1) .

1. Find 6(2u − v ). 2. Find u + v . 3. Find u u v . 4. Find

1 v. v

5. Find

1 v . v

6. Show that u + v ≤ u + v . 7. Show that u + v

2

2

2

≠ u + v .

8. Find a vector x such that 5x − 2 v = 2(u − 5x ) . 9. Find (6u) u v . 10. Show that (1, 0, 0), ( 0,1, 0), and ( 0, 0,1) are mutually orthogonal.

Vectors as matrices ⎡u1 ⎤ ⎢u2 ⎥ The vector u = (u1 , u2 ,. . . , un ) can be viewed as a column matrix u = ⎢ ⎥ or a row matrix ⎢% ⎥ ⎣⎢un ⎦⎥ u = ⎡⎣u1 u2 . . . un ⎤⎦ . In this manner, vectors can be incorporated into the matrix algebra so prevalent in linear algebra. It is a common practice to consider a 1 × 1 matrix of a single real number ⎡u1 ⎤ ⎡v1 ⎤ ⎢u2 ⎥ ⎢v ⎥ as the number itself. Adopting this convention, if u = ⎢ ⎥ and v = ⎢ 2 ⎥ , then u u v = uT v = ⎢% ⎥ ⎢% ⎥ ⎢⎣un ⎥⎦ ⎢⎣vn ⎥⎦ ⎣⎡u1v1 + u2v2 + $ + unvn ⎤⎦ = u1v1 + u2v2 + $ + unvn , which is a convenient alternate notation for the

dot product. Another definition that will be used in future chapters is that of a linear combination. This concept was presented earlier in the discussion of two-dimensional vectors, but the generalization is an important part of the study of linear algebra. If a1 , a2 ,. . . , an are scalars and u1 , u 2 ,. . . , un are vectors, then the vector v = a1u1 + a2 u 2 + $ + an un is a linear combination of the vectors u1 , u 2 ,. . . , un.

Vectors in R n

75

EXERCISE

6·4 For 1 and 2, respond as indicated.

1. If u = (2, 2, 2) and v = ( 0, 4 , −2), find u − v . 2. Let v = ( −2, 3, 0, 6). Find all scalars k such that k v = 5. For 3–5, determine whether the given vectors are orthogonal.

3. u = ( −1, 3, 2) and v = ( 4 , 2,1) 4. u = ( −4 , 6, −10,1) and v = (2,1, −2, 9) 5. u = ( a , b , c ) and v = ( 0, 0, 0) For 6 and 7, find k such that u and v are orthogonal.

6. u = (2,1, 3) and v = (1, 7, k ) 7. u = ( k , k ,1) and v = ( k , 5, 6) 8. Find a vector of norm 2 that is orthogonal to the three vectors, u = (2,1, 0), v = ( −1, −1, 0) , and w = ( 4 , 0, 0). 9. Find scalars a, b, and c such that a(2, 0,1) + b(1, 3, 2) + c (1, 0,1) = (7, 3, 4 ). 10. Show that u = ( 4 , 0,1) can be expressed as a linear combination of u1 = (2, 0, 0), u2 = ( 0, 3, 0), and u3 = ( 0, 0,1) . 11. If a(2, 2, 0) + b(2, −2, 0) + c ( 0, 0, 2) = 0, then solve for a, b, and c. 12. Show that ( 8, 0, 4 ) can be expressed as a linear combination of (2, 2, 0), (2, −2, 0), and ( 0, 0, 2). 13. Show that (3,1, 5) can be expressed as a linear combination of (2, 2, 0), (2, −2, 0), and ( 0, 0, 2). 14. Show that ( 4 , 0, 3, 7) can be expressed as a linear combination of (1, 0, 0, 0), ( 0,1, 0, 0), ( 0, 0,1, 0), and ( 0, 0, 0,1). 15. Find the vector that is the normalization of v = ( 4 , 0, 3, 7). 16. If u = (2, k ,1, 4 ) and v = (3, −1, 6, −3) , find k such that u − v = 6 . 17. If S = {(2, 2, 0),(2, −2, 0),( 0, 0, 2)}, find the set T of vectors that are the normalizations of the vectors in S. 18. If u = (1, 2, 3) and v = ( 0, 2, 2), show that u u v ≤ u v . (Note: This inequality is true for all vectors and is called the Cauchy-Schwarz inequality.) 19. If u = (1, 2, 0, 3) and v = ( 0, 3, 0, 2) , calculate (u u v )( 4 , 4 , 3,1) . 20. If u = (1, 2, 0, 3) and v = ( 0, 3, 0, 2), calculate u(v u ( 4 , 4 , 3,1)).

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Vector spaces

·7·

In this chapter, you will learn: ◆ ◆ ◆ ◆ ◆ ◆

Deﬁnitions and terminology of vector spaces Linear independence Basis Dimension Row space, column space, and null space Rank and nullity

Reaching this point in your study of linear algebra is a milestone. Because of your hard work and effort in previous chapters, you are prepared to plunge into a topic that is a key concept in the study of linear algebra—vector spaces. Vector spaces are the underlying structure of linear algebra.

Deﬁnitions and terminology of vector spaces Your study of vectors in Rn provides a basis for the more abstract concept of a general vector space. The term vector in this inquiry is expanded to include a multitude of mathematical objects such as polynomials, matrices, functions, and n-tuples. The study begins with an axiomatic groundwork. These axioms and the properties that are their consequences make vector spaces intriguing and important entities in linear algebra. A vector space over the real numbers is a nonempty set V of objects, called vectors, on which two operations, called vector addition and scalar multiplication, are defined and for which the following axioms hold:

For all vectors u, v, and w in V and all scalars a and b. i. The sum of u and v, denoted u + v, is in V . ii. u + v = v + u iii. u + ( v + w ) = (u + v ) + w iv. There exists a zero vector 0 in V such that 0 + u = u + 0 = u for all u in V. v. For each u in V, there is a vector −u such that u + (− u ) = (− u ) + u = 0. vi. The scalar multiple au is in V. vii. a(u + v ) = au + av viii. (a + b )u = au + bu ix. a(bu ) = (ab )u x. 1u = u

77

Four results that follow immediately from these axioms are the following: 0u = 0 ; a0 = 0 ; − u = (−1)u; and if ku = 0, then k = 0 or u = 0. Here are examples of vector spaces. ◆

◆

◆

◆

The spaces Rn for n ≥ 1 with the standard operations of vector addition and scalar multiplication are all vector spaces. The set M 22 of all 2 × 2 matrices of real numbers with the standard operations of matrix addition and scalar multiplication is a vector space as are all the spaces Mnn of such n × n matrices with the standard operations. The spaces Pn of all polynomials of the form a0 + a1 x + a2 x 2 + $ + an x n with the standard operations of polynomial addition and scalar multiplication are vector spaces. The set of all real-valued functions defined on the same set of real numbers with the standard operations of function addition and scalar multiplication is a vector space.

There are numerous other examples of vector spaces and this circumstance is exactly why there is a need to study the underlying mathematics of such spaces. PROBLEM

Show that R 2 with the standard operations of addition and scalar multiplication is a vector space.

SOLUTION

Proof (Refer to Chapter 6 for definitions and properties of addition and scalar multiplication in R 2.) Axioms i and vi are satisfied by the definitions of addition and scalar multiplication in R 2. Axioms ii-v and vii-x are satisfied by the properties of addition and scalar multiplication in R 2. All the axioms are satisfied, so R 2 is a vector space.

PROBLEM

SOLUTION

⎧⎡ x ⎤ ⎫ Show that T = ⎨ ⎢ ⎥ : x ≥ 0, y ≥ 0 ⎬ is not a vector space under the standard ⎪⎩ ⎣ y ⎦ ⎪⎭ operations of addition and scalar multiplication of matrices. ⎧⎡ x ⎤ ⎫ ⎡2⎤ If T = ⎨ ⎢ ⎥ : x ≥ 0, y ≥ 0 ⎬, then for u = ⎢ ⎥ in T and c = −3, the vector y ⎪⎭ ⎣3⎦ ⎩⎪ ⎣ ⎦ ⎡ 2 ⎤ ⎡ −6 ⎤ cu = −3 ⎢ ⎥ = ⎢ ⎥ is not in T because neither component is greater than ⎣ 3 ⎦ ⎣ −9 ⎦

or equal to 0. Thus, axiom vi of vector spaces is not satisfied, so T is not a vector space.

In many instances there are subsets of V that are themselves vector spaces and merit study in their own right. Consequently, criteria for such a subset is established and the result is called a subspace of V. A subspace of a vector space V is a subset W of V that has the following three properties: 1. The zero vector of V is in W. 2. For each u and v in W, u + v is in W. (W is closed under addition.) 3. For each u in W and each scalar k, ku is in W. (W is closed under scalar multiplication.)

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For any vector space V, V itself and the set consisting of only the zero vector {0} are subspaces of V. PROBLEM

⎧ ⎡0 ⎤ ⎫ ⎪⎢ ⎥ ⎪ bers ⎬ is a subspace of R 3. Show that W = ⎨ ⎢ s ⎥ , where s and t are real numb ⎪⎢ t ⎥ ⎪ ⎩⎣ ⎦ ⎭

SOLUTION

⎧ ⎡0 ⎤ ⎫ ⎪⎢ ⎥ ⎪ bers ⎬, then If W = ⎨ ⎢ s ⎥ , where s and t are real numb ⎪ ⎪⎢ t ⎥ ⎭ ⎩⎣ ⎦ ⎡0 ⎤ 1. 0 = ⎢⎢0 ⎥⎥ is of the form required because 0 is a real number, so 0 ∈W . ⎢⎣0 ⎥⎦ ⎡0 ⎤ ⎡ 0 ⎤ ⎡ 0 ⎤ ⎡0 ⎤ ⎡0⎤ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ 2. If u = ⎢ s ⎥ and v = ⎢m ⎥ are in W, then u + v = ⎢ s ⎥ + ⎢m ⎥ = ⎢ s + m ⎥ is in W ⎢⎣ t ⎥⎦ ⎢⎣ n ⎥⎦ ⎣⎢ t + n ⎥⎦ ⎢⎣ t ⎥⎦ ⎢⎣ n ⎥⎦

because s + m and t + n are real numbers. ⎡0 ⎤ ⎡0 ⎤ ⎡ 0 ⎤ ⎥ ⎢ 3. If u = ⎢ s ⎥ is in W and k is a real number, then ku = k ⎢⎢ s ⎥⎥ = ⎢⎢ ks ⎥⎥ is in W ⎢⎣ t ⎥⎦ ⎢⎣ t ⎥⎦ ⎢⎣ kt ⎥⎦

because ks and kt are real numbers. The three requirements of a subspace are met so W is a subspace of R 3. PROBLEM

Show that L = {( x , x ), where x ∈R} is a subspace of R 2 .

SOLUTION

If L = {( x , x ), where x ∈R}, then 1. (0, 0) = 0 is of the required form, so 0 ∈L. 2. If u = ( x , x ) and v = ( y , y ) are in L, then u + v = ( x + y , x + y ) is in L because x + y is a real number. 3. If u = ( x , x ) is in L and k is a real number, then ku = k( x , x ) = (kx , kx ) is in L because kx is a real number. The three requirements of a subspace are met so L is a subspace of R 2.

PROBLEM

Show that if v 1 , v 2 ,…, v p are vectors in a vector space V and S is the set of all linear combinations of v 1 , v 2 ,…, v p , then S is a subspace of V (see Chapter 6 for a definition of a linear combination). This subspace is the Span of S, denoted Span (S), and the vectors v 1 , v 2 ,…, v p are said to span (generate) S. Also, the set S is the smallest subspace of V that contains the vectors v 1 , v 2 ,…, v p .

SOLUTION

1. 0 = 0 v 1 + 0 v 2 + $ + 0 v p, thus 0 is in S. 2. If x = a1 v 1 + a2 v 2 + $ + a p v p and y = b1 v 1 + b2 v 2 + $ + bp v p are in S, then x + y = (a1 v 1 + a2 v 2 + $ + a p v p ) + (b1 v 1 + b2 v 2 + $ + bp v p ) = (a1 + b1 )v 1 + (a2 + b2 )v 2 + $ + (a p + bp )v p is a linear combination of v 1 , v 2 ,…, v p and thus is in S. Vector spaces

79

3. If u = a1 v 1 + a2 v 2 + $ + a p v p is in S and if c is a scalar, then cu = ca1 v 1 + ca2 v 2 + $ + ca p v p is a linear combination of v 1 , v 2 ,…, v p and thus is in S. The three requirements of a subspace are met so S is a subspace of V. PROBLEM

Show that if AX = 0 is a homogeneous system of linear equations of m equations and n unknowns, then the set W of all solution vectors of the system is a subspace of Rn. (This space is the null space of A. See “Row space, column space, and null space” in this chapter for a discussion of null space.)

SOLUTION

1. The vector X = 0 is a solution (the trivial solution); thus, 0 ∈W . 2. If X1 and X 2 are solutions, then AX1 = 0 and AX 2 = 0. It follows that A( X1 + X 2 ) = AX1 + AX 2 = 0 + 0 = 0. Thus, X1 + X 2 is a solution, so ( X1 + X 2 ) ∈W . 3. If X1 is a solution and k is a scalar, then A(kX1 ) = k( AX1 ) = k 0 = 0 . Thus, kX1 is a solution, so kX1 ∈W . The three requirements of a subspace are met so W is a subspace of Rn . EXERCISE

7·1 Respond as indicated.

1. Let V be the set of all real-valued functions with domain [0, 1]. Let addition, f + g, scalar multiplication, kf , and −f , be defined by ( f + g )( x ) = f ( x ) + g( x ), ( kf )( x ) = kf ( x ), and − f = ( −1)f , respectively. Prove that V is a vector space over the real numbers.

⎧⎡ x ⎤ ⎫ 2. Let H = ⎨ ⎢ ⎥ : x 2 + y 2 ≤ 1, where x , y ∈R ⎬ . Show that H is not a vector space under the ⎪⎩ ⎣ y ⎦ ⎪⎭ standard operations of addition and scalar multiplication. ⎫ ⎧ ⎡ 3t ⎤ ⎪ ⎪⎢ ⎥ 3. Let W = ⎨ ⎢ 0 ⎥ , where t ∈R ⎬ . Show that W is a subspace of R 3. ⎪ ⎪ ⎢ −7t ⎥ ⎦ ⎭ ⎩⎣ ⎧⎡ x ⎤ ⎫ 4. Let W = ⎨ ⎢ , where x ∈R ⎬ . Show that W is not a subspace of R 2 . ⎥ ⎩⎪ ⎣ x + 1⎦ ⎭⎪ 5. Let v1 = ( 0, 2, 3), v 2 = (111 , , ), and S = { v1 , v 2 }. Construct the Span of S. 6. Show that the Span of S , where S = {(1, 0),( 0,1)} , is R 2. 7. If a(1, 0, 0, 0) + b( 0,1, 0, 0) + c ( 0, 0,1, 0) + d ( 0, 0, 0,1) = 0 = ( 0, 0, 0, 0), then solve for a, b, c, and d.

, , ), u2 = (1, 2, 3), and 8. Express the vector v = (1, −2, 5) as a linear combination of u1 = (111 u3 = (2, −11 , ). ⎡3 1 ⎤ ⎡1 1 ⎤ as a linear combination of the matrices U1 = ⎢ 9. Express the matrix A = ⎢ ⎥ ⎥, ⎣ 1 −1⎦ ⎣1 0 ⎦ ⎡0 0 ⎤ ⎡0 2 ⎤ , and U3 = ⎢ U2 = ⎢ ⎥ ⎥. ⎣ 1 1⎦ ⎣ 0 −1⎦ 10. Show that W = {( a , b , 0), where a , b ∈R } is generated by the vectors u = (2,1, 0) and v = ( 0,1, 0).

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Linear Algebra

Linear independence If S = { v 1 , v 2 ,…, v p } is a nonempty set of vectors, then the set is linearly independent if and only if a1 v 1 + a2 v 2 + $ + a p v p = 0 implies that a1 = a2 = $ = a p = 0. That is, if the only possible solution to a1 v 1 + a2 v 2 + $ + a p v p = 0 is the trivial one, then the vectors v 1 , v 2 ,…, v p are linearly independent. If there is any other solution, the set is linearly dependent. PROBLEM

Determine whether the set of vectors is linearly independent. a. u = (2,1), v = (3, 0) in R 2 b. u = (3, −3, 6), v = (0,1,1), w = (3, −5, 4 ) in R 3 c. e1 = (1, 0, 0, 0), e 2 = (0,1, 0, 0), e 3 = (0, 0,1, 0) in R 4 d. u = (0,1), v = (2, 3), w = (3,1) in R 2 e. u = 1 − x, v = 5 + 3x − 2 x 2, w = 1 + 3x − x 2 in P2

SOLUTION 2 a. u = (2,1), v = (3, 0) in R Let au + bv = a(2,1) + b(3, 0) = 0, from which you get the system of equations 2a + 3b = 0 a=0 The only solution to this system is a = b = 0, so the set is linearly independent. b. u = (3, −3, 6), v = (0,1,1), w = (3, −5, 4 ) in R 3 Let au + bv + cw = a(3, −3, 6) + b(0,1,1) + c(3, −5, 4 ) = 0, from which you get the system of equations

3a + 3c = 0 ⎡ 3 0 3 0 ⎤ 1R + R2 −3a + b − 5c = 0 ⇒ ⎢⎢ −3 1 −5 0 ⎥⎥ 1 −2 R1 + R3 6a + b + 4c = 0 ⎢⎣ 6 1 4 0 ⎥⎦

⎡3 0 3 0 ⎤ ⎢0 1 −2 0 ⎥ ⎥ ⎢ ⎢⎣0 1 −2 0 ⎥⎦

Because in reduced form, two rows are the same there is a nontrivial solution to this system. For instance, a = −1, b = 2, and c = 1 is a nontrivial solution. Thus, the set is linearly dependent. c. e1 = (1, 0, 0, 0), e 2 = (0,1, 0, 0), e 3 = (0, 0,1, 0) in R 4 Let ae1 + be 2 + ce 3 = 0, then a = b = c = 0 is the only solution, so the set is linearly independent. d. u = (0,1), v = (2, 3), w = (3,1) in R 2 Let au + bv + cw = a(0,1) + b(2, 3) + c(3,1) = 0, from which you get the system of 2b + 3c = 0 equations . a + 3b + c = 0 Because there are more unknowns than equations, there are infinitely many solutions, so the set is linearly dependent. e. u = 1 − x , v = 5 + 3x − 2 x 2 , w = 1 + 3x − x 2 in P2 Let au + bv + cw = 0, then a(1 − x ) + b(5 + 3x − 2 x 2 ) + c (1 + 3x − x 2 ) = 0. Hence, (−2b − c )x 2 + (−a + 3b + 3c )x + (a + 5b + c ) = 0, from which you get the system of equations a + 5b + c = 0 ⎡ 1 5 1 0 ⎤ −a + 3b + 3c = 0 ⇒ ⎢⎢ −1 3 3 0 ⎥⎥ 1R 1 + R 2 2b + c = 0 ⎢⎣ 0 2 1 0 ⎥⎦

⎡1 5 1 0 ⎤ ⎢0 8 4 0 ⎥ ⎢ ⎥ ⎢⎣0 2 1 0 ⎥⎦ Vector spaces

81

⎡1 5 1 0 ⎤ 1 R2 ⎢⎢0 2 1 0 ⎥⎥ − 1R2 + R 3 4 ⎢ ⎣0 2 1 0 ⎥⎦

⎡1 5 1 0 ⎤ ⎢0 2 1 0 ⎥ ⎥ ⎢ ⎢⎣0 0 0 0 ⎥⎦

The reduced form indicates that there are infinitely many solutions, so the system is linearly dependent.

Some observations (theorems) that are helpful in determining linear independence or dependence are the following: ◆

◆

◆ ◆ ◆

◆

A set of vectors is linearly dependent if and only if at least one of the vectors in the set can be expressed as a linear combination of the other vectors in the set. A set of vectors is linearly independent if and only if no vector in the set can be expressed as a linear combination of the other vectors in the set. A finite set of vectors that contains the zero vector is a linearly dependent set. If S = { v 1 , v 2 ,…, v p } is a set of vectors in Rn and if p > n , S is a linearly dependent set. A set with exactly two vectors is linearly independent if and only if neither vector is a scalar multiple of the other. If one vector in a set is a multiple of another vector in the set, then the set is a linearly dependent set.

EXERCISE

7·2 For 1–6, determine whether the vectors are linearly independent or linearly dependent.

1. u = ( −1, 2, 4 ), v = (5, −10, −20) in R 3 ⎡ −3 4 ⎤ ⎡ 3 −4 ⎤ 2. A = ⎢ and B = ⎢ ⎥ ⎥ in M22 ⎣ 2 0⎦ ⎣ −2 0 ⎦ , , 3) in R 3 3. u = ( −3, 0, 4 ), v = (5, −1, 2), w = (11 4. u = 6 − x 2, v = 1+ x + 4 x 2 in P2 5. u = 2 − x + 4 x 2 , v = 3 + 6 x + 2 x 2, w = 2 + 10 x − 4 x 2 in P2 6. u = ( 0, 3,1, −1), v = ( 6, 0, 5,1), w = ( 4 , −7,1, 3) in R 4 7. For which value(s) of k are the following vectors linearly dependent?

1 1⎞ 1⎞ ⎛ ⎛ 1 ⎛ 1 1 ⎞ u = ⎜ k, − , − ⎟ , v = ⎜ − , k , − ⎟ , w = ⎜ − , − ,k ⎟ ⎝ ⎝ 2 ⎝ 2 2 ⎠ 2 2⎠ 2⎠ 8. Prove that for any vectors u, v, and w, the vectors u − v , v − w , and w − u form a linearly dependent set. 9. Prove that vectors e1 = (1, 0, 0), e2 = ( 0,1, 0), and e3 = ( 0, 0,1) are orthogonal to each other and each has norm 1. 10. Prove that vectors e1 = (1, 0, 0), e2 = ( 0,1, 0), and e3 = ( 0, 0,1) span (generate) R 3.

82

practice makes perfect

Linear Algebra

Basis If V is a vector space and S = { v 1 , v 2 ,…, v n } is a set of linearly independent vectors that span (generate) V, then S is a basis for V. A basis representation is unique because any vector u in V can be written as u = a1 v 1 + a2 v 2 + $ + an v n in exactly one way. If the coefficients of this linear combination are written as an ordered n-tuple, (a1 , a2 , # , an ), then this vector is the coordinate vector of u, denoted [u]S , relative to the basis S = { v 1 , v 2 ,…, v n } . So you see, a basis for a vector space is a generalization of the coordinate system concept with which you are familiar from two- and threedimensional geometry. A set of vectors S = { v 1 , v 2 ,…, v p } is a basis for a vector space V, if the following two conditions hold: 1. The vectors in S generate (span) V, written Span(S ) = V . That is, any vector in V is a linear

combination of v 1 , v 2 ,…, v p .

2. The set S is a linearly independent set. That is, if a1 v 1 + a2 v 2 + $ + a p v p = 0 , then

a1 = a2 = $ = a p = 0. PROBLEM

Determine if the given set is a basis for the indicated space. a. S = {(1, 0, 0),(0,1, 0),(0, 0,1)} for R 3 b. T = {(1,1,1),(1,1, 0),(−1, 0, −1)} for R 3 c. W = {(1, 0,1),(0,1,1),(2, 0, 2)} for R 3 ⎧ ⎡1 0 ⎤ ⎡ 2 0 ⎤ ⎡0 1⎤ ⎡ −3 0 ⎤ ⎫ d. U = ⎨ ⎢ ⎥,⎢ ⎥,⎢ ⎥,⎢ ⎥ ⎬ for M 22 ⎩⎪ ⎣0 0 ⎦ ⎣ −1 0 ⎦ ⎣0 1⎦ ⎣ 0 2 ⎦ ⎪⎭

SOLUTION

a. S = {(1, 0, 0),(0,1, 0),(0, 0,1)} for R 3 (1) If u = (u1 , u2 , u3 ) is any vector in R 3, then you can express u as u = (u1 , u2 , u3 ) = u1 (1, 0, 0) + u2 (0,1, 0) + u3 (0, 0,1). Consequently, Span(S ) = R 3. (2) Let x (1, 0, 0) + y (0,1, 0) + z (0, 0,1) = 0 , then ( x , y , z ) = (0, 0, 0). Thus, x = y = z = 0, which is the trivial solution, so the vectors in S are linearly independent. Hence, (1) and (2) are satisfied, so S = {(1, 0, 0),(0,1, 0),(0, 0,1)} is a basis for R 3. b. T = {(1,1,1),(1,1, 0),(−1, 0, −1)} for R 3 (1) If u = (a, b, c ) is any vector in R 3, then find coefficients, if any, such that (a, b, c ) = x (1,1,1) + y (1,1, 0) + z (−1, 0 − 1). That is, solve (a, b , c ) = ( x , x , x ) + ( y , y , 0) + (− z , 0 − z ).

This equality yields the system of equations: x+ y−z =a x+ y =b ⇒ x −z =c

⎡1 1 −1 a ⎤ ⎢1 1 0 b ⎥ 1R1 + R3 ⎥− ⎢ ⎣⎢1 0 −1 c ⎥⎦

⎡1 1 −1 a ⎤ ⎥ ⎢ − 1R1 + R2 ⎢0 0 1 b − a ⎥ 1R 3 + R 1 ⎢⎣0 −1 0 c − a ⎥⎦

⎡1 1 −1 a ⎤ ⎢1 1 0 b ⎥⎥ ⎢ ⎢⎣0 −1 0 c − a ⎥⎦

⎡1 0 −1 c ⎤ ⎢0 0 1 b − a ⎥ ⎥ ⎢ ⎢⎣0 −1 0 c − a ⎥⎦

Vector spaces

83

x =b+c−a ⎡1 0 0 b − a + c ⎤ ⎥ ⎢ 1R 0 1 b − a ⎥ , which has the solution: y = a − c 2 + R 1 ⎢0 ⎢⎣0 −1 0 c − a ⎥⎦ z =b−a

Thus, u can be expressed as (a, b, c ) = (b + c − a)(1,1,1) + (a − c )(1,1, 0) + (b − a)(−1, 0, −1)

so, Span(T ) = R 3 . (2) Suppose x (1,1,1) + y (1,1, 0) + z (−1, 0 − 1) = (0, 0, 0). Then x+ y−z =0 x+y =0 x −z =0

By the same row transformations in (1), you obtain ⎡1 1 1 0 ⎤ ⎡0 0 1 0 ⎤ ⇒ ⎢⎢1 1 0 0 ⎥⎥ ⇒ ⎢⎢1 0 0 0 ⎥⎥ whose only solution is ⎢⎣1 0 −1 0 ⎥⎦ ⎢⎣0 1 0 0 ⎥⎦

the trivial one, so the vectors in T are linearly independent. Hence, (1) and (2) are satisfied, so T = {(1,1,1),(1,1, 0),(−1, 0, −1)} is a basis for R 3 . c. W = {(1, 0,1),(0,1,1),(2, 0, 2)} for R 3 The vector (2, 0, 2) is a multiple of the vector (1, 0,1) so, the vectors in W are linearly dependent and thus cannot be a basis for R 3 . ⎧ ⎡1 0 ⎤ ⎡ 2 0 ⎤ ⎡0 1⎤ ⎡ −3 0 ⎤ ⎫ d. U = ⎨ ⎢ ⎥,⎢ ⎥,⎢ ⎥,⎢ ⎥ ⎬ for M 22 ⎩⎪ ⎣0 0 ⎦ ⎣ −1 0 ⎦ ⎣0 1⎦ ⎣ 0 2 ⎦ ⎪⎭ ⎡a b ⎤ (1) If ⎢ ⎥ is any vector in M 22, then find coefficients, if any, such that ⎣c d ⎦ ⎡a b ⎤ ⎡1 0 ⎤ ⎡ 2 0 ⎤ ⎡0 1⎤ ⎡ −3 0 ⎤ ⎢ c d ⎥ = x ⎢0 0 ⎥ + y ⎢ −1 0 ⎥ + z ⎢0 1⎥ + t ⎢ 0 2 ⎥ . That is, solve ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡a b ⎤ ⎡ x 0 ⎤ ⎡ 2 y 0 ⎤ ⎡0 z ⎤ ⎡ −3t ⎢ c d ⎥ = ⎢ 0 0 ⎥ + ⎢ − y 0 ⎥ + ⎢0 z ⎥ + ⎢ 0 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣

This equality yields the system of equations: x + 2 y − 3t = a z =b −y = c z + 2t = d

which reduces to

84

practice makes perfect

Linear Algebra

0⎤ . 2t ⎥⎦

⎡1 2 ⎢0 0 ⎢ ⎢0 −1 ⎢ ⎢⎣0 0

0 −3 1 0 0 0 1 2

⎡1 − R3 ⎢ 0 1 ⎢⎢ R4 0 2 ⎢0 ⎢⎣

0 0 1 0

a⎤ ⎡1 0 ⎥ b −1R2 + R4 ⎢⎢0 0 ⎥ c ⎥ 2 R3 + R1 ⎢0 −1 ⎢ ⎥ d⎦ ⎢⎣0 0

0 −3 1 0 0 0 0 1

a + 2c b −c

0 −3 a + 2c ⎤ b ⎥ 1 0 ⎥ c ⎥ 0 0 ⎥ 0 2 d −b ⎦

⎤ ⎡1 ⎥ ⎢0 ⎥ R4 + R1 ⎢ ⎥ 3 ⎢0 ⎥ ⎢ d − b ( ) ⎥ ⎢⎣0 2⎦

0 0 1 0

0 1 0 0

⎛ d −b⎞ ⎤ 0 a + 2c + 3 ⎜⎝ 2 ⎟⎠ ⎥ ⎥ 0 b ⎥ ⎥ 0 −c ⎥ 1 ⎥ (d − b ) ⎥⎦ 2

⎡a b ⎤ This system has a unique solution, from which you can express ⎢ ⎥ as ⎣c d ⎦ ⎡ ⎛ d − b ⎞ ⎤ ⎡1 0 ⎤ ⎡ 2 0 ⎤ ⎡0 1⎤ ⎛ d − b ⎞ ⎡ −3 0 ⎤ ⎢a + 2c + 3 ⎜⎝ 2 ⎟⎠ ⎥ ⎢0 0 ⎥ − c ⎢ −1 0 ⎥ + b ⎢0 1⎥ + ⎜⎝ 2 ⎟⎠ ⎢ 0 2 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎦ ⎣ ⎣ ⎦

So Span(U ) = M 22 . ⎡1 0 ⎤ ⎡ 2 0 ⎤ ⎡0 1⎤ ⎡ −3 0 ⎤ ⎡0 0 ⎤ + y⎢ (2) Suppose x ⎢ ⎥ ⎥+z ⎢ ⎥+t ⎢ ⎥=⎢ ⎥ , then ⎣0 0 ⎦ ⎣ −1 0 ⎦ ⎣0 1⎦ ⎣ 0 2 ⎦ ⎣0 0 ⎦ x + 2 y − 3t = 0 z =0 −y = 0 z + 2t = 0

Clearly, the only solution to this system is x = y = z = t = 0, so the vectors are linearly independent. Hence, (1) and (2) are satisfied, so U is a basis for M 22.

Any set of linearly independent vectors that span Rn is a basis for Rn (Note: For R 3 see part b. above); however, the vectors e1 = (1, 0, 0, # , 0), e 2 = (0,1, 0, # , 0), $ , e n = (0, 0, 0, # ,1) are the standard basis for R n. (Note: The proof of the standard basis for R 3 is part a. above.) The standard basis for M 22 , the set of all 2 × 2 matrices, is the set of vectors, ⎡1 0 ⎤ ⎡0 1 ⎤ ⎡0 0 ⎤ ⎡0 0 ⎤ M1 = ⎢ , M2 = ⎢ , M3 = ⎢ , and M 4 = ⎢ ⎥ ⎥ ⎥ ⎥. ⎣0 0 ⎦ ⎣0 0 ⎦ ⎣1 0 ⎦ ⎣0 1 ⎦

If Pn is the set of all polynomials of the form a0 + a1 x + a2 x 2 + $ + an x n , then the set S = {1, x , x 2 , x 3 , # , x n } is the standard basis for Pn. It is important to know that any two bases of the above vector spaces have the same number of elements. Therefore, for example, any basis for Rn has n vectors because the standard basis for Rn has n vectors. In general, if V has a basis of exactly n elements, then any other basis for V has exactly n elements.

Vector spaces

85

EXERCISE

7·3 For 1–4, by inspection, explain why the given vectors cannot be a basis for the indicated vector space.

1. u = (3,1), v = (1, 0), w = ( 0, 2) for R 2 2. u = ( 0,11 , ), v = ( 4 , 0, 0) for R 3

⎡ 1 1⎤ ⎡ 6 0⎤ ⎡5 1⎤ ⎡ 7 1⎤ ⎡0 3⎤ 3. A = ⎢ ,B=⎢ ,C=⎢ ,D=⎢ ,E=⎢ ⎥ ⎥ ⎥ ⎥ for M22 ⎥ ⎣2 3 ⎦ ⎣ −1 4 ⎦ ⎣3 2 ⎦ ⎣2 5 ⎦ ⎣ 7 1⎦ 4. u = ( 0, 0), v = (1, 2) for R 2 For 5–9, determine whether the given set is a basis for the indicated vector space.

5. {(2,1),(3, 0)} for R 2 6. {( 4 ,1),( −7, −8)} for R 2 7. {(1, 0, 0),(2, 0, 0),(3, 3, 3)} for R 3 8. {1+ x + x 2 , x + x 2 , x 2 } for P2 9. {1, x , x 3 } for P3

⎧ ⎡1 3 ⎤ ⎡ 0 1 ⎤ ⎡ 0 −2 ⎤ ⎡ 1 0 ⎤ ⎫ 10. ⎨ ⎢ ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎥ ⎬ for M22 ⎩⎪ ⎣1 −3 ⎦ ⎣ 1 0 ⎦ ⎣ −3 −1⎦ ⎣ −1 2 ⎦ ⎪⎭

Dimension A nonzero vector space V is finite-dimensional if it contains a finite set of basis vectors,

{ v 1 , v 2 ,…, v n }. In this case, the space V has dimension n, denoted dim(V ) = n, the number of vec-

tors in the basis. Any basis for an n-dimensional vector space has n vectors. If no such set exists, then it is an infinite dimensional vector space. The zero vector space, V = {0}, is defined to have dimension zero, and the empty set is regarded as a basis for it. Here are examples of dimensions of vector spaces.

86

◆

dim(Rn ) = n

◆

dim( M 22 ) = (2)(2) = 4

◆

dim( Mmn ) = mn

◆

dim(Pn ) = n +1

◆

The set of real-valued functions defined on [0,1] is an infinite dimensional vector space.

practice makes perfect

Linear Algebra

PROBLEM

Determine the dimension of the given subspace. a. The subspace of R 4 consisting of all vectors of the form (a, b, a − b, a + b). b. The subspace of P3 consisting of all the polynomials of the form a1 x + a2 x 3. ⎡ 2a ⎤ ⎥ ⎢ c. The subspace of M 31 of all vectors of the form ⎢ −4b ⎥ . ⎢⎣ −2a ⎥⎦

SOLUTION

a. The subspace of R 4 consisting of all vectors of the form (a, b, a − b, a + b). For any vector in the subspace, (a, b, a − b, a + b) = a(1, 0, 0, 0) + b(0,1, 0, 0) + (a − b)(0, 0,1, 0) + (a + b)(0, 0, 0,1) = ae1 + be2 + (a − b)e3 + (a + b)e4 = a(e1 + e3 + e4 ) + b(e2 − e3 + e4 ) = a(1, 0,1,1) + b(0,1, −1,1). Hence (1, 0,1,1),(0,1, −1,1) spans the subspace. Also, these vectors are linearly independent because m(1, 0,1,1) + n(0,1, −1,1) = (0, 0, 0, 0) yields m = n = 0 (the trivial solution). Therefore, {(1, 0,1,1),(0,1, −1,1)} is a basis for the subspace. Hence, the dimension of the subspace is 2. b. The subspace of P3 consisting of all the polynomials of the form a1 x + a2 x 3 . Because any vector in the subspace has the form a1 x + a2 x 3, it follows that the set {x , x 3 } spans the space. Also, {x , x 3 } is a linearly independent set because it is a subset of the standard basis for P3. Therefore, {x , x 3 } is a basis for the subspace. Hence, the dimension of the subspace is 2. ⎡ 2a ⎤ ⎥ ⎢ c. The subspace of M 31 of all vectors of the form ⎢ −4b ⎥ . ⎢⎣ −2a ⎥⎦ ⎡2⎤ ⎡0⎤ ⎡ 2a ⎤ ⎢ −4b ⎥ = a ⎢ 0 ⎥ + b ⎢ −4 ⎥ For any vector in the subspace, ⎢ ⎢ ⎥ ⎢ ⎥ . Thus, ⎥ ⎢⎣ −2 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ −2a ⎥⎦ ⎧⎡ 2 ⎤ ⎪⎢ ⎥ ⎨⎢ 0 ⎥ ⎪ ⎢ −2 ⎥ ⎩⎣ ⎦

⎡ 0 ⎤⎫ ⎪ , ⎢⎢ −4 ⎥⎥ ⎬ spans the subspace. Also these vectors are linearly ⎢⎣ 0 ⎥⎦ ⎪⎭

⎡2⎤ independent because x ⎢⎢ 0 ⎥⎥ + y ⎢⎣ −2 ⎥⎦

⎧⎡ 2 ⎤ ⎪⎢ ⎥ solution). Therefore, ⎨ 0 ⎢ ⎥ ⎪ ⎢ −2 ⎥ ⎩⎣ ⎦

⎡ 0 ⎤ ⎡0 ⎤ ⎢ −4 ⎥ = ⎢0 ⎥ yields x = y = 0 (the trivial ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣0 ⎥⎦

⎡ 0 ⎤⎫ ⎪ , ⎢⎢ −4 ⎥⎥ ⎬ is a basis for the subspace. Hence, the ⎢⎣ 0 ⎥⎦ ⎪⎭

dimension of the subspace is 2.

Normally, to show that a set of vectors is a basis for a vector space, you must show that the set is linearly independent and that the set spans the vector space. However, if the number of vectors in the set is the same as the dimension of the vector space, then you must show only either that the set is linearly independent or that the set spans the vector space. Thus, you have the following: If dim(V ) = n, then any set of n linearly independent vectors in V is a basis for V; and a set of n vectors that spans V is a basis for V. Vector spaces

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Also when dealing with subspaces, you have that if W is a subspace of V, then dim(W ) ≤ dim(V ). Furthermore, if dim(W ) = dim(V ), then W = V . EXERCISE

7·4 For 1 and 2, determine the dimension of the indicated subspace of R 4.

1. The set of all vectors of the form ( a , b , c , 0). 2. The set of all vectors of the form ( a , a , a , a ). 3. Determine the dimension of the subspace of P3 consisting of all polynomials of the form a1x + a2 x 2 + a3 x 3. 4. Determine the dimension of the subspace of M31 consisting of vectors of the form

⎡ a − 2b ⎤ ⎢ a + b ⎥. ⎥ ⎢ ⎢⎣ 3b ⎥⎦

⎡ 1⎤ 5. Find the dimension of the subspace of R spanned by ⎢5 ⎥ , ⎢ ⎥ ⎢⎣ 0 ⎥⎦ 3

⎡ −2 ⎤ ⎡ −3 ⎤ ⎢10 ⎥ , and ⎢15 ⎥ ⎢ ⎥ ⎢ ⎥. ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦

Row space, column space, and null space If A is an m × n matrix, then the row space of A is the subspace of Rn spanned by the row vectors of A. Similarly, the column space of A is the subspace of Rm spanned by the column vectors of A. The null space of A is the subspace of Rn that is the solution space of the homogenous system Ax = 0. The dimensions of the row space and column space of A are equal. If the null space of A contains nonzero vectors, then the dimension of the null space of A is the number of free variables in the solution of Ax = 0. Solutions to systems of equations can be expressed with these ideas. To wit, the system Ax = b is consistent if and only if b is in the column space of A. That is, if and only if you can express b as a linear combination of the column vectors of A. PROBLEM

⎡ −1 3 2 ⎤ ⎡ x ⎤ ⎡ 1 ⎤ ⎥⎢ ⎥ ⎢ ⎥ ⎢ Show that if Ax = b is the linear system ⎢ 1 2 −3 ⎥ ⎢ y ⎥ = ⎢ −9 ⎥ , then b is in ⎢⎣ 2 1 −2 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ −3 ⎥⎦ the column space of A and express b as a linear combination of the column vectors of A.

SOLUTION

⎡ −1 3 2 1 ⎤ ⎢ 1 2 −3 −9 ⎥ 1R1 + R2 ⎥ 2R + R ⎢ 1 3 ⎢⎣ 2 1 −2 −3 ⎥⎦

88

practice makes perfect

Linear Algebra

⎡ −1 3 2 1 ⎤ ⎢ 0 5 −1 −8 ⎥ 1R2 + R3 ⎥− ⎢ ⎢⎣ 0 7 2 −1 ⎥⎦

⎡ −1 3 2 1 ⎤ ⎢ 0 5 −1 −8 ⎥ ⎥ ⎢ ⎢⎣ 0 2 3 7 ⎥⎦

1 ⎤ ⎡ −1 3 2 ⎢ 0 1 −7 −22 ⎥ − 2 R3 + R2 ⎢ 2 R2 + R3 ⎥− ⎢⎣ 0 2 3 7 ⎥⎦ 1 ⎤ ⎡ −1 3 2 ⎢ 0 1 −7 −22 ⎥ 7 R + R 3 2 ⎥ ⎢ ⎢⎣ 0 0 1 3 ⎥⎦

1 ⎤ ⎡ −1 3 2 ⎢ 0 1 −7 −22 ⎥ 1 ⎥ 17 R3 ⎢ ⎢⎣ 0 0 17 51 ⎥⎦

⎡ −1 3 2 1 ⎤ ⎢ 0 1 0 −1⎥ −2 R3 + R1 ⎥ −3R + R ⎢ 2 1 ⎢⎣ 0 0 1 3 ⎥⎦

⎡ −1 0 0 −2 ⎤ ⎢ 0 1 0 −1 ⎥ ⎥ ⎢ ⎢⎣ 0 0 1 3 ⎥⎦

The solution is x = 2, y = −1, z = 3, so the system is consistent. Therefore, b is in the column space of A. Additionally, the solution yields the coefficients to express b as a linear combination of the column vectors of A: ⎡ −1⎤ ⎡ 3 ⎤ ⎡ 2 ⎤ ⎡1⎤ ⎥ ⎢ b = −9 = 2 ⎢⎢ 1 ⎥⎥ − 1 ⎢⎢ 2 ⎥⎥ + 3 ⎢⎢ −3 ⎥⎥ ⎢ ⎥ ⎢⎣ 2 ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣ −2 ⎥⎦ ⎢⎣ −3 ⎥⎦

If x 0 is any solution of a consistent nonhomogeneous linear system Ax = b and if { v 1 , v 2 ,…, v k } is a basis for the null space of A, then every solution, x, of Ax = b can be expressed in the form called the general solution of Ax = b as x = x 0 + a1 v 1 + a2 v 2 + $ + ak v k

where x 0 is a single particular solution of Ax = b and a1 v 1 + a2 v 2 + $ + ak v k is the general solution of Ax = 0 , the corresponding homogeneous system. Thus, you can specify the complete solution (meaning all solutions and only solutions) of a linear system Ax = b by finding a single particular solution x 0 to the equation system Ax = b and by finding a basis for the solution space of Ax = 0 . Then every solution to the original system can be obtained from the expression x = x 0 + a1 v 1 + a2 v 2 + $ + ak v k by selecting suitable values for the ai′ s; and, conversely, for all choices of the ai′ s, the vector x is a solution of Ax = b. PROBLEM

x + y + 2z = 1 Find the general solution to the linear system Ax = b given by x + z = 3. 2 x + y + 3z = 4

SOLUTION

x + y + 2z = 1 ⎡ −z + 3⎤ x + z = 3 has solution ⎢⎢ − z − 2 ⎥⎥ , From Chapter 1, recall that the system ⎢⎣ z ⎥⎦ 2 x + y + 3z = 4

where z is a free variable. A particular solution (when z = 0) for Ax = b is ⎡3⎤ ⎡ −1⎤ ⎥ ⎢ x 0 = ⎢ −2 ⎥ . The corresponding homogeneous system Ax = 0 has solution z ⎢⎢ −1⎥⎥ , ⎢⎣ 0 ⎥⎦ ⎢⎣ 1 ⎥⎦ where, again, z is a free variable. The dimension of the null space is 1 because there is only one free variable in the null space solution. A basis for the null ⎡ −1⎤ space is ⎢⎢ −1⎥⎥ . Therefore, the general solution of the system Ax = b can be ⎢⎣ 1 ⎥⎦

Vector spaces

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⎡ −1⎤ ⎡ 3 ⎤ ⎡ −1⎤ ⎡x ⎤ ⎥ ⎢ written ⎢ y ⎥ = x 0 + a ⎢⎢ −1⎥⎥ = ⎢⎢ −2 ⎥⎥ + a ⎢⎢ −1⎥⎥ , where a is arbitrary. ⎢⎣ 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ z ⎥⎦

Two row equivalent matrices have the same row space. As a matter of fact, elementary row operations do not change the row space or null space of a matrix; however, they can change the column space. If a matrix A is in row-echelon form, then the row vectors with the leading 1s form a basis for the row space of A. The column vectors with the leading 1s of the row vectors form a basis for the column space of A. Here is an example. ⎡1 −2 ⎢0 1 Given the matrix A = ⎢ ⎢0 0 ⎢ ⎣0 0

0⎤ 0⎥ ⎥ , the vectors u = [1 −2 5 0], u = [0 1 3 0], and 1 2 0⎥ ⎥ 0⎦ ⎡1 ⎤ ⎡5 ⎤ ⎡ −2 ⎤ ⎢0 ⎥ ⎢3⎥ ⎢1⎥ u 3 = [0 0 1 0] are basis vectors for the row space and the vectors v 1 = ⎢ ⎥ , v 2 = ⎢ ⎥ , and d v3 = ⎢ ⎥ ⎢0 ⎥ ⎢1 ⎥ ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣0 ⎦ ⎣0 ⎦ ⎣0⎦ are basis vectors for the column space. 5 3 1 0

Note: Brackets are used on the row vectors for clarity. PROBLEM

SOLUTION

Find a basis for the space spanned by u1 = (1, 0, 3), u 2 = (1,1, 0), and u 3 = (2,1, 3) . ⎡1 0 3 ⎤ The space spanned by these vectors is the row space of the matrix ⎢⎢1 1 0 ⎥⎥ . ⎢⎣ 2 1 3 ⎥⎦ Reduce this matrix to row-echelon form. ⎡1 0 3 ⎤ −1R + R ⎡1 0 3 ⎤ 1 2 ⎢ ⎥ −1R + R ⎢1 1 0 ⎥ 2 3 ⎥− ⎢ ⎢0 1 −3 ⎥ 2 + R R 1 3 ⎢⎣ 2 1 3 ⎥⎦ ⎢⎣0 1 −3 ⎥⎦

⎡1 0 3 ⎤ ⎢0 1 −3 ⎥. The basis for ⎥ ⎢ ⎢⎣0 0 0 ⎥⎦

the space is S = {[1 0 3],[0 1 −3]}.

Remark: The vectors in the null space of a matrix have no obvious relation to the entries of the matrix. Solving the equation Ax = 0 is equivalent to finding a description of the null space of A. PROBLEM

Find a set of vectors that spans the null space of the matrix ⎡ 1 −2 2 3 −1 ⎤ A = ⎢⎢ 2 −4 5 8 −4 ⎥⎥ . ⎢⎣ −3 6 −1 1 −7 ⎥⎦

90

practice makes perfect

Linear Algebra

SOLUTION

The null space of A is the solution space of the homogeneous system Ax = 0. Reduce the coefficient matrix to row-echelon form. ⎡ 1 −2 2 3 −1 ⎤ ⎢ 2 −4 5 8 −4 ⎥ −2 R1 + R2 ⎥ 3R + R ⎢ 1 3 ⎢⎣ −3 6 −1 1 −7 ⎥⎦

⎡1 −2 2 3 −1 ⎤ ⎢0 0 1 2 −2 ⎥ ⎥ ⎢ ⎢⎣0 0 5 10 −10 ⎥⎦

⎡1 −2 2 3 −1 ⎤ − 5 R2 + R3 ⎢⎢0 0 1 2 −2 ⎥⎥ − 2 R2 + R1 ⎢⎣0 0 0 0 0 ⎥⎦ ⇒

⎡1 −2 0 −1 3 ⎤ ⎢0 0 1 2 −2 ⎥ ⎥ ⎢ ⎢⎣0 0 0 0 0 ⎥⎦

x1 − 2 x 2 − x 4 + 3 x 5 = 0 x 3 + 2x 4 − 2x5 = 0 0=0

Thus, x 2 , x 4 , and x5 are free variables, yielding the vector equation ⎡ −3 ⎤ ⎡1⎤ ⎡2⎤ ⎡ x1 ⎤ ⎡ 2 x 2 + x 4 − 3 x 5 ⎤ ⎥ ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎢x ⎥ ⎢ 0 1 x2 ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ x = ⎢ x 3 ⎥ = ⎢ −2xx 4 + 2 x5 ⎥ = x 2 ⎢ 0 ⎥ + x 4 ⎢ −2 ⎥ + x5 ⎢ 2 ⎥ . Hence, any vector in the ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ x4 ⎢0⎥ ⎢1⎥ ⎥ ⎢0 ⎥ ⎢ x4 ⎥ ⎢ ⎢⎣ 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ x5 ⎥⎦ ⎢⎣ x5

null space can be expressed as a linear combination of the vectors in ⎧ ⎡ 2 ⎤ ⎡ 1 ⎤ ⎡ −3 ⎤ ⎫ ⎪ ⎢1 ⎥ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎪ ⎪⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪⎪ ⎨ ⎢ 0 ⎥ , ⎢ −2 ⎥ , ⎢ 2 ⎥ ⎬, so this set is a spanning set for the null space. These vectors ⎪⎢0 ⎥ ⎢ 1 ⎥ ⎢ 0 ⎥ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪⎩ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎪⎭ ⎡ 2 ⎤ ⎡ 1 ⎤ ⎡ −3 ⎤ ⎡0 ⎤ ⎢1 ⎥ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ are linearly independent because a ⎢ 0 ⎥ + b ⎢ −2 ⎥ + c ⎢ 2 ⎥ = ⎢0 ⎥ implies ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ 1 ⎥ ⎢ 0 ⎥ ⎢0 ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣0 ⎥⎦ a = b = c = 0. Therefore, they are a basis for the null space. EXERCISE

7·5 For 1–4, determine whether b is in the column space of A, and if so, express b as a linear combination of the column vectors of A.

⎡1 3 ⎤ 1. A = ⎢ ⎥; ⎣2 −3 ⎦ ⎡1 1 2⎤ 2. A = ⎢⎢ 1 0 1⎥⎥ ; ⎢⎣2 1 3 ⎥⎦

⎡ −1⎤ b=⎢ ⎥ ⎣5⎦ ⎡ −1⎤ b = ⎢⎢ 0 ⎥⎥ ⎢⎣ 4 ⎥⎦ Vector spaces

91

⎡ 1 −1 1⎤ ⎥ ⎢ 3. A = ⎢9 3 1⎥ ; ⎢⎣ 1 1 1⎥⎦

⎡5⎤ b = ⎢⎢ 1 ⎥⎥ ⎢⎣ −1⎥⎦

⎡ −1 −1 1 ⎤ ⎥ ⎢ 4. A = ⎢ 1 1 −1⎥ ; ⎢⎣ 1 −1 1 ⎥⎦

⎡ 1⎤ b = ⎢⎢ 0 ⎥⎥ ⎢⎣ 0 ⎦⎥

For 5 and 6, find the vector form of the general solution of the given linear system Ax = b. Use that solution to find the vector form of the general solution of Ax = 0.

5.

x1 − 3 x 2 = 1 2 x1 − 6 x 2 = 2

6.

x + y + 2z = 5 x + z = −2 2 x + y + 3z = 3

For 7 and 8, find a basis for the null space of A.

⎡ 2 0 −1⎤ ⎥ ⎢ 7. A = ⎢ 4 0 −2 ⎥ ⎢⎣ 0 1 0 ⎥⎦ ⎡ 1 4 5 2⎤ 8. A = ⎢⎢ 2 1 3 0 ⎥⎥ ⎢⎣ −1 3 2 2 ⎥⎦ For 9 and 10, by inspection, find bases for the row and column spaces of A.

⎡1 0 2⎤ 9. A = ⎢⎢ 0 0 1 ⎥⎥ ⎢⎣ 0 0 0 ⎥⎦ ⎡1 ⎢0 10. A = ⎢ ⎢0 ⎢ ⎣0

2 −1 5 ⎤ 1 4 3⎥ ⎥ 0 1 −7 ⎥ ⎥ 0 0 1⎦

Rank and nullity For any given matrix, Am×n , there are four fundamental vector spaces associated with Am×n : ◆ ◆ ◆ ◆

92

The row space of Am×n , which is a subspace of Rn. The null space of Am×n , which is a subspace of Rn. The column space of Am×n , which is a subspace of Rm. The null space of AnT×m , which is a subspace of Rm.

practice makes perfect

Linear Algebra

The row and column spaces of any matrix A have the same dimension, and this dimension is called the rank of A, denoted rank(A). The dimension of the null space of A is the nullity of A, denoted nullity(A). Basic results that are important to know about rank and nullity are the following: ◆ ◆ ◆

◆

◆

◆

For any matrix A, rank (A) = rank (AT ). If A is an m × n matrix, then rank (A) + nullity(A) = n. If A is an m × n matrix, then the nullity(A) equals the number of free variables in the general solution of Ax = 0 ; and the rank(A) equals the number of leading (non-free) variables in the solution of Ax = 0. If A is an m × n matrix, then (because the row and column spaces have the same rank) rank (A) ≤ minimum(m, n). For example, for A5×3 , the rank of A is at most 3, and so the five row vectors are linearly dependent. Similarly, for A2×4 , the rank of A is at most 2, and so the four column vectors are linearly dependent. If the rank of the coefficient matrix A and the rank of the augmented matrix [ A | b] are the same, then the linear system Ax = b is consistent. If A is an n × n matrix, then A is nonsingular if and only if rank(A) = n . PROBLEM

Respond as indicated. ⎡1 0 1 2 ⎤ a. Verify that the rank (A) = rank (A ) for A = ⎢⎢0 1 1 −2 ⎥⎥ . ⎢⎣1 3 1 0 ⎥⎦ T

⎡2 0 2⎤ b. Find the rank and nullity of A = ⎢⎢ 4 0 4 ⎥⎥ . ⎢⎣ 0 2 0 ⎥⎦ ⎡ −1 −3 t ⎤ c. Discuss how the rank of A = ⎢⎢ 3 6 −2 ⎥⎥ varies with t. ⎢⎣ t 3 −1 ⎥⎦ SOLUTION

⎡1 0 1 2 ⎤ a. Verify that the rank (A) = rank (A ) for A = ⎢⎢0 1 1 −2 ⎥⎥ . ⎢⎣1 3 1 0 ⎥⎦ T

⎡1 0 ⎡1 0 1 2 ⎤ ⎢0 1 A = ⎢⎢0 1 1 −2 ⎥⎥ and AT = ⎢ ⎢1 1 ⎢⎣1 3 1 0 ⎥⎦ ⎢ ⎣ 2 −2

1⎤ 3⎥ ⎥ 1⎥ ⎥ 0⎦

⎡1 ⎡1 0 1 2 ⎤ ⎢ A = ⎢⎢0 1 1 −2 ⎥⎥ is row equivalent to ⎢0 ⎢ ⎢⎣1 3 1 0 ⎥⎦ ⎢0 ⎣ 1 1 0 1 ⎡ 0 ⎡ ⎤ ⎢0 1 ⎢0 1 3⎥ ⎥ is row equivalent to ⎢ AT = ⎢ ⎢0 0 ⎢1 1 1 ⎥ ⎢ ⎢ ⎥ ⎣0 0 ⎣ 2 −2 0 ⎦

0 0 10 3 ⎤ ⎥ 1 0 − 2 3 ⎥ . Thus, rank(A) = 3. ⎥ 0 1 − 4 3 ⎥⎦ 0⎤ 0⎥ ⎥ . Thus, rank(AT ) = 3. 1⎥ ⎥ 0⎦ Vector spaces

93

Because rank A = rank (AT ) = 3 verification is established. ⎡2 0 2⎤ b. Find the rank and nullity of A = ⎢⎢ 4 0 4 ⎥⎥ . ⎢⎣ 0 2 0 ⎥⎦ ⎡2 0 2⎤ ⎡1 0 1 ⎤ ⎥ ⎢ A = ⎢ 4 0 4 ⎥ is row equivalent to ⎢⎢0 1 0 ⎥⎥ . Thus, the rank(A) = 2. ⎢⎣ 0 2 0 ⎥⎦ ⎢⎣0 0 0 ⎥⎦ The solution of the homogenous system Ax = 0 is x1 = − x 3 and x 2 = 0 with x 3 a free variable. Because there is only one free variable, nullity(A) = 1. ⎡ −1 −3 t ⎤ c. Discuss how the rank of A = ⎢⎢ 3 6 −2 ⎥⎥ varies with t. ⎢⎣ t 3 −1 ⎥⎦ ⎡ −1 −3 t ⎤ A = ⎢⎢ 3 6 −2 ⎥⎥ ⎢⎣ t 3 −1 ⎥⎦ ⎡ −1 −3 t ⎤ 3R1 + R2 A = ⎢⎢ 3 6 −2 ⎥⎥ tR1 + R3 ⎢⎣ t 3 −1 ⎥⎦

t ⎤ ⎡ −1 −3 ⎥ −1R2 + R1 ⎢0 3 3 2 − − t ⎥ 1R2 + R3 ⎢ ⎢⎣ 0 3 − 3t t 2 − 1 ⎥⎦

2 − 2t ⎤ −1R1 ⎡ −1 0 ⎢ − tR2 + R3 ⎢ 0 −3 3t − 2 ⎥⎥ 1 − R2 3 ⎢⎣ 0 0 −2t 2 + 5t − 3 ⎥⎦

2 − 2t ⎤ ⎡ −1 0 ⎢ 0 −3 3t − 2 ⎥⎥ ⎢ ⎢⎣ 0 −3t t 2 + 3t − 3 ⎥⎦

2t − 2 ⎤ ⎡1 0 ⎢ 2 − 3t ⎥ ⎢0 1 ⎥ 3 ⎢ ⎥ ⎢0 0 −2t 2 + 5t − 3 ⎥ ⎣ ⎦

Hence, A will have rank 3 if −2t 2 + 5t − 3 ≠ 0 and will have rank 2 if −2t 2 + 5t − 3 = 0. 3 Therefore, given that −2t 2 + 5t − 3 = −(2t 2 − 5t + 3) = −(t − 1)(2t − 3): If t = 1 or t = , 2 A will have rank 2; otherwise, A will have rank 3.

A linear system of equations with more equations than unknowns is an overdetermined linear system. In an overdetermined system, the number of rows exceeds the number of columns in the matrix equation Ax = b; that is, for A an m × n matrix, m > n . In this case, the equation Ax = b cannot have a solution for every possible vector b. A linear system of equations with more unknowns than equations is an underdetermined linear system. In an underdetermined system, the number of columns exceeds the number of rows in the matrix equation Ax = b; that is, for A an m × n matrix, m < n . In this case, if the system Ax = b has a solution, then it has infinitely many solutions; and, furthermore, the general solution must have at least one free variable. In particular, the corresponding homogeneous system Ax = 0 always has infinitely many solutions. PROBLEM

Determine the condition(s) under which the following system is consistent. x1 − 2 x 2 = b1 x1 − x 2 = b2 x1 + x 2 = b3 x1 + 2 x 2 = b4

94

practice makes perfect

Linear Algebra

SOLUTION

⎡1 −2 b1 ⎤ x1 − 2 x 2 = b1 ⎢1 −1 b ⎥ −1R1 + R2 x1 − x 2 = b2 2 ⎥ −1R1 + R3 ⇒ ⎢⎢ 1 1 b3 ⎥ x1 + x 2 = b3 1R1 + R4 ⎢ ⎥ − ⎢⎣1 2 b4 ⎦ x1 + 2 x 2 = b4

⎡1 −2 b1 ⎤ ⎢0 1 b − b ⎥ −3R2 + R3 2 1 ⎢ ⎥ −4 R2 + R4 ⎢0 3 b3 − b1 ⎥ R2 + R1 ⎢ ⎥ 2 ⎢⎣0 4 b4 − b1 ⎦

⎡1 ⎢0 ⎢ ⎢0 ⎢ ⎢⎣0

0 2b2 − b1 ⎤ b2 − b1 ⎥ 1 ⎥ 0 b3 − 3b2 + 2b1 ⎥ ⎥ 0 b4 − 4b2 + 3b1 ⎦

This system is consistent if and only if b3 − 3b2 + 2b1 = 0 and b4 − 4b2 + 3b1 = 0. Thus, the system is consistent if b1 = 4 s − 3t , b2 = 3s − 2t, b3 = s, and b4 = t , where s and t are arbitrary.

The following list summarizes main points so far. If A is an n × n matrix then the following are equivalent statements. 1. A is nonsingular. 2. det(A) ≠ 0. 3. rank(A) = n. 4. nullity(A) = 0. 5. Ax = 0 has only the trivial solution. 6. In×n is the reduced row-echelon form of A. 7. For every n ×1 matrix b, Ax = b is consistent and has only one solution. 8. The row and column vectors of A each span Rn. 9. The row and column vectors of A are each a linearly independent set. 10. The row and column vectors of A are each a basis for Rn.

EXERCISE

7·6 For 1 and 2, respond as indicated.

1. Determine the condition(s) under which the following system is consistent.

x1 − 3 x 2 = b1 x1 − 2 x 2 = b2 x1 + x 2 = b3 2. Verify that rank(A) = rank(AT ) for the following matrix.

⎡ 1 2 4 0⎤ A = ⎢⎢ −3 1 5 2 ⎥⎥ ⎢⎣ −2 3 9 2 ⎥⎦ For 3–5, find the rank and nullity of the indicated matrix.

⎡ 1 −1 3 ⎤ 3. A = ⎢⎢5 −4 −4 ⎥⎥ ⎢⎣7 −6 2 ⎥⎦ Vector spaces

95

⎡ 2 0 1⎤ 4. A = ⎢⎢ 4 0 2 ⎥⎥ ⎢⎣ 0 1 0 ⎥⎦ ⎡ 1 4 5 2⎤ 5. A = ⎢⎢ 2 1 3 0 ⎥⎥ ⎢⎣ −1 3 2 2 ⎥⎦ ⎡1 1 t ⎤ 6. Discuss how the rank of the matrix A = ⎢⎢1 t 1⎥⎥ varies with t. ⎢⎣1 1 t ⎥⎦ ⎡2 0 2 ⎤ 7. Solve the homogeneous system Ax = 0 if matrix A = ⎢⎢ 1 1 1⎥⎥ . ⎢⎣ 1 0 1⎥⎦ 8. If A is an m × n matrix, what are the largest possible value for its rank and the smallest possible value for its nullity? 9. Express ( 4 a , a − b , a + 2b ) as a linear combination of ( 4 ,11 , ) and ( 0, −1, 2). 10. Find a basis for the vector space of all 2 × 2 symmetric matrices.

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Inner product spaces

·8·

In this chapter, you will learn: ◆ ◆ ◆ ◆ ◆

Definition and terminology for inner product spaces Norm of a vector in an inner product space Cauchy-Schwarz inequality and properties of the norm Orthogonality in inner product spaces Gram-Schmidt procedure

This chapter extends the idea of the Euclidean inner product that was discussed in Chapter 6 to the more general notion of an inner product on a vector space V.

Definition and terminology for inner product spaces From your work in Chapter 6, you are familiar with the Euclidean inner product in Rn . The following definition generalizes this idea to the concept of inner product on a vector space V. Given a real vector space V, an inner product is a function that assigns a real number, denoted 〈u, v 〉, to each pair of vectors, u and v in V so that the following axioms are satisfied for all vectors, u, v, and w ∈V , and all scalars a, b ∈ R : i. Positivity 〈u , u 〉 ≥ 0 with 〈u , u 〉 = 0 if and only if u = 0. ii. Symmetry 〈u , v 〉 = 〈 v , u 〉 iii. Linearity 〈a u + b v , w 〉 = a〈u , w 〉 + b〈 v , w 〉 Notice that in axiom iii, the + symbol in au + bv indicates vector addition, but the + symbol in a u , w + b v , w indicates scalar addition. It is customary practice to use the same symbol for both operations when no confusion will arise. Also, note that the multiplication in a u + b v is scalar multiplication of a matrix, but the multiplication in a〈u , w 〉 + b〈 v , w 〉 is strictly scalar multiplication. A vector space V with an inner product , is an inner product space. Here are examples of inner product spaces.

97

Note: This discussion is limited to real inner product spaces. ◆

◆

◆

Euclidean vector space Rn with inner product 〈u , v 〉 = u u v = u1v1 + u2v2 + $ + unvn , where u = (u1 , u2 ,$ , un ) and v = (v1 , v2 ,$ , vn ) are vectors in Rn (as described in Chapter 6) is an inner product space. Note: Hereafter, unless explicitly stated otherwise, assume the inner product for Rn is the usual inner (dot) product as defined in this example. The vector space M 22 of all 2 × 2 matrices with (trace) inner product 〈 A, B〉 = tr(BT A) = tr(ATB)= a11b11 + a21b21 + a12b12 + a22b22 , where A = ⎡aij ⎤ and B = ⎡bij ⎤ are vectors in ⎣ ⎦2 × 2 ⎣ ⎦2 × 2 M 22 and tr( ) is the trace (the sum of the diagonal elements). The vector space P2 of all polynomials of degree n ≤ 2 with inner product 〈p, q 〉 = a0b0 + a1b1 + a2b2 , when p = a0 + a1x + a2 x 2 and q = b0 + b1x + b2 x 2 are vectors in P2 , is an inner product space. PROBLEM

SOLUTION

Let u = (1, 5, −2), v = (−2, 0, 6), and w = (4 , −1, −3) be vectors in R 3 . Verify that 〈2u + 5v , w 〉 = 2〈u , w 〉 + 5〈 v , w 〉 First find 〈2u + 5v , w 〉: = 〈2(1, 5, −2) + 5( − 2, 0, 6),(4 , −1, −3)〉 = 〈(2, 10, −4 ) + (−10, 0, 30),(4 , −1, −3)〉 = 〈(−8, 10, 26),(4 , −1, −3)〉 = −32 − 10 − 78 = −120 Then find 2〈u , w 〉 + 5〈 v , w 〉: = 2〈(1, 5, −2),(4 , −1, −3)〉 + 5〈(−2, 0, 6),(4 , −1, −3)〉 = 2(4 − 5 + 6) + 5(−8 + 0 − 18) = 2(5) + 5(−26) = 10 − 130 = −120 Therefore, 〈2u + 5v , w 〉 = 2〈u , w 〉 + 5〈 v , w 〉 = −120

PROBLEM

Suppose that M 22 is the inner product space of all 2 × 2 matrices with inner product 〈 A, B〉 = tr(BT A) = tr(AT B) = a11b11 + a21b21 + a12b12 + a22b22 , where A = ⎡⎣aij ⎤⎦

⎡1 −2 ⎤ ⎡ −4 6 ⎤ and B = ⎡⎣bij ⎤⎦ are vectors in M 22 . For A = ⎢ ⎥ and B = ⎢ 2 5 ⎥ , find 2× 2 5 3 ⎣ ⎣ ⎦ ⎦ a. 〈 A, B〉 b. 〈 B, A〉 c. 〈 A, A〉 SOLUTION

⎛ ⎡1 −2 ⎤ T ⎡ −4 6 ⎤⎞ ⎛ a. 〈 A, B〉 = tr( AT B) = tr ⎜ ⎢ ⎥ ⎢ 2 5 ⎥⎟ = tr ⎜ 5 3 ⎝ ⎦ ⎣ ⎦⎠ ⎝⎣

⎡ 1 5 ⎤ ⎡ −4 6 ⎤⎞ ⎢ −2 3 ⎥ ⎢ 2 5 ⎥⎟ = ⎣ ⎦⎣ ⎦⎠

(1)(−4 ) + (5)(2) + (−2)(6) + (3)(5) = 9

b. 〈 B, A〉; because of the symmetric axiom, 〈 B, A〉 = 〈 A, B〉 = 9. ⎛ ⎡1 −2 ⎤ T ⎡1 −2 ⎤⎞ ⎛ ⎡ 1 5 ⎤ ⎡1 −2 ⎤⎞ c. 〈 A, A〉 = tr( AT A) = tr ⎜ ⎢ ⎥ ⎢5 3 ⎥⎟ = tr ⎜ ⎢ ⎥⎢ ⎥ = 5 3 ⎝ ⎣ −2 3 ⎦ ⎣5 3 ⎦⎟⎠ ⎦ ⎣ ⎦⎠ ⎝⎣ (1)(1) + (5)(5) + (−2)(−2) + (3)(3) = 39

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2× 2

The following are additional properties of inner products. For all vectors, u, v, and w in a real inner product space (V , 〈 ,〉) and all scalars k ∈ R , (a) (b) (c) (d)

〈0 , u 〉 = 〈u , 0 〉 = 0 〈k u , w 〉 = 〈u , k w 〉 = k 〈u , w 〉 〈u , v ± w 〉 = 〈u , v 〉 ± 〈u , w 〉 〈u ± v , w 〉 = 〈u , w 〉 ± 〈 v , w 〉 PROBLEM

SOLUTION

Suppose u and w are vectors in a real inner product space (V , 〈 , 〉) and k ∈R . Prove that 〈ku , w 〉 = k 〈u , w 〉 . 〈ku , w 〉 = 〈ku + 0 v , w 〉

By properties of vectors.

= k 〈 u , w 〉 + 0〈 v , w 〉

By the linearity axiom of inner products.

= k〈u , w 〉 + 0

By the zero property of scalar multiplication.

= k 〈u , w 〉

By the identity property of scalar addition.

Remark: The concept of inner product can also be defined in a complex vector space (see Prob. 15 in Exercise 8.1 that follows).

EXERCISE

8·1 For 1–4, suppose u, v, and w are vectors in a real inner product space (V , 〈 , 〉) , and a, b, and k ∈R .

1. Prove that 〈u, kw 〉 = k 〈u, w 〉. 2. Prove that 〈ku, kw 〉 = k 2 〈u, w 〉 . 3. Prove that 〈u,av + bw 〉 = a 〈u, v 〉 + b 〈u, w 〉. 4. Prove that 〈u − v , w 〉 = 〈u, w 〉 − 〈 v , w 〉. For 5 and 6, let u = ( x1 , x 2 ), v = ( y1 , y 2 ) and w = (w1 , w 2 ).

5. Verify that the following is an inner product in R 2 : 〈u, v 〉 = x1y1 − x1y 2 − x 2 y1 + 2 x 2 y 2 . 6. Show that 〈u, v 〉 = x1y1x 2 y 2 is not an inner product in R 2 . For 7–11, let u = (1, 4 ), v = (3, 5) , and w = (2, 3).

7. Find 〈u, v 〉 with respect to the usual inner product (dot product) of R 2 . 8. Find 〈 v , w 〉 with respect to the inner product of Prob. 5. 9. Find 〈u, w 〉 with respect to the usual inner product of R 2 . 10. Find 〈u, w 〉 with respect to the inner product of Prob. 5. 11. Find 〈3u + 2 v , w 〉 with respect to the inner product of Prob. 5.

Inner product spaces

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⎡ −1 0 ⎤ ⎡1 −1⎤ ⎡1 2⎤ T T For 12–14, let A = ⎢ ,B=⎢ , and C = ⎢ ⎥ ⎥ and let 〈 A, B 〉 = tr(B A) = tr(A B ) = ⎥ 3 7 0 2 1 1 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ a11b11 + a21b21 + a12 b12 + a22 b22 .

12. Find 〈 A, B 〉 . 13. Find 〈 A, C 〉 . 14. Find 〈 A, A〉 . 15. In the realm of complex numbers an inner product on C n, the space of ordered n-tuples of complex numbers, can be defined as follows: If z = ( z1 , z2 ,#, z n ) and w = (w1 , w 2 ,#, w n ), n then 〈z, w 〉 = ∑ z k w k . Use this definition to find the inner product of z = (3, i , 2 − i ) and k =1

w = (5, −i ,1) in C 3.

Norm of a vector in an inner product space The norm of a vector u, denoted u , is the nonnegative real number given by u = 〈u , u 〉 . 2

Note, that from u = 〈u , u 〉 , you have u = 〈u , u 〉, a relationship that is useful in later work. PROBLEM

Let u = (1, 5, −2), v = (−4 , 0, 3) , and z = (1, 0, 0) be vectors in R 3. Find a. u b. v c. z

SOLUTION

a. u u = 〈u , u 〉 = u u u = (1, 5, −2) u (1, 5, −2) = 12 + 52 + (−2)2 = 30

b. v v = 〈 v , v 〉 = v u v = (−4 , 0, 3) u (−4 , 0, 3) = (−4 )2 + 02 + 32 = 25 = 5

c. z z = 〈z , z 〉 = z u z = (1, 0, 0) u (1, 0, 0) = 12 + 02 + 02 = 1 = 1

(Note that z is a unit vector in R3.) PROBLEM

Suppose that M 22 is the inner product space of all 2 × 2 matrices with inner product 〈 A, B〉 = tr(AT B) = a11b11 + a21b21 + a12b12 + a22b22 , where A = ⎡⎣aij ⎤⎦ and B = ⎡⎣bij ⎤⎦ 2× 2

⎡0 0 ⎤ ⎡1 −2 ⎤ ⎡ −4 6 ⎤ and B = ⎢ , and C = ⎢ are vectors in M 22 . For A = ⎢ ⎥ , find ⎥ ⎥ ⎣1 0 ⎦ ⎣5 3 ⎦ ⎣ 2 5⎦ a. A

b. B c. C

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2× 2

SOLUTION

⎛ a. A = 〈 A, A〉 = tr( AT A) = tr ⎜ ⎝

T ⎛ ⎡ 1 5 ⎤ ⎡1 −2 ⎤ ⎞ ⎡1 −2 ⎤ ⎡1 −2 ⎤ ⎞ ⎢5 3 ⎥ ⎢5 3 ⎥ ⎟ = tr ⎜ ⎢ −2 3 ⎥ ⎢5 3 ⎥ ⎟ = ⎝⎣ ⎣ ⎦ ⎣ ⎦⎠ ⎦⎣ ⎦⎠

12 + 52 + (−2)2 + 32 = 39 ⎛ b. B = 〈 B, B〉 = tr( BT B) = tr ⎜ ⎝

T ⎛ ⎡ −4 2 ⎤ ⎡ −4 6 ⎤ ⎞ ⎡ −4 6 ⎤ ⎡ −4 6 ⎤ ⎞ ⎢ 2 5 ⎥ ⎢ 2 5 ⎥ ⎟ = tr ⎜ ⎢ 6 5 ⎥ ⎢ 2 5 ⎥ ⎟ = ⎝⎣ ⎣ ⎦ ⎣ ⎦⎠ ⎦⎣ ⎦⎠

(−4 )2 + 22 + 62 + 52 = 81 = 9 ⎛ c. C = 〈C , C 〉 = tr(C T C ) = tr ⎜ ⎝

T ⎛ ⎡0 1 ⎤ ⎡0 0 ⎤ ⎞ ⎡0 0 ⎤ ⎡0 0 ⎤ ⎞ ⎢1 0 ⎥ ⎢1 0 ⎥ ⎟ = tr ⎜⎝ ⎢0 0 ⎥ ⎢1 0 ⎥ ⎟⎠ = ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎠

02 + 12 + 02 + 02 = 1 = 1

(Note that C is a unit vector in M 22.) A vector z for which z = 1 is a unit vector and is said to be normalized. You can normalize any nonzero vector u by multiplying u by the reciprocal of u to obtain the unit vector. That is, 1 u ⋅u = is the normalized form of u. u u PROBLEM

Prove that

u is a unit vector. u

SOLUTION

u u

PROBLEM

=

u u , u u

By definition of norm

=

1

By Prob. 2, Exercise 8.1

u

2

〈u , u 〉

=

1 u

=

1 ⋅ u =1 u

〈u , u 〉

By principal square root of a scalar By scalar multiplication

Let u = (1, 5, −2), v = (−4 , 0, 3), and z = (1, 0, 0) be vectors in R 3. Normalize a. u b. v c. z

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101

SOLUTION

a. u 1 1 1 (1, 5, −2) = ⋅u = ⋅ (1, 5, −2) = (1, 5, −2) u (1, 5, −2) u (1, 5, −2) 1 1 5 −2 ⎞ ⎛ 1 ⋅ (1, 5, −2) = ⎜ ⋅ (1, 5, −2) = , , ⎟ 2 2 2 ⎝ 30 30 30 30 ⎠ 1 + 5 + (−2) b. v 1 1 1 (−4 , 0, 3) = ⋅v = ⋅ (−4 , 0, 3) = 3 u ( −4 , 0, 3) ( − 4 , 0 , ) v (−4 , 0, 3) 1 (−4 ) + 0 + 3 2

2

2

⋅ (−4 , 0, 3) =

3⎞ 1 ⎛ 4 ⋅ (−4 , 0, 3) = ⎜ − , 0, ⎟ ⎝ ⎠ 5 5 25

c. z Because z = 〈z , z 〉 = z u z = (1, 0, 0) u (1, 0, 0) = 12 + 02 + 02 = 1 = 1, z = (1, 0, 0) is a unit vector, so it is already normalized. PROBLEM

Suppose that M 22 is the inner product space of all 2 × 2 matrices with inner product 〈 A, B〉 = tr( AT B) = a11b11 + a21b21 + a12b12 + a22b22 , where A = ⎡⎣aij ⎤⎦ and B = ⎡⎣bij ⎤⎦ 2× 2

2× 2

⎡1 −2 ⎤ ⎡ −4 6 ⎤ ⎡0 0 ⎤ are vectors in M 22 . For A = ⎢ ,B=⎢ , and C = ⎢ ⎥ ⎥ ⎥ in M 22 , normalize ⎣5 3 ⎦ ⎣ 2 5⎦ ⎣1 0 ⎦ a. A

b. B c. C SOLUTION

a. A 1 1 1 ⋅A = ⋅A = 2 2 A 〈 A, A〉 1 + 5 + (−2)2 + 32 ⎡ ⎢ ⎢ ⎢ ⎢⎣ b. B

1 39 5 39

−

1 ⎡1 −2 ⎤ ⎡1 −2 ⎤ = ⋅⎢ ⋅⎢ ⎥= ⎥ 39 ⎣5 3 ⎦ ⎣5 3 ⎦

2 ⎤ 39 ⎥ ⎥ 3 ⎥ 39 ⎥⎦

1 ⎡ −4 6 ⎤ 1 1 1 ⎡ −4 6 ⎤ = ⋅⎢ ⋅B = ⋅B = ⋅⎢ ⎥= ⎥ B 81 ⎣ 2 5 ⎦ 〈 B, B〉 (−4 )2 + 22 + 62 + 52 ⎣ 2 5 ⎦ ⎡ 4 − 1 ⎡ −4 6 ⎤ ⎢ 9 = ⋅⎢ ⎢ 9 ⎣ 2 5 ⎥⎦ ⎢ 2 ⎢⎣ 9 c. C

2⎤ 3⎥ ⎥ 5⎥ 9 ⎥⎦

⎛ ⎡0 0 ⎤ T ⎡0 0 ⎤⎞ ⎛ ⎡0 1 ⎤ ⎡0 0 ⎤⎞ Because C = 〈C , C 〉 = tr(C C ) = tr ⎜ ⎢ ⎥ ⎢1 0 ⎥⎟ = tr ⎜ ⎢0 0 ⎥ ⎢1 0 ⎥⎟ = 1 0 ⎝⎣ ⎦ ⎣ ⎦⎠ ⎦⎣ ⎦⎠ ⎝⎣ T

⎡0 0 ⎤ 02 + 12 + 02 + 02 = 1 = 1, C = ⎢ ⎥ is a unit vector, so it is already normalized. ⎣1 0 ⎦

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EXERCISE

8·2 Respond as indicated.

1. Find the norm of u = (5, 6,1) with respect to the usual inner product (dot product) of R 3.

⎡1 0 ⎤ 2. If A = ⎢ ⎥ , then find the norm of A with respect to the trace inner product. ⎣3 4 ⎦ 3. Let u = ( x1 , x 2 ) = ( 4 , 3), find the norm of u with respect to the inner product 〈u, v 〉 = x1y1 − x1y 2 − x 2 y1 + 2 x 2 y 2, where u = ( x1 , x 2 ) and v = ( y1 , y 2 ), as defined in Prob. 5, Exercise 8.1. 4. Find the norm of u = ( 4 , 3) with respect to the usual inner product (dot product) of R 2 . 5. Normalize u = ( 4 , 3) with respect to the inner product 〈u, v 〉 = x1y1 − x1y 2 − x 2 y1 + 2 x 2 y 2, where u = ( x1 , x 2 ) and v = ( y1 , y 2 ), as defined in Prob. 5, Exercise 8.1.

⎡1 0 ⎤ 6. Normalize A = ⎢ ⎥ with respect to the trace inner product. ⎣3 4 ⎦ 7. Find the norm of z = (i ,1− i , 5i ) with respect to the inner product of Prob. 15 in Exercise 8.1. 8. Normalize z = (i ,1− i , 5i ) with respect to the inner product of Prob. 15 in Exercise 8.1. 9. Find the norm of u = (3, 5, 0) with respect to the usual inner product (dot product) of R 3. 10. Normalize u = (3, 5, 0) with respect to the usual inner product (dot product) of R 3.

Cauchy-Schwarz inequality and properties of the norm The Cauchy-Schwarz Inequality is a useful theorem that states the following: For any vectors u and v in a real inner product space (V , 〈 , 〉), 〈u , v 〉 ≤ u v . PROBLEM

Let u = (1, 5, −2) and v = (−4 , 0, 3) be vectors in R 3. Verify that 〈u , v 〉 ≤ u v .

SOLUTION

First calculate 〈u , v 〉 : = 〈(1, 5, −2),(−4 , 0, 3)〉 = (1, 5, −2) u (−4 , 0, 3) = −4 + 0 − 6 = −10 = 10 Then calculate u v : = (1, 5, −2) (−4 , 0, 3) = 12 + 52 + (−2)2 (−4 )2 + 02 + 32 = 30 25 = 5 30 ≈ 27.4 Because 10 ≤ 5 30 ≈ 27.4, 〈u , v 〉 ≤ u v is verified.

PROBLEM

Suppose that M 22 is the inner product space of all 2 × 2 matrices with inner product 〈 A, B〉 = tr( AT B) = a11b11 + a21b21 + a12b12 + a22b22 , where A = ⎡⎣aij ⎤⎦

2× 2

and B = ⎡⎣bij ⎤⎦

2× 2

⎡1 −2 ⎤ ⎡ −4 6 ⎤ are vectors in M 22. For A = ⎢ and B = ⎢ ⎥ ⎥ in M 2 ,2, verify that ⎣5 3 ⎦ ⎣ 2 5⎦ 〈 A, B〉 ≤ A B . Inner product spaces

103

First calculate 〈 A, B〉 :

SOLUTION

⎛ ⎡1 −2 ⎤ T ⎡ −4 6 ⎤⎞ ⎛ ⎡ 1 5 ⎤ ⎡ −4 6 ⎤⎞ = tr( A B) = tr ⎜ ⎢ ⎟ = tr ⎜ ⎢ −2 3 ⎥ ⎢ 2 5 ⎥⎟ = ⎥ ⎢ ⎥ ⎝⎣ ⎦⎣ ⎦⎠ ⎝ ⎣5 3 ⎦ ⎣ 2 5 ⎦ ⎠ T

(1)(−4 ) + (5)(2) + (−2)(6) + (3)(5) = 9 = 9

Then calculate A B : = 〈 A, A〉 ⋅ 〈 B, B〉 = 12 + 52 + (−2)2 + 32 (−4 )2 + 22 + 62 + 52 = 39 81 = 9 39 ≈ 56.2

Because 9 ≤ 9 39 ≈ 56.2, 〈 A, B〉 ≤ A B is verified.

Suppose u and v are vectors in a real inner product space (V , 〈 , 〉), and k ∈ R, then the norm has the following properties: (a) u ≥ 0 (b) u = 0 if and only if u = 0 (c) ku = k u (d) u + v ≤ u + v You likely recognize that property (d) is the triangle inequality that was discussed in Chapter 6. The name derives from the geometric concept that the sum of two sides of a triangle must be greater than the third side. In the current context, this notion is abstracted to a wide variety of entities, not just geometric vectors. EXERCISE

8·3 For 1 and 2, let u = (5, 5) and v = (3, −2).

1. Verify that −4u = 4 u . 2. Verify that u + v ≤ u + v . ⎡ 1 −2 ⎤ ⎡3 3 ⎤ and B = ⎢ For 3–5, let A = ⎢ ⎥ ⎥. ⎣5 3 ⎦ ⎣3 3 ⎦

3. Calculate 〈 A, B 〉 . 4. Calculate 〈 A, A〉 . 5. Calculate 〈B , B 〉 .

Orthogonality in inner product spaces Suppose u and v are vectors in a real inner product space (V , 〈 , 〉), then u and v are orthogonal, denoted u ⊥ v , if and only if 〈u , v 〉 = 0.

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PROBLEM

Let u = 4 − 7 x − x 2, v = 6 x 2 , and r = −2x in the vector space P2 with inner product 〈p , q 〉 = a0b0 + a1b1 + a2b2 , when p = a0 + a1 x + a2 x 2 and q = b0 + b1 x + b2 x 2 are vectors in P2. a. Are u and v orthogonal? Justify your answer. b. Are u and r orthogonal? Justify your answer. c. Are v and r orthogonal? Justify your answer.

SOLUTION

a. Are u and v orthogonal? Justify your answer. 〈u , v 〉 = 〈4 − 7 x − x 2 , 0 + 0 x + 6 x 2 〉 = 4 ⋅ 0 + (−7 ) ⋅ 0 + (−1) ⋅ 6 = −6

The vectors u and v in P2 are not orthogonal because 〈u , v 〉 = −6 ≠ 0 . b. Are u and r orthogonal? Justify your answer. 〈u , r 〉 = 〈4 − 7 x − x 2 , 0 − 2 x + 0 x 2 〉 = 4 ⋅ 0 + (−7 ) ⋅ (−2)(+(−1) ⋅ 0 = 14

The vectors u and r in P2 are not orthogonal because 〈u , r 〉 = 14 ≠ 0. c. Are v and r orthogonal? Justify your answer. 〈 v , r 〉 = 〈0 + 0 x + 6 x 2 , 0 − 2 x + 0 x 2 〉 = 0 ⋅ 0 + 0 ⋅ (−2) + 6 ⋅ 0 = 0

The vectors v and r in P2 are orthogonal because 〈 v , r 〉 = 0. PROBLEM

Suppose that M 33 is the inner product space of all 3 × 3 matrices with inner product ⎡0 −1 0 ⎤ 〈 A, B〉 = tr( BT A) = tr( AT B), where A and B are vectors in M 33. For A = ⎢1 0 0 ⎥ ⎢ ⎥ ⎢⎣0 0 1 ⎥⎦ ⎡ −1 1 0 ⎤ ⎡5 0 0 ⎤ and B = ⎢ 0 −1 0 ⎥ , and C = ⎢0 3 0 ⎥ . ⎥ ⎢ ⎥ ⎢ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣0 0 −5 ⎥⎦ a. Are A and B orthogonal? Justify your answer. b. Are A and C orthogonal? Justify your answer. c. Are B and C orthogonal? Justify your answer.

SOLUTION

a. Are A and B orthogonal? Justify your answer. ⎛ ⎡0 −1 0 ⎤ T ⎡ −1 1 0 ⎤⎞ ⎛ ⎡0 −1 0 ⎤⎞ ⎟ ⎜⎢ ⎥ ⎢ ⎥ T 〈 A, B〉 = tr( A B) = tr ⎜ ⎢1 0 0 ⎥ ⎢ 0 −1 0 ⎥⎟ = tr ⎜ ⎢⎢1 −1 0 ⎥⎥⎟ = 0 − 1 + 1 = 0 ⎜ ⎟ ⎜⎝ ⎢0 0 1 ⎥⎟⎠ ⎜⎝ ⎢⎣0 0 1 ⎥⎦ ⎢⎣ 0 0 1 ⎦⎥⎟⎠ ⎣ ⎦

The vectors A and B in M 33 are orthogonal because 〈 A, B〉 = 0 . b. Are A and C orthogonal? Justify your answer. ⎛ ⎡0 −1 0 ⎤ T ⎡5 0 0 ⎤⎞ ⎛ ⎡ 0 3 0 ⎤⎞ ⎟ ⎜⎢ ⎢ ⎥ ⎥ ⎥ ⎢ T 〈 A, C 〉 = tr( A C ) = tr ⎜ ⎢1 0 0 ⎥ ⎢0 3 0 ⎥⎟ = tr ⎜ ⎢ −5 0 0 ⎥⎟ = 0 + 0 − 5 = −5 ⎜ ⎟ ⎜⎝ ⎢ 0 0 −5 ⎥⎟⎠ ⎜ ⎢⎣0 0 1 ⎥⎦ ⎢⎣0 0 −5 ⎥⎦⎟ ⎣ ⎦ ⎠ ⎝

The vectors A and C in M 33 are not orthogonal because 〈 A, C 〉 = −5 ≠ 0. Inner product spaces

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c. Are B and C orthogonal? Justify your answer. ⎛ ⎡ −1 1 0 ⎤ T ⎡5 0 0 ⎤⎞ ⎛ ⎡ −5 0 0 ⎤⎞ ⎟ ⎜⎢ T ⎥ ⎥ ⎢ 〈 B, C 〉 = tr( B C ) = tr ⎜ 0 −1 0 = tr ⎜ ⎢⎢ 5 −3 0 ⎥⎥⎟ = ⎟ ⎜ ⎢ ⎥ ⎢0 3 0 ⎥⎟ ⎜⎝ ⎢ 0 0 −5 ⎥⎟⎠ ⎜⎝ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣0 0 −5 ⎥⎦⎟⎠ ⎣ ⎦ −5 − 3 − 5 = −13

The vectors B and C in M 33 are not orthogonal because 〈 B, C 〉 = −13 ≠ 0.

Caution: Orthogonality is tied to an inner product defined on a vector space, meaning that two vectors in a vector space might be orthogonal with respect to one inner product, but not to a differently defined inner product. PROBLEM

If x = ( x1 , x 2 ,#, xn ) and u = (a1 , a2 ,#, an ) are vectors in Rn, then x ⊥ u if a1 x1 + a2 x 2 + $ + an xn = 0. In other words, x ⊥ u if x = ( x1 , x 2 ,#, xn ) satisfies the homogeneous equation whose coefficients are the components of u. Let u = (3, 2, −1) and w = (2, −4 , −2). a. Find x = ( x1 , x 2 , x 3 ) so that x ⊥ v and x ⊥ w . b. Normalize the solution.

SOLUTION

a. Find x = ( x1 , x 2 , x 3 ) so that x ⊥ v and x ⊥ w . For x ⊥ v and x ⊥ w , the vector x = ( x1 , x 2 , x 3 ) must satisfy the following system: 3 x1 + 2 x 2 − x 3 = 0 ⎡ 3 2 −1 0 ⎤ . The augmented matrix for this system is ⎢ ⎥ 2 x1 − 4 x 2 − 2 x 3 = 0 ⎣ 2 −4 −2 0 ⎦ ⎡1 0 −0.5 0 ⎤ with reduced row-echelon form ⎢ ⎥ , which yields x1 = 0.5 x 3 ⎣0 1 0.25 0 ⎦ and x 2 = −0.25 x 3 , with x 3 as a free variable. Letting x 3 = 4 gives the non-unique solution x = (2, −1, 4 ). b. Normalize the solution. 1 4 ⎞ 1 1 1 ⎛ 2 ,− , ⋅x = ⋅ (2, −1, 4 ) = ⋅ (2, −1, 4 ) = ⎜ ⎟ ⎝ 21 x 21 21 ⎠ 21 〈(2, −1, 4 ),(2, −1, 4 )〉

If u ⊥ v in an inner product space (V , 〈 , 〉), then you have the following generalization of the Pythagorean theorem: u + v 2 = u 2 + v 2 . PROBLEM

Suppose that M 22 is the inner product space of all 2 × 2 matrices with inner product 〈 A, B〉 = tr( AT B) = a11b11 + a21b21 + a12b12 + a22b22 , where A = ⎡⎣aij ⎤⎦ and

⎡0 2 ⎤ ⎡1 0 ⎤ and B = ⎢ B = ⎡⎣bij ⎤⎦ are vectors in M 22. Let A = ⎢ ⎥ ⎥ in M 2 ,2. 2× 2 ⎣0 0 ⎦ ⎣1 1 ⎦ a. Verify that A ⊥ B. b. Verify that A + B 2 = A 2 + B 2 . SOLUTION

a. Verify that A ⊥ B. ⎡0 2 ⎤ ⎡1 0 ⎤ 〈 A, B〉 = ⎢ ⎥,⎢ ⎥ = 0 ⋅1 + 0 ⋅1 + 2 ⋅ 0 + 0 ⋅1 = 0, so A ⊥ B. ⎣0 0 ⎦ ⎣1 1 ⎦

b. Verify that A + B 2 = A 2 + B 2 .

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2× 2

First calculate A + B 2 : ⎡0 2 ⎤ ⎡1 0 ⎤ = ⎢ ⎥+ ⎢ ⎥ ⎣0 0 ⎦ ⎣1 1 ⎦

2

⎡1 2 ⎤ ⎡1 2 ⎤ 2 2 2 2 = ⎢ ⎥ = 1 +1 + 2 +1 = 7 ⎥,⎢ ⎣1 1 ⎦ ⎣1 1 ⎦

2

2

Then calculate A + B : ⎡0 2 ⎤ = ⎢ ⎥ ⎣0 0 ⎦

2

⎡1 0 ⎤ + ⎢ ⎥ ⎣1 1 ⎦ 2

2

⎡1 0 ⎤ ⎡1 0 ⎤ ⎡0 2 ⎤ ⎡0 2 ⎤ 2 2 2 2 = ⎢ + ⎢ ,⎢ ⎥ ⎥,⎢ ⎥ = 2 +1 +1 +1 = 7 ⎥ ⎣1 1 ⎦ ⎣1 1 ⎦ ⎣0 0 ⎦ ⎣0 0 ⎦

2

2

2

2

2

Because A + B = A + B = 7, A + B = A + B is verified for A and B.

A set of vectors P in an inner product space (V , 〈 , 〉) is an orthogonal set of vectors if and only if 〈u , v 〉 = 0 for every pair of distinct vectors u and v in V. An orthogonal set of vectors P is orthonormal if and only if u = 1 for every vector u ∈P . Additionally, if P is an orthogonal set of nonzero vectors, then P is a linearly independent set. PROBLEM

Which of the following sets of vectors in R 2 is orthogonal? If the set is orthogonal, state and justify whether it also is orthonormal? a. {(0, 0),(1, 2)} b. {(1, 0),(0,1)} c. {(1, 0),(0,1),(1,1)} d. {(−10, 0),(0, 50)}

SOLUTION

a. {(0, 0),(1, 2)} This set is orthogonal because 〈(0, 0),(1, 2)〉 = 0. It is not orthonormal because (0, 0) = 0 ≠ 1. b. {(1, 0),(0,1)} This set is orthogonal because 〈(1, 0),(0,1)〉 = 0, and it also is orthonormal because (1, 0) = (0,1) = 1. c. {(1, 0),(0,1),(1,1)} This set is not orthogonal because 〈(1, 0),(1,1)〉 = 1 ≠ 0 d. {(−10, 0),(0, 50)} This set is orthogonal because 〈(−10, 0),(0, 50)〉 = 0. It is not orthonormal because (−10, 0) = 10 ≠ 1. EXERCISE

8·4 For 1 and 2, prove as indicated.

1. Prove that if u ⊥ v , then ku ⊥ v for every real number k. 2. Prove or disprove: If u ⊥ v and v ⊥ w then u ⊥ w . For 3–6, determine if the given set is an orthogonal set.

3. S = {(1, 0),( 0, 2),(11 , )} with respect to the Euclidean inner product of R 2.

, , 0),(1, −1, 0),( −11 , , 0)} with respect to the Euclidean inner product of R3. 4. S = {(11 Inner product spaces

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5. S = {(11 , , 0),( −11 , , 0),( 0, 0,1)} with respect to the Euclidean inner product of R 3 .

⎧ ⎡ 1 0 ⎤ ⎡ 0 1⎤ ⎡ 0 0 ⎤ ⎡ 0 0 ⎤ ⎫ 6. S = ⎨ ⎢ ⎥ ⎬ , with respect to the trace inner product. ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎪⎩ ⎣ 0 0 ⎦ ⎣ 0 0 ⎦ ⎣ 1 1 ⎦ ⎣ 1 −1⎦ ⎪⎭ For 7 and 8, determine if the given orthogonal set is also an orthonormal set.

7. S = {(1, 0),( 0,1)} with respect to the Euclidean inner product of R 2.

⎧ ⎡2 0 ⎤ ⎡ 0 1⎤ ⎡ 0 0 ⎤ ⎡ 0 0 ⎤ ⎫ 8. S = ⎨ ⎢ ⎥ ⎬ , with respect to the trace inner product. ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎩⎪ ⎣ 0 0 ⎦ ⎣ 0 0 ⎦ ⎣ 2 2 ⎦ ⎣ 1 −1⎦ ⎭⎪ For 9 and 10, normalize the given orthogonal set.

9. S = {(2, 0, 0),( 0, 2, 0),( 0, 0, 3)} with respect to the Euclidean inner product of R3.

⎧ ⎡2 0 ⎤ ⎡ 0 1⎤ ⎡ 0 0 ⎤ ⎡ 0 0 ⎤ ⎫ 10. S = ⎨ ⎢ ⎥ ⎬ , with respect to the trace inner product. ⎥, ⎢ ⎥, ⎢ ⎥, ⎢ ⎪⎩ ⎣ 0 0 ⎦ ⎣ 0 0 ⎦ ⎣ 2 2 ⎦ ⎣ 1 −1⎦ ⎪⎭

Gram-Schmidt procedure The Gram-Schmidt procedure is an orthogonalization process that produces an orthonormal basis from an arbitrary basis. Suppose { v 1 , v 2 ,$ , v n } is a basis for an inner product space (V , 〈 , 〉). You can construct an orthogonal basis {w1 , w 2 ,# , w n } of V as follows: w1 = v 1 ; w2 = v2 −

〈 v 2 , w1 〉 w; 〈w1 , w1 〉 1

〈 v 3 , w1 〉 〈v , w 〉 w − 3 2 w ; 〈w1 , w1 〉 1 〈w 2 , w 2 〉 2 ……………………………………………………….………;

w3 = v3 −

wn = vn −

〈 v n , w1 〉 〈v , w 〉 〈 v n , w n−1 〉 w w − n 2 w −$− 〈 w n−1 , w n−1 〉 n−1 〈w1 , w1 〉 1 〈w 2 , w 2 〉 2

When the vectors w1 , w 2 ,# , w n are normalized, the resulting set of vectors forms an orthonormal basis for (V , 〈 , 〉). PROBLEM

Let T = {(1,1,1),(1,1, 0),(−1, 0, −1)} be a basis for R 3. a. Apply the Gram-Schmidt process to transform T into an orthogonal basis. b. Normalize the solution to yield an orthonormal basis for R 3.

SOLUTION

a. Apply the Gram-Schmidt process to transform T into an orthogonal basis. Let v 1 = (1,1,1), v 2 = (1,1, 0), and v 3 = (−1, 0, −1). Then w 1 = v 1 = (1,1,1) ;

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w2 = v2 −

〈 v 2 , w1 〉 2 〈(1,1, 0),(1,1,1)〉 w 1 = (1,1, 0) − (1,1,1) = (1,1, 0) − (1,1,1) = 3 〈w1 , w1 〉 〈(1,1,1),(1,1,1)〉

⎛ 1 1 2⎞ 1 ⎜⎝ , , − ⎟⎠ = (1,1, −2) ; 3 3 3 3 w3 = v3 −

〈v , w 〉 〈 v 3 , w1 〉 〈(−1, 0, −1),(1,1,1)〉 w 1 − 3 2 w 2 = (−1, 0, −1) − (1,1,1) 〈w 2 , w 2 〉 〈w1 , w1 〉 〈(1,1,1),(1,1,1)〉

1 〈(−1, 0, −1),(1,1, −2)〉 1 , 1 , −2 −3 3 3 3 1 〈(1,1, −2),(1,1, −2)〉〉 9 2 1 ⎛ 1 1 ⎞ = (−1, 0, −1) + (1,1,1) − (1,1, −2) = ⎜ − , , 0⎟ ⎝ 2 2 ⎠ 3 6

(

)

⎛ 1 1 2⎞ ⎛ 1 1 ⎞ Thus, w 1 = (1,1,1), w 2 = ⎜ , , − ⎟ , and w 3 = ⎜ − , , 0⎟ form an orthogonal ⎝ 3 3 3⎠ ⎝ 2 2 ⎠ basis for R 3.

b. Normalize the solution to yield an orthonormal basis for R 3. 6 1 , and w 3 = . The norms of these vectors are w 1 = 3 , w 2 = 3 2 ⎛ 1 1 ⎞ 2 ⎞ ⎛ 1 1 1 ⎞ ⎛ 1 1 , , , 0⎟ form an Therefore, ⎜ , , , ,− ⎟ , and ⎜ − ⎝ ⎝ 3 3 3 ⎟⎠ ⎜⎝ 6 6 ⎠ 2 2 ⎠ 6

orthonormal basis for R 3 . EXERCISE

8·5 For 1–3, fill in the blank with the best response.

1. The Gram-Schmidt procedure is an ____________ process that produces an ____________ basis from an arbitrary basis. 2. If S = { v1 , v 2 } is a basis for R 2 , then in the Gram-Schmidt process, w1 = ____________ and w 2 = ____________. 3. The final step in the Gram-Schmidt process is to ____________ the w vectors. 4. Use the Gram-Schmidt process to compute an orthonormal basis for R 2 from the basis S = {(2, 2),(3, 0)} . Hint: It is okay to clear fractions when forming the w vectors. 5. Use the Gram-Schmidt process to compute an orthonormal basis for R 3 from the basis S = {(0,1,1),(0,0,1),(1,1,1)}.

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·9 ·

Linear transformations In this chapter, you will learn: ◆ ◆ ◆ ◆ ◆ ◆

Deﬁnition and terminology for linear transformations Kernel and image of a linear transformation Matrix representations of linear transformations Change of basis Algebra of linear transformations Linear operators on R 2 and R 3

This chapter provides a discussion of linear transformations. Linear transformations play a central role in linear algebra and have important applications in science, engineering, technology, and other branches of mathematics.

Deﬁnition and terminology for linear transformations A linear transformation T from a vector space V to a vector space W, denoted T : V → W , is a function that assigns to each vector v ∈V a vector T ( v ) ∈W such that the following two properties are satisfied: (a) T (u + v ) = T (u ) + T ( v ) (b) T (kv ) = kT ( v ) for all u , v ∈V and all scalars k. Clearly, T : V → W is linear if it preserves vector addition and scalar multiplication, the two fundamental operations of vector spaces. In the case that W = V, T is called a linear operator on V. Note: Linear transformations also are called linear mappings. PROBLEM

If T : V → W is a linear transformation, prove that T(0 ) = 0.

SOLUTION

T (0 ⋅ v ) = 0 ⋅ T ( v ) By property of linear transformations. Therefore, T(0 ) = 0 By scalar multiplication of a vector by 0. This problem result means that the image of the zero vector (0 ∈V ) under a linear transformation is the zero vector (0 ∈W ). PROBLEM

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Let T : V → W be a linear transformation. Prove that T (au + bv ) = aT (u ) + bT ( v ) for all u , v ∈V and all scalars a and b.

SOLUTION

T (au + bv )

= T (au ) + T (bv )

by property (a) of linear transformations

= aT (u ) + bT ( v )

by property (b) of linear transformations

Remark: Some texts use the condition T (a u + b v ) = aT (u ) + bT ( v ) as the definition for a linear transformation. In general, if T : V → W is a linear transformation, T (a1 v 1 + a2 v 2 + $ + an v n ) = T(a1v1) + T (a2 v 2 ) + $ + T (an v n ) = a1T ( v 1 ) + a2T ( v 2 ) + $ + anT ( v n ), for all v i ∈V and all scalars ai. PROBLEM

SOLUTION

Show that the function F : R 3 → R defined by F ( v ) = F ((v1 , v 2 , v3 )) = 3v1 − 2v 2 + 5v3 is a linear transformation. (a) Let u = (u1 , u2 , u3 ) and v = (v1 , v 2 , v3 ) ∈V , show F(u + v ) = F (u ) + F ( v ) . F(u + v ) = F ((u1 , u2 , u3 ) + (v1 , v 2 , v3 )) = F ((u1 + v1 , u2 + v 2 , u3 + v3 )) = 3(u1 + v1) − 2(u2 + v 2 ) + 5(u3 + v3 ) = 3u1 + 3v1 − 2u2 − 2v 2 + 5u3 + 5v3 = (3u1 − 2u2 + 5u3 ) + (3v1 − 2v 2 + 5v3 ) =F (u ) + F ( v )

Thus, F(u + v ) = F (u ) + F ( v ). (b) Let v = (v1 , v 2 , v3 ) ∈V and k ∈R , show F (kv ) = kF ( v ). F (kv ) = F (k(v1 , v 2 , v3 )) = F ((kv1 , kv 2 , kv3 )) = 3(kv1 ) − 2(kv 2 ) + 5(kv3 ) = k(3v1 − 2v 2 + 5v3 ) = kF( v )

Thus, F (kv ) = kF ( v ). Because (a) F(u + v ) = F (u ) + F ( v ) and (b) F (kv ) = kF ( v ), F is a linear transformation from R 3 to R. PROBLEM

SOLUTION

Show that the function G : R 3 → R 2 defined by G( v ) = G((v1 , v 2 , v3 )) = (v1 + 2, v 2 + 5) is not a linear transformation. G(0 ) = G((0, 0, 0)) = (0 + 2, 0 + 5) = (2, 5) ≠ 0 Because under G the image of 0 is not 0, G is not a linear transformation.

EXERCISE

9·1 For 1–3, fill in the blank with the best response.

1. A linear transformation is a function that preserves vector ____________ and ____________ multiplication, the two fundamental operations of vector spaces. 2. The image of 0 under a linear transformation is ____________. 3. If T : V → W is a linear transformation, then T ( au + bv ) = ____________ for all u, v ∈V and all scalars a and b. For 4–6, let T : V → W be a linear transformation.

4. Prove that T ( − v ) = −T ( v ) for all v ∈V . 5. Prove that T (u − v ) = T (u) − T ( v ) for all u, v ∈V . 6. Prove that T ( au − bv ) = aT (u) − bT ( v ) for all u, v ∈V and all scalars a and b.

Linear transformations

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For 7–11, show that the given function is a linear transformation.

7. T : R 2 → R 2 defined by T (( x , y )) = (3 x − y , x ) 8. T : R 3 → R 3 defined by T (( x , y , z )) = ( y , 0, z ) 9. T : R → R 2 defined by T ( x ) = (5 x , − x ) 10. T : R 2 → R 3 defined by T (( x , y )) = ( x + y , y , x ) 11. T : R 3 → R defined by T (( x , y , z )) = x − 2 y + 3 z For 12–15, show that the given function is not a linear transformation.

12. G : R → R defined by G( x ) = 3 x + 2 13. G : R 2 → R 3 defined by G( v ) = G((v1 , v2 )) = (v1 , v1v2 , v2 ) 14. G : R 2 → R 2 defined by G( v ) = G((v1 , v2 )) = (v12 , v22 ) 15. G : R 3 → R 2 defined by G( v ) = G((v1 , v2 , v3 )) = (v1 − 5, v2 + v3 )

Kernel and image of a linear transformation Given a linear transformation T : V → W , the kernel of T , denoted Ker T , is the set of all vectors v ∈V that map into 0 ∈W . That is, Ker T = { v ∈V | T ( v ) = 0} . PROBLEM

SOLUTION

PROBLEM

SOLUTION

Given the linear transformation T : R 3 → R 3 defined by T( v ) = T ((v1 , v 2 , v3 )) = (v1 , 0, v3 ). Describe the kernel of T. Suppose T ( v ) = T ((v1 , v 2 , v3 )) = (v1 , 0, v3 ). Then, v ∈Ker T if and only if (v1 , 0, v3 ) = (0, 0, 0); that is, when v1 = 0 and v3 = 0 with no restriction on v 2. Thus, letting a ∈R be arbitrary, Ker T = { v ∈V | v = (0, a, 0) ∈R 3 } . (Note: In an xyz coordinate system, the Ker T under this transformation is the y-axis.) Given a linear transformation T : V → W , the image (or range) of T , denoted Im T , is the set of all image vectors w ∈W that correspond to at least one v ∈V . That is Im T = {w ∈W | T ( v ) = w for at least one v ∈V } . Given the linear transformation T : R 3 → R 3 defined by T( v ) = T ((v1 , v 2 , v3 )) = (v1 , 0, v3 ). Describe the image of T. Suppose w ∈R 3 is in Im T , then for some v ∈R 3 , T ( v ) = (v1 , 0, v3 ) = (w1 , w 2 , w3 ); this occurs when w1 = v1, w 2 = 0, and w3 = v3. Thus, letting a, b ∈R be arbitrary, Im T = {w ∈W | w = (a, 0, b) ∈R 3 } . (Note: In an xyz coordinate system, the image of T under this transformation is the xz-plane.)

From your work in Chapter 7 with vector spaces, you might have conjectured that in the above problems Ker T = { v ∈V | v = (0, a, 0) ∈R 3 } and Im T = {w ∈W| w = (a, 0, b) ∈R 3 } are subspaces of R 3. In general, you have the following: If T : V → W is a linear transformation, then Ker T is a subspace of V and Im T is a subspace of W. PROBLEM SOLUTION

Let T : V → W be a linear transformation. Prove that Ker T is a subspace of V. 1. Show 0 ∈Ker T . Because T(0 ) = 0, 0 ∈Ker T .

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2. For u , v ∈Ker T , show u + v ∈Ker T . For u , v ∈Ker T , T (u + v ) = T (u ) + T ( v ) = 0 + 0 = 0. Thus, u + v ∈Ker T . 3. For v ∈Ker T and any scalar k, show kv ∈Ker T . For v ∈Ker T , T (kv ) = kT ( v ) = k 0 = 0. Thus, kv ∈Ker T . Therefore, Ker T is a subspace of V.

If V is finite dimensional, then both Ker T and Im T are finite dimensional and dim(V ) = dim (Ker T ) + dim (Im T ). Additionally, in Chapter 7, you learned that the row and column spaces of any matrix A have the same dimension and this dimension is the rank of A, denoted rank( A), and that the dimension of the null space of A is the nullity of A, denoted nullity( A). Extending these definitions to linear transformations, you have the following: If T : V → W is a linear transformation, then the dimension of Im T is the rank of T, denoted rank(T ), and the dimension of Ker T is the nullity of T, denoted nullity(T ). Using this terminology, you have dim (V ) = dim (Ker T ) + dim (Im T ) = nullity(T ) + rank(T ). A linear transformation T for which Ker T = {0} is nonsingular; otherwise T is singular. Nonsingular linear transformations map n-dimensional vector spaces onto n-dimensional vector spaces. Obviously, the kernel of a nonsingular linear transformation has dimension 0. Note: Recall that T : V → W is onto if for every w in W there exists a v in V such that T( v ) = w .

EXERCISE

9·2 For 1–5, fill in the blank with the best response.

1. Given a linear transformation T : V → W , the kernel of T is the set of all vectors v ∈V such that T ( v ) is the ____________ vector in W. 2. Given a linear transformation T : V → W , the image of T is the set of all image vectors w ∈W such that ____________ equals w for at least one v ∈V . 3. If T : V → W is a linear transformation, then Ker T is a subspace of ____________ and Im T is a subspace of ____________. 4. If T : V → W is a linear transformation and V is finite dimensional, dim (KerT ) + dim (Im T ) = ____________ ( dim(V ) or dim(W ) ?). 5. If T : V → W is a linear transformation, then the dimension of Im T is the ____________ (rank or nullity?) of T, and the dimension of the Ker T is the ____________ (rank or nullity?) of T. For 6–11, let T : R 2 → R 2 be the linear transformation defined by T ( v ) = T ((v1 , v2 )) = ( −3v1 + 6v2 , v1 − 2v2 )

6. Is the vector (12, 6) in the kernel of T ? Explain. 7. Is the vector (5, 3) in the kernel of T ? Explain. 8. Is 0 ∈R 2 in the kernel of T ? Explain. 9. Is the vector (14 , 6) in the image of T ? Explain. 10. Is the vector ( −15, 5) in the image of T ? Explain. 11. Is 0 ∈R 2 in the image of T ? Explain. Linear transformations

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For 12–14, use the information given to find the nullity of T.

12. T : R 4 → R 3 has rank 1. 13. T : M22 → M22 has rank 4. 14. The image of T : R 3 → P2 is P2. 15. Let T : V → W be a linear transformation. Prove that Im T is a subspace of W. For 16–20, determine whether the linear transformation T is singular or nonsingular.

16. T : R 4 → R 3 has rank 1. 17. T : M22 → M22 has rank 4. 18. The image of T : R 3 → P2 is P2.

⎛ ⎡ x1 ⎤ ⎞ ⎡ x1 − 2 x 2 ⎤ 19. T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥. ⎝ ⎣ x 2 ⎦ ⎠ ⎣ x1 − x 2 ⎦ ⎡ a0 ⎤ 20. T : P2 → R defined by T ( a0 + a1x + a2 x ) = ⎢⎢ a1 ⎥⎥ . ⎢⎣ a2 ⎥⎦ 3

2

Matrix representations of linear transformations Using matrices to represent linear transformations is a clever and practical strategy. For a given linear transformation T from a vector space V to a vector space W, the process involves finding a matrix representation relative to ordered bases of V and W, where an ordered basis is one whose vectors are placed in a specific order. Suppose T : V → W is a linear transformation and BV = { v 1 , v 2 ,$, v n } is an ordered basis for vector space V and BW = {w1 , w 2 ,$ , w m } is an ordered basis for vector space W. For each of the basis vectors v j of V, you can represent T ( v j ) as a unique linear combination of the basis vectors w i of W. In other words, there exists unique numbers aij , 1 ≤ i ≤ m , 1 ≤ j ≤ n such that T ( v 1 ) = a11w1 + a21w 2 + + am1w m T ( v 2 ) = a12 w1 + a22 w 2 + + am 2 w m T ( v n ) = a1n w1 + a2n w 2 + + amn w m

Arranging the coefficients as an m × n matrix A = ⎡⎣aij ⎤⎦, where the columns of A are the coordinates of the vectors T ( v j ) relative to the basis {w1 , w 2 ,$ , w m }, then ⎡a11 a12 … a1n ⎤ ⎢a a22 … a2n ⎥ A = ⎢ 21 ⎥⎥ ⎢ ⎣⎢am1 am 2 … amn ⎥⎦

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is the matrix for the transformation T relative to the bases BV and Bw and is denoted [T ]BW , BV . In the case that V = W and B is a common basis, the matrix for T with respect to B is denoted [T ]B. What’s more, you can use the simple notation [T ] when the basis or bases are understood. You will find that your calculator is an indispensable tool in determining [T ]BW , BV . One important note is that you must enter vectors as column vectors, not row vectors. Here is the process. 1. Apply the transformation T to each of the basis vectors v j of V. 2. Construct an augmented matrix that has the form

[[w i ]| T ( v j )]

where the m columns of [w i ] are the basis vectors w i of W and the n columns of T ( v j ) are the vectors obtained by applying T to the n basis vectors v j of V. 3. Convert [[w i ]| T ( v j )] to reduced row-echelon form to transform [w i ] to the m × m identity

matrix and T ( v j ) to [T ]BW , BV . The columns of [T ]BW , BV , denoted [T ( v j )]BW , are the coordinate vectors for T ( v j ) in W; that is, these are the coefficients for T ( v j ) as a linear combination of the basis vectors w i of W. Note: Hereafter, for notational convenience and clarity, vectors will be written as column vectors. PROBLEM

SOLUTION

⎛ ⎡ x1 ⎤ ⎞ Suppose T : R 2 → R 3 is a linear transformation defined by T( x ) = T ⎜ ⎢ ⎥⎟ = ⎝ ⎣ x 2 ⎦⎠ ⎡ x2 ⎤ ⎢ 2 x − x ⎥ . Find the matrix for the transformation T relative to the bases ⎢ 2 1⎥ ⎢⎣ 4 x1 ⎥⎦ ⎧ ⎡ −1⎤ ⎡ 1 ⎤ ⎡ 0 ⎤ ⎫ ⎧⎡2 ⎤ ⎡3 ⎤ ⎫ ⎪ ⎪ BV = { v 1 , v 2 } = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ for R 2 and BW = {w 1 , w 2 , w 3 } = ⎨ ⎢⎢ 2 ⎥⎥ , ⎢⎢ 0 ⎥⎥ , ⎢⎢1 ⎥⎥ ⎬ for R 3. 1 2 ⎩⎪ ⎣ ⎦ ⎣ ⎦ ⎭⎪ ⎪ ⎢ 2 ⎥ ⎢ −1⎥ ⎢ 2 ⎥ ⎪ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭

1. Apply the transformation T to each of the basis vectors v 1 and v 2 of V. ⎡2⎤ ⎡1 ⎤ ⎛ ⎡ 2 ⎤⎞ ⎢ ⎥ ⎛ ⎡ 3 ⎤⎞ ⎢ ⎥ T( v 1 ) = T ⎜ ⎢ ⎥⎟ = ⎢0 ⎥ , T( v 2 ) = T ⎜ ⎢ ⎥⎟ = ⎢ 1 ⎥ ⎝ ⎣1 ⎦⎠ ⎝ ⎣ 2 ⎦⎠ ⎢⎣12 ⎥⎦ ⎢⎣8 ⎥⎦

2. Construct an augmented matrix that has the form [[w i ]| T ( v j )]. ⎡ −1 1 0 1 2 ⎤ ⎢ ⎥ ⎢2 0 1 0 1⎥ ⎢ 2 −1 2 8 12 ⎥ ⎣ ⎦

3. Convert [[w i ] T ( v j )] to reduced row-echelon form to transform [w i ] to the m × m identity matrix and T ( v i ) to [T ]BW ,BV . ⎡1 0 0 −3 −4 ⎤ ⎢ ⎥ ⎢0 1 0 −2 −2 ⎥ ⎢0 0 1 6 9 ⎥ ⎣ ⎦

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Then, [T ]BW , BV

⎡ −3 −4 ⎤ = ⎢⎢ −2 −2 ⎥⎥ . ⎢⎣ 6 9 ⎥⎦

Thus, [T ( v 1 )]BW

[T ( v 2 )]BW

⎡ −3 ⎤ = ⎢⎢ −2 ⎥⎥ , so T( v 1 ) = −3w 1 − 2 w 2 + 6 w 3 and ⎢⎣ 6 ⎥⎦

⎡ −4 ⎤ = ⎢⎢ −2 ⎥⎥ , so T( v 2 ) = −4 w 1 − 2 w 2 + 9 w 3 . ⎢⎣ 9 ⎥⎦

The matrix [T ]BW ,BV has the property that [T ]BW ,BV ⋅[x ]BV = [T ( x )]BW . Therefore, you can get the coordinate vector for T (x ) ∈W relative to the basis BW by doing the following: 1. Determine the coordinate vector for [x ]BV by expressing x as a linear combination of the

basis vectors of V.

2. Compute [T ]BW ,BV ⋅[x ]BV to obtain [T ( x )]BW .

[T ( x )]BW is the coordinate vector for T( x ). That is, these are the coefficients you would use to express T( x ) as a linear combination of the basis vectors of W. PROBLEM

SOLUTION

Suppose T : R 2 → R 3 is the linear transformation defined as in the previous ⎡ −3 −4 ⎤ problem that has transformation matrix [T ]BW ,BV = ⎢⎢ −2 −2 ⎥⎥ . Verify that ⎣⎢ 6 9 ⎦⎥ ⎡ −1⎤ [T ]BW ,BV ⋅[x ]BV = [T ( x )]BW for x = ⎢ ⎥ . ⎣5⎦ First, find T( x ) by computing [T ]BW ,BV ⋅[x ]BV = [T ( x )]BW : 1. Determine the coordinate vector for [x ]BV by expressing x as a linear combination of the basis vectors of V. ⎡ −1⎤ ⎡2⎤ ⎡3⎤ Let x = ⎢ ⎥ = a ⎢ ⎥ + b ⎢ ⎥ . Equating corresponding entries yields the system ⎣5⎦ ⎣1 ⎦ ⎣ 2 ⎦ ⎡ 2 3 −1⎤ 2a + 3b = −1 of equations that has augmented matrix : ⎢ ⎥ , which is a + 2b = 5 ⎣1 2 5 ⎦ ⎡1 0 −17 ⎤ ⎢0 1 11 ⎥ in reduced row-echelon form. Thus, a = −17 and b = 11, so ⎣ ⎦ − 17 ⎡ ⎤ [x ]BV = ⎢ ⎥. 11 ⎣ ⎦

2. Compute [T ]BW ,BV ⋅[x ]BV to obtain [T ( x )]BW . ⎡ −3 −4 ⎤ ⎡7⎤ ⎡ −17 ⎤ ⎢ ⎥ 12 = [T ( x )]BW = ⎢⎢ −2 −2 ⎥⎥ ⎢ 11 ⎥⎦ ⎢ ⎥ ⎣ ⎢⎣ 6 9 ⎥⎦ ⎢⎣ −3 ⎥⎦

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So as a linear combination of the basis vectors of W, ⎡ −1⎤ ⎡ 1 ⎤ ⎡0 ⎤ ⎡ 5 ⎤ ⎢ ⎥ T( x ) = 7 w 1 + 12 w 2 − 3w 3 = 7 ⎢ 2 ⎥ + 12 ⎢⎢ 0 ⎥⎥ − 3 ⎢⎢1 ⎥⎥ = ⎢⎢ 11 ⎥⎥ ⎢⎣ 2 ⎥⎦ ⎢⎣ −1⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ −4 ⎥⎦ ⎡ x2 ⎤ ⎛ ⎡ x1 ⎤ ⎞ ⎢ Next, obtain T( x ) directly by using T( x ) = T ⎜ ⎢ ⎥⎟ = ⎢ 2 x 2 − x1 ⎥⎥ : ⎝ ⎣ x 2 ⎦⎠ ⎢⎣ 4 x1 ⎥⎦ ⎡5⎤ ⎛ ⎡ −1⎤⎞ ⎢ ⎥ T( x ) = T ⎜ ⎢ ⎥⎟ = ⎢ 11 ⎥ ⎝ ⎣ 5 ⎦⎠ ⎢⎣ −4 ⎥⎦

⎡ −1⎤ The results are the same, so [T ]BW ,BV ⋅[x ]BV = [T ( x )]BW is verified for x = ⎢ ⎥ . ⎣5⎦

As shown in the following problem, computations are simplified when standard bases for V and W are used. PROBLEM

⎛ ⎡ x1 ⎤ ⎞ ⎡ x1 ⎤ Given the linear transformation T : R → R defined by T( x ) = T ⎜ ⎢⎢ x 2 ⎥⎥⎟ = ⎢⎢ 0 ⎥⎥ . ⎜ ⎟ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ x ⎥ ⎣ 3⎦ ⎣ 3⎦ 3

3

a. Find the matrix representation for T relative to the standard basis ⎧ ⎡0 ⎤ ⎫ ⎡0 ⎤ ⎡1 ⎤ ⎪ ⎢ ⎥⎪ ⎥ ⎢ ⎥ ⎢ 3 ⎨e1 = ⎢0 ⎥ , e 2 = ⎢1 ⎥ , e 3 = ⎢0 ⎥ ⎬ of R . ⎪ ⎢⎣1 ⎥⎦ ⎪⎭ ⎢⎣0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎩

b. Verify that [T ]B ⋅[x ]B = [T ( x )]B for every vector x in R 3. SOLUTION

a. Find the matrix representation for T relative to the standard basis ⎧ ⎡0 ⎤ ⎫ ⎡0 ⎤ ⎡1 ⎤ ⎪ ⎢ ⎥⎪ ⎥ ⎢ ⎥ ⎢ 3 ⎨e1 = ⎢0 ⎥ , e 2 = ⎢1 ⎥ , e 3 = ⎢0 ⎥ ⎬ of R . ⎪ ⎢⎣1 ⎥⎦ ⎪⎭ ⎢⎣0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎩ 1. Apply the transformation T to each of the basis vectors e1, e 2, and e 3 . ⎡0 ⎤ ⎡1 ⎤ ⎡0 ⎤ ⎥ ⎥ ⎢ ⎢ T(e1 ) = ⎢0 ⎥ , T(e 2 ) = ⎢0 ⎥ , and T(e1 ) = ⎢⎢0 ⎥⎥ . ⎢⎣1 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣0 ⎥⎦

2. Construct an augmented matrix that has the form ⎡⎣[w i ] T ( v j )⎤⎦ . ⎡1 0 0 1 0 0 ⎤ ⎢ ⎥ ⎢0 1 0 0 0 0 ⎥ ⎢0 0 1 0 0 1 ⎥ ⎣ ⎦ ⎡1 0 0 ⎤ 3. By inspection, [T ]B = ⎢⎢0 0 0 ⎥⎥ . ⎢⎣0 0 1 ⎥⎦ Linear transformations

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b. Verify that [T ]B ⋅[x ]B = [T ( x )]B for every vector x in R 3 . First, compute T( x ) by using ( [T ]B ⋅[x ]B = [T ( x )]B . ⎡ x1 ⎤ Suppose x = ⎢⎢ x 2 ⎥⎥ is a vector in R 3, then by inspection x = x1e1 + x 2 e 2 + x 3 e 3, so ⎢⎣ x 3 ⎥⎦ ⎡ 1 0 0 ⎤ ⎡ x1 ⎤ ⎡ x1 ⎤ ⎡ x1 ⎤ ⎥ ⎢ [x ]B = ⎢ x 2 ⎥ and [T ]B ⋅[x ]B = ⎢⎢0 0 0 ⎥⎥ ⎢⎢ x 2 ⎥⎥ = ⎢⎢ 0 ⎥⎥ . ⎢⎣0 0 1 ⎥⎦ ⎢⎣ x 3 ⎥⎦ ⎣⎢ x 3 ⎥⎦ ⎢⎣ x 3 ⎥⎦ ⎡ x1 ⎤ Thus, T ( x ) = x1e1 + 0e 2 + x 3 e 3 = ⎢⎢ 0 ⎥⎥ . ⎢⎣ x 3 ⎥⎦ ⎛ ⎡ x1 ⎤ ⎞ ⎡ x1 ⎤ Next, compute T( x ) using T( x ) = T ⎜ ⎢⎢ x 2 ⎥⎥⎟ = ⎢⎢ 0 ⎥⎥ : ⎜ ⎟ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ x ⎥ ⎣ 3⎦ ⎣ 3⎦ ⎛ ⎡ x1 ⎤ ⎞ ⎡ x1 ⎤ T ( x ) = T ⎜ ⎢⎢ x 2 ⎥⎥⎟ = ⎢⎢ 0 ⎥⎥ ⎟ ⎜ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ x ⎥ ⎣ 3⎦ ⎣ 3⎦

The results are the same, so [T ]B ⋅[x ]B = [T ( x )]B is verified for every vector x in R 3.

EXERCISE

9·3 For 1–5, calculate the matrix representation for the given transformation relative to the given bases.

⎧ ⎡2 ⎤ ⎡ 3 ⎤ ⎫ ⎧ ⎡1⎤ ⎡ 1⎤ ⎫ ⎛ ⎡ x ⎤⎞ ⎡ x + y ⎤ 1. T ⎜ ⎢ ⎥⎟ = ⎢ relative to the bases BV = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ and BW = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ x − 2 y ⎦ ⎪⎩ ⎣ 1⎦ ⎣2 ⎦ ⎪⎭ ⎪⎩ ⎣1⎦ ⎣2 ⎦ ⎪⎭ ⎧ ⎡ 1 ⎤ ⎡ −1⎤ ⎡ 0 ⎤ ⎫ y ⎤ ⎡ ⎧ ⎡2 ⎤ ⎡ 3 ⎤ ⎫ ⎛ ⎡ x ⎤⎞ ⎢ ⎪ ⎪ ⎥ 2. T ⎜ ⎢ ⎥⎟ = ⎢13 y − 5 x ⎥ relative to the bases BV = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ and BW = ⎨ ⎢⎢ 0 ⎥⎥ , ⎢⎢ 2 ⎥⎥ , ⎢⎢ 1 ⎥⎥ ⎬ 1 2 ⎝ ⎣ y ⎦⎠ ⎩⎪ ⎣ ⎦ ⎣ ⎦ ⎭⎪ ⎪ ⎢ −1⎥ ⎢ 2 ⎥ ⎢ 1 ⎥ ⎪ ⎢⎣16 y − 7 x ⎥⎦ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ ⎧⎡ 1⎤ ⎡ 0 ⎤ ⎫ ⎧ ⎡2 ⎤ ⎡ 3 ⎤ ⎫ ⎛ ⎡ x1 ⎤ ⎞ ⎡ x1 − x 2 ⎤ 3. T ⎜ ⎢ ⎥⎟ = ⎢ relative to the bases BV = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ and BW = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎥ ⎝ ⎣ x 2 ⎦ ⎠ ⎣ x1 + x 2 ⎦ ⎪⎩ ⎣ 0 ⎦ ⎣ 1 ⎦ ⎭⎪ ⎩⎪ ⎣ 1⎦ ⎣2 ⎦ ⎭⎪ ⎧ ⎡1⎤ ⎡ 3 ⎤ ⎡ 2 ⎤ ⎫ ⎡ x + 2 x2 ⎤ ⎛ ⎡ x1 ⎤ ⎞ ⎢ 1 ⎧ ⎡ 1⎤ ⎡ −1⎤ ⎫ ⎪ ⎪ ⎥ 4. T ⎜ ⎢ ⎥⎟ = ⎢ 0 ⎥ relative to the bases BV = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ and BW = ⎨ ⎢⎢1⎥⎥ , ⎢⎢ 0 ⎥⎥ , ⎢⎢ 2 ⎥⎥ ⎬ ⎝ ⎣ x 2 ⎦⎠ ⎪⎩ ⎣3 ⎦ ⎣ 2 ⎦ ⎪⎭ ⎪ ⎢1⎥ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎪ ⎢⎣ − x1 ⎥⎦ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ ⎛ ⎡ x ⎤⎞ ⎡ x + 2 y − z ⎤ ⎧ ⎡ 1 ⎤ ⎡ 1 ⎤ ⎡1⎤ ⎫ ⎧ ⎡ 1⎤ ⎡2 ⎤ ⎡ 1⎤ ⎫ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢ ⎥ ⎥ ⎢ ⎜ ⎟ 5. T ⎜ ⎢ y ⎥⎟ = ⎢ y + z ⎥ relative to the bases BV = ⎨ ⎢ 1 ⎥ , ⎢ 0 ⎥ , ⎢1⎥ ⎬ and BW = ⎨ ⎢⎢ 1⎥⎥ , ⎢⎢2 ⎥⎥ , ⎢⎢2 ⎥⎥ ⎬ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ x + y − 2 z ⎥ ⎪ ⎢ 0 ⎥ ⎢ 1 ⎥ ⎢1⎥ ⎪ ⎪ ⎢2 ⎥ ⎢ 1⎥ ⎢2 ⎥ ⎪ ⎦ ⎣ ⎦ ⎣ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭

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For 6–10, determine the matrix representation of the given linear transformation relative to the standard basis.

⎛ ⎡ x ⎤⎞ ⎡2 x − 3 y + 4 z ⎤ 6. T ⎜ ⎢⎢ y ⎥⎥⎟ = ⎢⎢ 5 x − y + 2 z ⎥⎥ ⎟ ⎜ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ 4 x + 7 y ⎥ ⎦ ⎣ ⎦ ⎣ ⎛ ⎡ x ⎤⎞ ⎡ 2 y + z ⎤ 7. T ⎜ ⎢⎢ y ⎥⎥⎟ = ⎢⎢ x − 4 y ⎥⎥ ⎜ ⎟ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ 3 x ⎥ ⎦ ⎣ ⎦ ⎣ ⎛ ⎡ x ⎤⎞ ⎡3 x − y ⎤ 8. T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ 2 y ⎦ ⎛ ⎡ x ⎤⎞ ⎡ x + 4 y ⎤ 9. T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣3 x − 2 y ⎦ ⎛ ⎡ x ⎤⎞ ⎡ x ⎤ 10. T ⎜ ⎢⎢ y ⎥⎥⎟ = ⎢⎢ y ⎥⎥ ⎟ ⎜ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ z ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ −3 −4 ⎤ For 11–13, suppose [T ]BW ,BV = ⎢⎢ −2 −2 ⎥⎥ is the matrix representation of a linear transformation T : R 2 → R 3 ⎢⎣ 6 9 ⎥⎦ ⎧ ⎡ −1⎤ ⎡ 1 ⎤ ⎡ 0 ⎤ ⎫ ⎧ ⎡2 ⎤ ⎡ 3 ⎤ ⎫ ⎪ ⎪ relative to the bases BV = { v1 , v 2 } = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ for R 2 and BW = { w1 , w 2 , w 3 } = ⎨ ⎢⎢ 2 ⎥⎥ , ⎢⎢ 0 ⎥⎥ , ⎢⎢ 1 ⎥⎥ ⎬ for R 3 . Find 1 2 ⎩⎪ ⎣ ⎦ ⎣ ⎦ ⎭⎪ ⎪ ⎢ 2 ⎥ ⎢ −1⎥ ⎢ 2 ⎥ ⎪ T ( x ) for the given vector x. ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭

⎡2⎤ 11. ⎢ ⎥ ⎣ −3 ⎦ ⎡4⎤ 12. ⎢ ⎥ ⎣ −3 ⎦ ⎡1⎤ 13. ⎢ ⎥ ⎣ −5 ⎦ ⎡3 1 ⎤ 2 2 For 14 and 15, suppose ⎢ ⎥ is the matrix representation for the linear transformation T : R → R relative ⎣5 −2 ⎦ ⎧ ⎡ −1⎤ ⎡ 1⎤ ⎫ to the basis B = { v1 , v 2 } = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬. ⎩⎪ ⎣ 4 ⎦ ⎣3 ⎦ ⎭⎪

14. Find [T ( v1 )]B and [T ( v 2 )]B . 15. Find T ( v1 ) and T ( v 2 ).

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Change of basis Suppose B = {u1 , u 2 ,# , un } and B′ = { v 1 , v 2 ,#, v n } are two different (ordered) bases for the same vector space V. Because B is a basis, you can write each vector v i ∈ B′ as a unique linear combination of the vectors in B as shown here. v 1 = c11u1 + c21u 2 + + cn1un v 2 = c12 u1 + c u 2 + + cn 2 un v n = c1n u1 + c2n u 2 + + cnn un

The matrix C = [cij ] of the above coefficients is the change-of-basis matrix (or transition matrix) from the first basis B to the second basis B′. Notice that the columns of C are, respectively, the coordinate column vectors of the second basis vectors v i ∈ B′ relative to the first basis B; that is, C =[[ v 1 ]B ,[ v 2 ]B ,#,[ v n ]B ]

Likewise, there is a change-of-basis matrix C ′ from the second basis B′ back to the first basis B; and, similarly, the columns of C ′ are, respectively, the coordinate column vectors of the first basis vectors ui ∈ B relative to the second basis B′; that is, C ′ = [[u1 ]B′ ,[u 2 ]B′ ,#,[un ]B′ ]

Because the vectors ui ∈ B and v i ∈ B′ are linearly independent, then both C and C ′ have inverses; and, not surprisingly, C ′ = C −1 and C = (C ′ )−1. PROBLEM

⎧ ⎡ −2 ⎤ ⎡ −1⎤ ⎫ ⎧ ⎡5 ⎤ ⎡ 2 ⎤ ⎫ Let B = {u1 , u 2 } = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ and B′ = { v 1 , v 2 } = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ be bases of R 2. ⎩⎪ ⎣ 1 ⎦ ⎣ 1 ⎦ ⎭⎪ ⎩⎪ ⎣3 ⎦ ⎣1 ⎦ ⎭⎪

a. Find C, the change-of-basis matrix from the basis B to the basis B′. b. Find C ′ , the change-of-basis matrix back to the basis B from the basis B′. SOLUTION

a. Find C, the change-of-basis matrix from the basis B to the basis B′. Determine the coordinate vectors [ v 1 ]B and [ v 2 ]B by expressing each of v 1 and v 2 as a linear combination of the basis vectors u1 and u 2 of B. ⎡ −2 ⎤ ⎡5 ⎤ ⎡ 2 ⎤ v 1 = ⎢ ⎥ = a ⎢ ⎥ + b ⎢ ⎥ . Equating corresponding entries yields the system ⎣1⎦ ⎣3⎦ ⎣1 ⎦ ⎡5 2 −2 ⎤ 5a + 2b = −2 that has augmented matrix: ⎢ of equations ⎥ , which 3a + b = 1 ⎣3 1 1 ⎦ ⎡1 0 4 ⎤ is ⎢ ⎥ in reduced row-echelon form. Thus, a = 4 and b = −11, so ⎣0 1 −11⎦ ⎡ 4 ⎤ [ v 1 ]B = ⎢ ⎥. ⎣ −11⎦

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⎡ −1⎤ ⎡5 ⎤ ⎡ 2 ⎤ v 2 = ⎢ ⎥ = c ⎢ ⎥ + d ⎢ ⎥ . Equating corresponding entries yields the system of ⎣ 1 ⎦ ⎣3 ⎦ ⎣1 ⎦ ⎡5 2 −1⎤ 5c + 2d = −1 equations that has augmented matrix: ⎢ ⎥ , which is 3c + d = 1 ⎣3 1 1 ⎦ ⎡1 0 3 ⎤ ⎡3⎤ ⎢0 1 −8 ⎥ in reduced row-echelon form. Thus, c = 3 and d = −8, so [ v 2 ]B = ⎢ −8 ⎥ . ⎣ ⎦ ⎣ ⎦ 3⎤ v 1 = 4 u1 − 11u 2 ⎡ 4 and C = ⎢ Therefore, ⎥. v 2 = 3u1 − 8u 2 ⎣ −11 −8 ⎦

b. Find C ′ , the change-of-basis matrix back to the basis B from the basis B′. −1

3⎤ ⎡ 4 ⎡ −8 −3 ⎤ =⎢ C ′ = C −1 = ⎢ ⎥ ⎥ ⎣ −11 −8 ⎦ ⎣ 11 4 ⎦

Again, when you have a standard basis involved, the computations are simplified as shown in the following problem. PROBLEM

⎧⎡2 ⎤ ⎡ 1 ⎤ ⎫ ⎧ ⎡1 ⎤ ⎡ 0 ⎤ ⎫ Let E = {e1 , e 2 } = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ and S = { v 1 , v 2 } = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ be bases of R 2. 0 1 ⎩⎪ ⎣ 0 ⎦ ⎣ −1⎦ ⎭⎪ ⎩⎪ ⎣ ⎦ ⎣ ⎦ ⎭⎪

a. Find C, the change-of-basis matrix from the basis E to the basis S. b. Find C ′ , the change-of-basis matrix back to the basis E from the basis S. SOLUTION

a. Find C, the change-of-basis matrix from the basis E to the basis S. Determine the coordinate vectors [ v 1 ]E and [ v 2 ]E by expressing each of v 1 and v 2 as a linear combination of the basis vectors e1 and e 2 of S. ⎡2⎤ ⎡1⎤ ⎡2 1 ⎤ By inspection, v 1 = ⎢ ⎥ = 2e1 + 0e 2 and v 2 = ⎢ ⎥ = e1 − e 2 . Therefore, C = ⎢ ⎥. ⎣ 0 −1⎦ ⎣0 ⎦ ⎣ −1⎦ b. Find C ′ , the change-of-basis matrix back to the basis E from the basis S. −1 ⎡1 ⎡2 1 ⎤ =⎢ 2 C ′ = C −1 = ⎢ ⎥ ⎣ 0 −1⎦ ⎣⎢ 0

1 ⎤ 2⎥ −1 ⎥⎦

EXERCISE

9·4 For 1–3, find the coordinate vector for v relative to the given basis B = {u1 , u2 } of R 2 .

⎧⎡3 ⎤ ⎡ 1 ⎤ ⎫ ⎡8⎤ 1. v = ⎢ ⎥ ; B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎪⎩ ⎣ 0 ⎦ ⎣ −2 ⎦ ⎪⎭ ⎣ −4 ⎦ ⎧ ⎡ −2 ⎤ ⎡ 1 ⎤ ⎫ ⎡0 ⎤ 2. v = ⎢ ⎥ ; B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎪⎩ ⎣ 3 ⎦ ⎣ −1⎦ ⎪⎭ ⎣ 1⎦ ⎧ ⎡ −1⎤ ⎡ 0 ⎤ ⎫ ⎡20 ⎤ 3. v = ⎢ ⎥ ; B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎣14 ⎦ ⎩⎪ ⎣ 5 ⎦ ⎣ −1⎦ ⎭⎪

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For 4 and 5, find the coordinate vector for v relative to the given basis B = {u1 , u2 , u3 } of R 3 .

⎧ ⎡ 2 ⎤ ⎡ 0 ⎤ ⎡ −1⎤ ⎫ ⎡5 ⎤ ⎪ ⎪ ⎥ ⎢ 4. v = ⎢5 ⎥ ; B = ⎨ ⎢⎢ 0 ⎥⎥ , ⎢⎢ 0 ⎥⎥ , ⎢⎢ 3 ⎥⎥ ⎬ ⎪ ⎢ −1⎥ ⎢ 5 ⎥ ⎢ 2 ⎥ ⎪ ⎢⎣5 ⎥⎦ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ ⎧ ⎡ −3 ⎤ ⎡ 5 ⎤ ⎡ 0 ⎤ ⎫ ⎡0 ⎤ ⎪ ⎪ ⎥ ⎢ 5. v = ⎢ 1 ⎥ ; B = ⎨ ⎢⎢ 1 ⎥⎥ , ⎢⎢ 0 ⎥⎥ , ⎢⎢ 1 ⎥⎥ ⎬ ⎪ ⎢ −1⎥ ⎢ 5 ⎥ ⎢ −2 ⎥ ⎪ ⎢⎣ 0 ⎥⎦ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ For 6–10, Find (a) C, the change-of-basis matrix from the basis B to the basis B ′, and (b) C ′ , the change-ofbasis matrix back to the basis B from the basis B ′.

⎧ ⎡ −2 ⎤ ⎡ 1 ⎤ ⎫ ⎧ ⎡ −1⎤ ⎡ 0 ⎤ ⎫ 6. B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬, B ′ = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎪⎩ ⎣ 3 ⎦ ⎣ −1⎦ ⎪⎭ ⎪⎩ ⎣ 5 ⎦ ⎣ −1⎦ ⎪⎭ ⎧ ⎡1⎤ ⎡ −1⎤ ⎫ ⎧⎡ 0 ⎤ ⎡2 ⎤ ⎫ 7. B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬, B ′ = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎩⎪ ⎣1⎦ ⎣ 0 ⎦ ⎭⎪ ⎩⎪ ⎣ 3 ⎦ ⎣ 0 ⎦ ⎭⎪ ⎧ ⎡2 ⎤ ⎡ 0 ⎤ ⎫ ⎧ ⎡ 1⎤ ⎡ −1⎤ ⎫ 8. B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬, B ′ = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ 3 4 ⎩⎪ ⎣2 ⎦ ⎣ −1⎦ ⎭⎪ ⎩⎪ ⎣ ⎦ ⎣ ⎦ ⎭⎪ ⎧ ⎡ 1⎤ ⎡ 3 ⎤ ⎫ ⎧⎡ 1 ⎤ ⎡ 3 ⎤ ⎫ 9. B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬, B ′ = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎪⎩ ⎣3 ⎦ ⎣ 8 ⎦ ⎪⎭ ⎪⎩ ⎣ −2 ⎦ ⎣ −4 ⎦ ⎪⎭ ⎧ ⎡ 1⎤ ⎡ −2 ⎤ ⎡ 2 ⎤ ⎫ ⎧ ⎡ 1 ⎤ ⎡ −1⎤ ⎡ 0 ⎤ ⎫ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ 10. B = ⎨ ⎢ 0 ⎥ , ⎢ 2 ⎥ , ⎢ 1 ⎥ ⎬, B ′ = ⎨ ⎢⎢2 ⎥⎥ , ⎢⎢ −3 ⎥⎥ , ⎢⎢ 6 ⎥⎥ ⎬ ⎪ ⎢ 1⎥ ⎢ 1 ⎥ ⎢7 ⎥ ⎪ ⎪ ⎢ −1⎥ ⎢ 2 ⎥ ⎢ 2 ⎥ ⎪ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭

Algebra of linear transformations Because linear transformations are functions, you can combine them to form other functions. However, when combining linear transformations, you are not always guaranteed that the result will be a linear transformation, so it is important to check whether the requirements for a linear transformation are satisfied. PROBLEM

SOLUTION

Addition of linear transformations: If F : V → W and G : V → W are linear transformation, show that the sum F + G defined as the function from V into W that assigns to each vector v ∈V the vector ( F + G )( v ) = F (v ) + G( v ) in W is a linear transformation. Suppose u , v ∈V and k is any scalar. (a) Show ( F + G )(u + v ) = ( F + G )(u ) + ( F + G )( v ) . By definition, ( F + G )(u + v ) = F (u + v ) + G(u + v ) = F (u ) + F ( v ) + G(u ) + G( v ) = [F (u ) + G(u )] + [F ( v ) + G( v )] = ( F + G )(u ) + ( F + G )( v )

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(b) Show ( F + G )(kv ) = k( F + G )( v ). By definition, ( F + G )(kv ) = F (kv ) + (G )(kv ) = kF ( v ) + k(G )( v ) = k[F ( v ) + (G )( v )] = k( F + G )( v ) Because (a) and (b) are satisfied, F + G : V → W is a linear transformation. PROBLEM

SOLUTION

Scalar multiplication of linear transformations: If F : V → W is a linear transformation, show that the scalar product kF defined as the function from V into W that assigns to each vector v ∈V the vector (kF )( v ) = k[F ( v )] in W is a linear transformation. Suppose u , v ∈V and k, k ′ are scalars. (a) Show (kF )(u + v ) = (kF )(u ) + (kF )( v ). By definition, (kF )(u + v ) = k[F (u + v )] = k[F (u ) + F ( v )] = kF (u ) + kF ( v ) (b) Show (kF )(k ′v ) = k ′[(kF )( v )]. By definition, (kF )(k′v ) = = k[F (k ′v )] = k[k ′F ( v )] = kk ′F ( v ) = k ′kF( v ) = k ′[(kF )( v )] Because (a) and (b) are satisfied, kF : V → W is a linear transformation.

Let V and W be vector spaces over R and let ᐀(V ,W ) be the set of all linear transformations from V to W, then ᐀(V ,W ) is a vector space in which vectors are linear transformations, vector addition is addition of linear transformations (as defined above), and vector multiplication is scalar multiplication of linear transformations (as defined above). It is a straightforward verification that axioms i through x given in Chapter 7 for vector spaces are satisfied by ᐀(V ,W ) under addition and scalar multiplication of linear transformations. For instance, axiom iv is satisfied by the zero transformation in ᐀(V ,W ) defined as O : V → W such that O( v ) = 0 for every v in V; and axiom x is satisfied by the identity transformation in ᐀(V ,V ) defined as I : V → V such that I( v ) = v for every v in V. Note: See Probs. 1 and 2 in Exercise 9.5, for proofs that these transformations are linear. Also, because ᐀(V ,W ) is a vector space if G ∈᐀(V ,W ), you have −G ∈᐀(V ,W ) ; and you can easily establish that ( F − G )( v ) = F ( v ) − G( v ) = F(v) + (−G(v)) for all F , G ∈᐀(V ,W ) . PROBLEM

Let F : R 3 → R 3 and G : R 3 → R 3 be linear transformations defined as follows: ⎛ ⎡ x1 ⎤ ⎞ ⎡ x 2 ⎤ ⎛ ⎡ x1 ⎤ ⎞ ⎡ x1 + x 2 + x 3 ⎤ ⎢ ⎥ ⎢ ⎥ ⎥ ⎜ ⎟ F ⎢ x 2 ⎥ = ⎢ x 3 ⎥ and G ⎜ ⎢⎢ x 2 ⎥⎥⎟ = ⎢⎢ 0 ⎥ ⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ x ⎥ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ ⎥⎦ 0 ⎣ 3⎦ ⎣ 1⎦ ⎣ 3⎦ ⎣

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Calculate ⎛ ⎡1 ⎤ ⎞ a. ( F + G ) ⎜ ⎢⎢1 ⎥⎥⎟ ⎟ ⎜ ⎜⎝ ⎢0 ⎥⎟⎠ ⎣ ⎦ ⎛ ⎡1 ⎤ ⎞ b. ( F − G ) ⎜ ⎢⎢1 ⎥⎥⎟ ⎜ ⎟ ⎜⎝ ⎢0 ⎥⎟⎠ ⎣ ⎦ ⎛ ⎡1 ⎤ ⎞ c. ( F + I ) ⎜ ⎢⎢1 ⎥⎥⎟ ⎟ ⎜ ⎜⎝ ⎢0 ⎥⎟⎠ ⎣ ⎦ ⎛ ⎡1 ⎤ ⎞ d. (G + O ) ⎜ ⎢⎢1 ⎥⎥⎟ ⎟ ⎜ ⎜⎝ ⎢0 ⎥⎟⎠ ⎣ ⎦ SOLUTION

⎛ ⎡1 ⎤ ⎞ ⎛ ⎡1 ⎤ ⎞ ⎛ ⎡1 ⎤ ⎞ ⎡1 ⎤ ⎡ 2 ⎤ ⎡ 3 ⎤ ⎢ ⎥ ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ a. ( F + G ) ⎢1 ⎥ = F ⎢1 ⎥ + G ⎜ ⎢⎢1 ⎥⎥⎟ = ⎢⎢0 ⎥⎥ + ⎢⎢ 0 ⎥⎥ = ⎢⎢0 ⎥⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎢0 ⎥⎟⎠ ⎜⎝ ⎢0 ⎥⎟⎠ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢1 ⎥ ⎢ 0 ⎥ ⎢1 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎛ ⎡1 ⎤ ⎞ ⎛ ⎡1 ⎤ ⎞ ⎛ ⎡1 ⎤⎞ ⎡1 ⎤ ⎡ 2 ⎤ ⎡ −1⎤ ⎢ ⎥ ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ b. ( F − G ) ⎢1 ⎥ = F ⎢1 ⎥ − G ⎜ ⎢⎢1 ⎥⎥⎟ = ⎢⎢0 ⎥⎥ − ⎢⎢ 0 ⎥⎥ = ⎢⎢ 0 ⎥⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎢0 ⎥⎟⎠ ⎜⎝ ⎢0 ⎥⎟⎠ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢1 ⎥ ⎢ 0 ⎥ ⎢ 1 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎛ ⎡1 ⎤ ⎞ ⎛ ⎡1 ⎤ ⎞ ⎛ ⎡1 ⎤ ⎞ ⎡1 ⎤ ⎡1 ⎤ ⎡ 2 ⎤ ⎢ ⎥ ⎟ ⎜ c. ( F + I ) ⎢1 ⎥ = F ⎜ ⎢⎢1 ⎥⎥⎟ + I ⎜ ⎢⎢1 ⎥⎥⎟ = ⎢⎢0 ⎥⎥ + ⎢⎢1 ⎥⎥ = ⎢⎢1 ⎥⎥ ⎟ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎢0 ⎥⎟⎠ ⎜⎝ ⎢0 ⎥⎟⎠ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢1 ⎥ ⎢0 ⎥ ⎢1 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎛ ⎡1 ⎤ ⎞ ⎛ ⎡1 ⎤ ⎞ ⎛ ⎡1 ⎤ ⎞ ⎡ 2 ⎤ ⎡ 0 ⎤ ⎡ 2 ⎤ ⎢ ⎥ ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ d. (G + O ) ⎢1 ⎥ = G ⎢1 ⎥ + O ⎜ ⎢⎢1 ⎥⎥⎟ = ⎢⎢ 0 ⎥⎥ + ⎢⎢0 ⎥⎥ = ⎢⎢ 0 ⎥⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎢0 ⎥⎟⎠ ⎜⎝ ⎢0 ⎥⎟⎠ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢ 0 ⎥ ⎢0 ⎥ ⎢ 0 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Given the linear transformations F : V → U and G : U → W , then the composition F F G (read F circle G) defined by ( F F G )(u ) = F (G(u )), where u ∈U , is also a linear transformation. Compositions of linear transformations are used in showing that two vector spaces are isomorphic; that is, that they are, in some sense, equivalent. This notion is formalized as follows: A linear transformation F : V → W is an isomorphism and V and W are isomorphic, denoted V ≅ W , if there exists a linear transformation G : W → V such that (G F F )( v ) = v , for all v ∈V and (F F G) ( w ) = w, for all w ∈W . Showing that a linear transformation is an isomorphism is simplified by the following consequence: A linear transformation T : V → W is an isomorphism if and only if Ker T = {0} and Im T = W .

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PROBLEM

SOLUTION

⎡ a0 ⎤ Show that T : P2 → R defined by T (a0 + a1 x + a2 x ) = ⎢⎢ a1 ⎥⎥ is an isomorphism. ⎢⎣a2 ⎥⎦ First show Ker T = {0}. 3

2

⎡0 ⎤ ⎡ a0 ⎤ Suppose T (a0 + a1 x + a2 x 2 ) = ⎢⎢ a1 ⎥⎥ = 0 = ⎢⎢0 ⎥⎥ , then a0 = 0, a1 = 0, and a3 = 0. Thus, ⎢⎣0 ⎥⎦ ⎢⎣a2 ⎥⎦ Ker T = {0}.

Next, show Im T = R 3. Recall the following established assertions: ◆ ◆ ◆

For a linear transformation T : V → W , if V is finite dimensional, then both Ker T and Im T are finite dimensional and dim (V ) = dim (Ker T ) + dim (Im T ). If T : V → W is a linear transformation, then Ker T is a subspace of V and Im T is a subspace of W. If U is a finite dimensional subspace of Rn and dim(U ) = n , then U = Rn.

Therefore, to show Im T = R 3 , you can reason like this: 3 = dim ( P2 ) = dim (Ker T ) + dim (Im T ) = 0 + dim (Im T ) = dim(Im T ); and, thus, because Im T is a subspace of R 3 and dim(Im T ) = 3 = dim( R 3 ), Im T = R 3. Hence, T is an isomorphism.

EXERCISE

9·5 For 1–6, prove as indicated.

1. Prove that the zero transformation is linear; that is, prove that the function O : V → W defined by O( v ) = 0 for every v in V is linear. 2. Prove that the identity transformation is linear; that is, prove that the function I : V → V defined by I( v ) = v for every v in V is linear. 3. Prove that −G( v ) = ( −1)G( v ) for every G ∈᐀(V ,W ) . 4. Prove that ( F − G )( v ) = F ( v ) − G( v ) for all F , G ∈᐀(V ,W ) . 5. Prove that F + O = F for every F ∈᐀(V ,W ) . 6. Prove that the composition of two linear transformations is linear; that is, prove that if F : V → U and G : U → W , then F F G defined by ( F F G )(u) = F (G(u)) , where u ∈U , is a linear transformation. For 7–14, let F : R 3 → R 3 and G : R 3 → R 3 be linear transformations defined as follows:

⎛ ⎡ x1 ⎤ ⎞ ⎡ x1 ⎤ ⎛ ⎡ x 1 ⎤ ⎞ ⎡ x1 + x 2 ⎤ ⎢ ⎥ ⎥ ⎢ ⎜ ⎟ F ⎢ x 2 ⎥ = ⎢ 0 ⎥ and G ⎜ ⎢⎢ x 2 ⎥⎥⎟ = ⎢⎢ x 2 ⎥⎥ calculate as indicated. ⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ x ⎥ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ 0 ⎥ ⎦ ⎣ 3⎦ ⎣ 3⎦ ⎣ 3⎦ ⎣ ⎛ ⎡ x1 ⎤ ⎞ 7. ( F + G ) ⎜ ⎢⎢ x 2 ⎥⎥⎟ ⎟ ⎜ ⎜⎝ ⎢ x ⎥⎟⎠ ⎣ 3⎦

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⎛ ⎡ x1 ⎤ ⎞ 8. ( F − 2G ) ⎜ ⎢⎢ x 2 ⎥⎥⎟ ⎟ ⎜ ⎜⎝ ⎢ x ⎥⎟⎠ ⎣ 3⎦ ⎛ ⎡ x1 ⎤ ⎞ 9. (2F + 3G ) ⎜ ⎢⎢ x 2 ⎥⎥⎟ ⎟ ⎜ ⎜⎝ ⎢ x ⎥⎟⎠ ⎣ 3⎦ ⎛ ⎡ x1 ⎤ ⎞ 10. ( F F G ) ⎜ ⎢⎢ x 2 ⎥⎥⎟ ⎟ ⎜ ⎜⎝ ⎢ x ⎥⎟⎠ ⎣ 3⎦ ⎛ ⎡ x1 ⎤ ⎞ 11. (G F F ) ⎜ ⎢⎢ x 2 ⎥⎥⎟ ⎟ ⎜ ⎜⎝ ⎢ x ⎥⎟⎠ ⎣ 3⎦ ⎛ ⎡ x1 ⎤ ⎞ 12. (G F kF ) ⎜ ⎢⎢ x 2 ⎥⎥⎟ ⎜ ⎟ ⎜⎝ ⎢ x ⎥⎟⎠ ⎣ 3⎦ ⎛ ⎡ x1 ⎤ ⎞ 13. ( F F F ) ⎜ ⎢⎢ x 2 ⎥⎥⎟ ⎟ ⎜ ⎜⎝ ⎢ x ⎥⎟⎠ ⎣ 3⎦ ⎛ ⎡ x1 ⎤ ⎞ 14. ((G F F ) F F ) ⎜ ⎢⎢ x 2 ⎥⎥⎟ ⎟ ⎜ ⎜⎝ ⎢ x ⎥⎟⎠ ⎣ 3⎦ ⎛ ⎡ x1 ⎤ ⎞ ⎡ x1 − 2 x 2 ⎤ 15. Show that T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ is an isomorphism. ⎝ ⎣ x 2 ⎦ ⎠ ⎣ x1 − x 2 ⎦

Linear operators on R2 and R3 This chapter concludes with a discussion of the familiar geometric concepts of projections, reflections, rotations, dilations, and contractions in the context of linear operators on R 2 and R3 and their corresponding matrix representations (Note: Recall that T : V → V is called a linear operator on V). For the linear operator T : Rn → Rn , the transformation matrix, denoted [T ], relative to the standard basis for Rn is referred to as the standard matrix for T, and T ( x ) = [T ]x , where [T ] = [T (e1 ),T (e 2 ),#,T (en )]. PROBLEM

⎛ ⎡ x ⎤⎞ ⎡ x cosθ − y sinθ ⎤ Given T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ that rotates a vector ⎝ ⎣ y ⎦⎠ ⎣ x sinθ + y cosθ ⎦ counterclockwise through an angle θ (see Fig. 9.1),

a. Find the standard matrix representation for T. ⎡2⎤ b. Compute [T ]x for x = ⎢ ⎥ when θ = 30°. ⎣ −6 ⎦

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y

(x', y')

(x, y)

x

Figure 9.1 Rotation through an Angle q from (x, y) to (x′, y′). SOLUTION

a. Find the standard matrix representation for T. ⎛ ⎡0 ⎤⎞ ⎡ − sinθ ⎤ ⎛ ⎡1 ⎤⎞ ⎡cosθ ⎤ and T (e 2 ) = T ⎜ ⎢ ⎥⎟ = ⎢ T (e1 ) = T ⎜ ⎢ ⎥ ⎟ = ⎢ ⎥ ⎥ ⎝ ⎣1 ⎦⎠ ⎣ cosθ ⎦ ⎝ ⎣0 ⎦⎠ ⎣ sinθ ⎦ ⎡cosθ Thus, [T ] = ⎢ ⎣ sinθ

− sinθ ⎤ . cosθ ⎥⎦

⎡2⎤ b. Compute [T ]x for x = ⎢ ⎥ when θ = 30°. ⎣ −6 ⎦ ⎡ 3 1 ⎤ ⎡cos 30° − sin 30° ⎤ ⎡ 2 ⎤ ⎢ 2 − 2 ⎥ ⎡ 2 ⎤ ⎡ 3 + 3 ⎤ [T ]x = ⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ = 3 ⎥ ⎣ −6 ⎦ ⎢⎣1 − 3 3 ⎥⎦ ⎣ sin 30° cos 30° ⎦ ⎣ −6 ⎦ ⎢⎢ 1 ⎥ ⎣ 2 2⎦

PROBLEM

⎛ ⎡ x ⎤⎞ ⎡ x ⎤ Given T : R → R defined by T ⎜ ⎢⎢ y ⎥⎥⎟ = ⎢⎢ 0 ⎥⎥ that maps a vector into its orthogonal ⎟ ⎜ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ z ⎥ ⎣ ⎦ ⎣ ⎦ projection on the xz-plane. (See Fig. 9.2.) 3

3

a. Find the standard matrix representation for T. ⎡5⎤ b. Compute [T ]x for x = ⎢⎢ −3 ⎥⎥ ⎢⎣ 2 ⎥⎦ z

(x, 0, z)

(x, y, z)

y

x

Figure 9.2 Orthogonal Projection on the xz-plane. Linear transformations

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SOLUTION

a. Find the standard matrix representation for T. ⎛ ⎡1 ⎤ ⎞ ⎡1 ⎤ ⎛ ⎡0 ⎤⎞ ⎡0 ⎤ ⎛ ⎡0 ⎤⎞ ⎡0 ⎤ ⎥ ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ ⎢ ⎟ ⎟ ⎜ ⎜ T (e1 ) = T ⎢0 ⎥ = ⎢0 ⎥ , T (e 2 ) = T ⎢1 ⎥ = ⎢0 ⎥ , and T (e 3 ) = T ⎜ ⎢⎢0 ⎥⎥⎟ = ⎢⎢0 ⎥⎥ ⎟ ⎟ ⎟ ⎜ ⎜ ⎜ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢0 ⎥ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢0 ⎥ ⎜⎝ ⎢1 ⎥⎟⎠ ⎢1 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡1 0 0 ⎤ Thus, [T ] = ⎢⎢0 0 0 ⎥⎥ . ⎢⎣0 0 1 ⎥⎦ ⎡5⎤ b. Compute [T ]x for x = ⎢⎢ −3 ⎥⎥ ⎢⎣ 2 ⎥⎦ ⎡1 0 0 ⎤ ⎡ 5 ⎤ ⎡ 5 ⎤ [T ]x = ⎢⎢0 0 0 ⎥⎥ ⎢⎢ −3 ⎥⎥ = ⎢⎢ 0 ⎥⎥ ⎢⎣0 0 1 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ 2 ⎥⎦ EXERCISE

9·6 For 1-9, find [T ]x and T ( x ).

⎛ ⎡ x ⎤⎞ ⎡ − x ⎤ ⎡7⎤ 1. T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ that is a reflection about the y-axis; x = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ y ⎦ ⎣ −3 ⎦ ⎛ ⎡ x ⎤⎞ ⎡ y ⎤ ⎡4⎤ 2. T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ that is a reflection about the line y = x ; x = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ x ⎦ ⎣ 1⎦ ⎛ ⎡ x ⎤⎞ ⎡ 0 ⎤ ⎡3⎤ 3. T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ that is an orthogonal projection on the y-axis; x = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ y ⎦ ⎣ −3 ⎦ ⎛ ⎡ x ⎤⎞ ⎡ x ⎤ ⎡4⎤ 4. T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ that is a reflection about the x-axis; x = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ − y ⎦ ⎣ 1⎦ ⎡ 2 ⎤ ⎛ ⎡ x ⎤⎞ ⎡5 x ⎤ 5 ⎥ 5. T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ that is a dilation with factor 5; x = ⎢ ⎢ −3 ⎥ ⎝ ⎣ y ⎦⎠ ⎣5 y ⎦ ⎣ 10 ⎦ ⎛ ⎡ x ⎤⎞ ⎡ x ⎤ ⎡ 1⎤ ⎢ ⎥ ⎥ ⎢ ⎜ ⎟ 6. T : R → R defined by T ⎢ y ⎥ = ⎢ − y ⎥ that is a reflection about the xz-plane; x = ⎢⎢2 ⎥⎥ ⎜ ⎟ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ z ⎥ ⎢⎣ 3 ⎥⎦ ⎣ ⎦ ⎣ ⎦ 3

3

⎛ ⎡ x ⎤⎞ ⎡ 0 ⎤ 7. T : R → R defined by T ⎜ ⎢⎢ y ⎥⎥⎟ = ⎢⎢ y ⎥⎥ that is an orthogonal projection on the yz-plane; ⎜ ⎟ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ z ⎥ ⎣ ⎦ ⎦ ⎣ ⎡ −3 ⎤ ⎥ ⎢ x=⎢ 2 ⎥ ⎢⎣ 4 ⎥⎦ 3

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⎛ ⎡ x ⎤⎞ ⎡ x ⎤ ⎥ ⎢ ⎢ ⎟ ⎜ 8. T : R → R defined by T ⎢ y ⎥ = ⎢ y cosθ − z sinθ ⎥⎥ that is a counterclockwise rotation about ⎟ ⎜ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ y sinθ + z cosθ ⎥ ⎣ ⎦ ⎣ ⎦ 3

3

⎡5⎤ the positive x-axis; x = ⎢⎢ 2 ⎥⎥ , θ = 45° ⎢⎣ −2 ⎥⎦ ⎛ ⎡ x ⎤⎞ ⎡ 0.2 x ⎤ ⎡6⎤ ⎢ ⎥ ⎥ ⎢ ⎟ ⎜ 9. T : R → R defined by T ⎢ y ⎥ = ⎢ 0.2 y ⎥ that is a contraction with factor 0.2; x = ⎢⎢ −3 ⎥⎥ ⎟ ⎜ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ 0.2 z ⎥ ⎢⎣ 8 ⎥⎦ ⎦ ⎣ ⎦ ⎣ 3

3

⎛ ⎡ x ⎤⎞ ⎡ x + 3 ⎤ 10. Show that T : R → R defined by T ⎜ ⎢⎢ y ⎥⎥⎟ = ⎢⎢ y + 3 ⎥⎥ is not a linear transformation. ⎟ ⎜ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ z + 3 ⎥ ⎦ ⎣ ⎦ ⎣ 3

3

Linear transformations

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·10·

·10 ·

Eigenvalues and eigenvectors In this chapter, you will learn: ◆ The eigenvalue problem ◆ Useful properties of eigenvalues ◆ Diagonalization

Many applications of linear algebra involve determining eigenvalues and their corresponding eigenvectors. This chapter provides a discussion of this useful topic.

The eigenvalue problem Suppose A =[aij ] is a square matrix of size n. A scalar l is an eigenvalue (or characteristic value) of A if there exists a nonzero n ×1 vector x such that Ax = l x . Such a vector x is an eigenvector (or characteristic vector) corresponding to the eigenvalue l. The equation Ax = l x can be rewritten as (lI − A)x = 0, where I is the identity matrix of the same size as A. This homogeneous system has a nontrivial solution if and only if ⎛ ⎡l − a11 ⎜ ⎢ −a det(l I − A) = det ⎜ ⎢ 21 ⎜⎢ a − ⎢ ⎝ ⎣ n1

−a12 l − a22 −an 2

−a1n ⎤⎞ −a2n ⎥⎟ =0 ⎥⎥⎟ ⎟ … l − ann ⎥⎦⎠ … …

The equation det(lI − A) = 0 is the characteristic equation of A and the eigenvalues of A are the scalar values for l that satisfy it. Note: Many authors write the characteristic equation as the equivalent equation det( A − l I ) = 0. When you expand det(lI − A), you obtain the characteristic polynomial of A, which is an nth-degree polynomial p in the variable l, with leading coefficient 1, that has the form p(l ) = l n + c1l n−1 + $ + cn . Using this form for det(lI − A), you can write the characteristic equation as l n + c1l n−1 + $ + cn = 0

You know from the Fundamental Theorem of Algebra that this equation has at most n distinct roots. Therefore, an n × n matrix has at most n eigenvalues. Remark: Actually over the complex numbers the equation p( x ) = 0 has exactly n roots, some or all of which might be nonreal complex numbers. Thus,

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you can have nonreal complex eigenvalues. For purposes of this discussion, only real eigenvalues will be considered. PROBLEM

⎡1 2 ⎤ Find the eigenvalues of A = ⎢ ⎥. ⎣3 2 ⎦

SOLUTION

⎛ ⎡ l − 1 −2 ⎤⎞ 2 det(l I − A) = det ⎜ ⎢ ⎥⎟ = (l − 1)(l − 2) − 6 = l − 3l − 4 , so the ⎝ ⎣ −3 l − 2 ⎦⎠

characteristic equation is l 2 − 3l − 4 = 0. Solving for l,

λ 2 − 3λ − 4 = 0 (λ − 4 )(λ + 1) = 0 Thus, the eigenvalues of A are λ1 = 4 and λ2 = −1 .

The next step in solving an eigenvalue problem is to find the eigenvectors for the matrix. Each of the eigenvalues of the matrix A obtained in the problem above has corresponding eigenvectors. Because A is a 2 × 2 matrix, an eigenvector x for A is a nonzero 2 × 1 vector that satisfies the equation (λ I − A) x = 0. The set of nonzero vectors that satisfy (λ I − A)x = 0 , together with the zero vector, is the eigenspace of A corresponding to λ . PROBLEM

For the matrix A in the previous problem, a. Find the eigenvectors corresponding to the eigenvalues, λ1 = 4 and λ2 = −1. b. Find bases for the eigenspaces of A.

SOLUTION

a. Find the eigenvectors corresponding to the eigenvalues, λ1 = 4 and λ2 = −1. ⎡1 2 ⎤ ⎡ x1 ⎤ When A = ⎢ and x = ⎢ ⎥ , the equation (λ I − A)x = 0 becomes ⎥ ⎣3 2 ⎦ ⎣ x2 ⎦ ⎡ λ − 1 −2 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢ −3 λ − 2 ⎥ ⎢ x ⎥ = ⎢0 ⎥ ⎣ ⎦⎣ 2 ⎦ ⎣ ⎦ ⎡a ⎤ Let ⎢ ⎥ be an eigenvector corresponding to λ1 = 4 , then substituting into ⎣b ⎦ ⎡ 3 −2 ⎤ ⎡a ⎤ ⎡0 ⎤ (λ I − A)x = 0 yields ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ , which has augmented matrix ⎣ −3 2 ⎦ ⎣b ⎦ ⎣0 ⎦ ⎡ 3 −2 0 ⎤ ⎢ −3 2 0 ⎥ with reduced ⎣ ⎦ ⎡2 ⎤ ⎡1 − 2 3 0 ⎤ . So, a = 2 t , b = t , and ⎡a ⎤ = ⎢ 3 t ⎥ , where t is row-echelon form ⎢ ⎢ ⎥ ⎥ 0⎦ 3 0 ⎣b ⎦ ⎢⎣ t ⎥⎦ ⎢⎣0 ⎡2 ⎤ arbitrary. Thus, any nonzero vector that has the form t ⎢ 3 ⎥ is a eigenvector ⎢⎣ 1 ⎥⎦ ⎡2⎤ corresponding to λ1 = 4. For instance, when t = 3, ⎢ ⎥ is a eigenvector ⎣3⎦ corresponding to λ1 = 4.

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⎡c ⎤ Let ⎢ ⎥ be an eigenvector corresponding to λ2 = −1, then substituting into ⎣d ⎦ ⎡ −2 −2 ⎤ ⎡ c ⎤ ⎡0 ⎤ (λ I − A)x = 0 yields ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ , which has augmented matrix ⎣ −3 −3 ⎦ ⎣d ⎦ ⎣0 ⎦ ⎡ −2 −2 0 ⎤ ⎡1 1 0 ⎤ ⎢ −3 −3 0 ⎥ with reduced row-echelon form ⎢ ⎥ . So, c = −t , d = t , and ⎣ ⎦ ⎣0 0 0 ⎦ ⎡ c ⎤ ⎡ −t ⎤ ⎢d ⎥ = ⎢ t ⎥ where t is arbitrary. Thus, any nonzero vector that has the form ⎣ ⎦ ⎣ ⎦ ⎡ −1⎤ t ⎢ ⎥ is a eigenvector corresponding to λ2 = −1. For instance, when t = 1, ⎣1⎦ ⎡ −1⎤ ⎢ 1 ⎥ is a eigenvector corresponding to λ2 = −1. ⎣ ⎦

b. Find bases for the eigenspaces of A. ⎡− 2 ⎤ The eigenvectors corresponding to λ1 = 4 have the form t ⎢ 3 ⎥ , t ≠ 0. ⎢⎣ 1 ⎥⎦ ⎡− 2 ⎤ Therefore, ⎢ 3 ⎥ is a basis for the eigenspace of A corresponding to λ1 = 4 . ⎢⎣ 1 ⎥⎦ ⎡ −1⎤ The eigenvectors corresponding to λ2 = −1 have the form t ⎢ ⎥ , t ≠ 0. ⎣1⎦ ⎡ −1⎤ Therefore, ⎢ ⎥ is a basis for the eigenspace of A corresponding to λ2 = −1. ⎣1⎦

PROBLEM

⎡ −2 −6 19 ⎤ Given the matrix A = ⎢⎢ 0 −2 5 ⎥⎥ ⎢⎣ 1 0 −4 ⎥⎦ a. Find the eigenvalues of A. b. Find the eigenvectors for the eigenvalues of A. c. Find bases for the eigenspaces of A.

SOLUTION

a. Find the eigenvalues of A. ⎛ ⎡λ + 2 6 −19 ⎤⎞ ⎢ ⎜ det(λ I − A) = det ⎢ 0 λ + 2 −5 ⎥⎥⎟ = λ 3 + 8λ 2 + λ + 8, so the characteristic ⎜ ⎟ ⎜⎝ ⎢ −1 0 λ + 4 ⎥⎦⎟⎠ ⎣

equation is λ 3 + 8λ 2 + λ + 8 = 0. Solving for λ ,

λ 3 + 8λ 2 + λ + 8 = 0 (λ + 8)(λ 2 + 1) = 0

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Thus, the eigenvalues of A are λ1 = −8, λ2 = i , and λ3 = −i. Because the discussion is limited to real-valued eigenvalues, parts b. and c. will deal only with λ1 = −8. b. Find the eigenvectors for the eigenvalues of A. ⎡ x1 ⎤ When x = ⎢⎢ x 2 ⎥⎥ , the equation (λ I − A)x = 0 becomes ⎢⎣ x 3 ⎥⎦ 6 −19 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡λ + 2 ⎢ 0 λ + 2 −5 ⎥⎥ ⎢⎢ x 2 ⎥⎥ = ⎢⎢0 ⎥⎥ ⎢ ⎢⎣ −1 0 λ + 4 ⎥⎦ ⎢⎣ x 3 ⎦⎥ ⎢⎣0 ⎥⎦ ⎡a ⎤ ⎢ ⎥ Let ⎢b ⎥ be an eigenvector corresponding to λ1 = −8, then substituting into ⎢⎣ c ⎥⎦ ⎡ −6 6 −19 ⎤ ⎡a ⎤ ⎡0 ⎤ ⎥⎢ ⎥ ⎢ ⎥ ⎢ (λ I − A)x = 0 yields ⎢ 0 −6 −5 ⎥ ⎢b ⎥ = ⎢0 ⎥ , which has augmented matrix ⎢⎣ −1 0 −4 ⎥⎦ ⎢⎣ c ⎥⎦ ⎢⎣0 ⎥⎦ ⎡1 0 4 ⎡ −6 6 −19 0 ⎤ ⎢ 0 −6 −5 0 ⎥ with reduced row-echelon form ⎢ ⎢0 1 5 6 ⎥ ⎢ ⎢ ⎢⎣ −1 0 −4 0 ⎥⎦ ⎣0 0 0

0⎤ ⎥ 0 ⎥ . So, ⎥ 0⎦

⎡a ⎤ ⎡ −4t ⎤ 5 ⎥ ⎢ a = −4t , b = − t , and c = t , and ⎢b ⎥ = ⎢ −5 t ⎥ , where t is arbitrary. Thus, any 6 ⎥ ⎢ 6 ⎢⎣ c ⎥⎦ ⎢ t ⎥ ⎦ ⎣ ⎡ −4 ⎤ ⎢ ⎥ nonzero vector that has the form t ⎢ −5 6 ⎥ is a eigenvector corresponding to ⎢ 1 ⎥ ⎣ ⎦ ⎡ −24 ⎤ λ1 = −8. For instance, when t = 6 , ⎢⎢ −5 ⎥⎥ is a eigenvector corresponding to ⎢⎣ 6 ⎥⎦ λ1 = −8.

c. Find bases for the eigenspaces of A. ⎡ −4 ⎤ ⎥ ⎢ The eigenvectors corresponding to λ1 = −8 have the form t ⎢ −5 6 ⎥ , t ≠ 0. ⎢ 1 ⎥ ⎦ ⎣ ⎡ −4 ⎤ ⎥ ⎢ Therefore, ⎢ −5 6 ⎥ is a basis for the eigenspace of A corresponding to λ1 = −8. ⎢ 1 ⎥ ⎦ ⎣

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As you can see from the above problems, calculating eigenvalues and eigenvectors can be a long and laborious process. As the size of the matrix increases, doing the computations by hand becomes impractical. As a result, sophisticated calculators and software programs are now available to ease the chore. Nevertheless, having a strong understanding of the theory and process is critical to using these tools appropriately and efficiently. EXERCISE

10·1 For 1–5, fill in the blank with the best response. 1. A scalar λ is an eigenvalue if there exists a nonzero n ×1 vector x such that Ax = ________. 2. A vector that corresponds to an eigenvalue is called a(n) ________ corresponding to the eigenvalue λ. 3. The equation ( λ I − A)x = 0 has a nontrivial solution if and only if det ( λ I − A) = ________. 4. The characteristic polynomial of an n × n matrix is an nth-degree polynomial p in the variable λ that has the form p( λ ) = ________. 5. An n × n matrix has at most ________ eigenvalues. ⎡10 4 ⎤ For 6–9, let A = ⎢ ⎥. ⎣ −9 −2 ⎦

6. Find the characteristic polynomial for A. 7. Find the eigenvalues of A. 8. Find eigenvectors corresponding to the eigenvalues of A. 9. Find bases for the eigenspaces of A. ⎡3 2 2⎤ For 10–13, let B = ⎢⎢ −1 0 −1⎥⎥ ⎢⎣ −2 −2 −1⎥⎦

10. Find the characteristic polynomial for B. 11. Find the eigenvalues of B. 12. Find eigenvectors corresponding to the eigenvalues of B. 13. Find bases for the eigenspaces of B. For 14–20, find the eigenvalues of the matrix.

⎡ −4 8 ⎤ 14. ⎢ ⎥ ⎣ 0 3⎦ ⎡ −4 0 ⎤ 15. ⎢ ⎥ ⎣ 8 3⎦ ⎡ −4 0 ⎤ 16. ⎢ ⎥ ⎣ 0 3⎦

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⎡4 0 ⎤ 17. ⎢ ⎥ ⎣ 81 −2 ⎦ ⎡1 0 0⎤ ⎢ 2 ⎥ 18. ⎢ −3 5 0⎥ ⎢ ⎥ ⎢⎣ 6 4 9 −8 ⎥⎦ 0 ⎤ ⎡ −1 5 ⎢ 0 11 −10 ⎥ 19. ⎢ ⎥ ⎢⎣ 0 0 6 ⎥⎦ ⎡1 0 ⎢ 0 −3 20. ⎢ ⎢0 0 ⎢ ⎣0 0

0 0⎤ 0 0⎥ ⎥ 5 0⎥ ⎥ 0 −1⎦

Useful properties of eigenvalues Some useful properties of the eigenvalues of a square matrix A are the following: (a) tr ( A) is the sum of the eigenvalues of A. (b) det ( A) is the product of the eigenvalues of A. 1 is an eigenvalue of A−1. λ (d) If λ is an eigenvalue of A with corresponding eigenvector x, then λ k is an eigenvalue of Ak with corresponding eigenvector x for k a positive integer.

(c) If A is nonsingular and λ is an eigenvalue of A, then

(e) A and AT have the same eigenvalues. (f) If A is a triangular matrix, the eigenvalues of A are the same as the elements on its

diagonal. (g) If A is a diagonal matrix, the eigenvalues of A are the same as the elements on its

diagonal. PROBLEM

⎡2 2⎤ Given the matrix A = ⎢ ⎥ ⎣1 3 ⎦ a. Find the eigenvalues of A. b. Verify that tr ( A) is the sum of the eigenvalues of A. c. Verify that det ( A) is the product of the eigenvalues of A. 1 d. Verify that if l is an eigenvalue of A, then is an eigenvalue of A−1. l e. Verify that if l is an eigenvalue of A, then l3 is an eigenvalue of A3 . f. Verify that A and AT have the same eigenvalues.

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SOLUTION

a. Find the eigenvalues of A. ⎛ ⎡ l − 2 −2 ⎤⎞ 2 det(lI − A) = det ⎜ ⎢ ⎥⎟ = (l − 2)(l − 3) − 2 = l − 5l + 4 , so the ⎝ ⎣ −1 λ − 3 ⎦⎠ 2 characteristic equation is λ − 5λ + 4 = 0 . Solving for l,

λ 2 − 5λ + 4 = 0 (λ − 4 )(λ − 1) = 0

Thus, the eigenvalues of A are l1 = 4 and λ2 = 1. b. Verify that tr(A) is the sum of the eigenvalues of A. ⎛ ⎡ 2 2 ⎤⎞ tr( A) = tr ⎜ ⎢ ⎥ = 2+3= 5 ⎝ ⎣1 3 ⎦⎟⎠

λ1 + λ2 = 4 + 1 = 5 The results are the same, so verification is established. c. Verify that det( A) is the product of the eigenvalues of A. ⎛ ⎡ 2 2 ⎤⎞ det( A) = det ⎜ ⎢ ⎥ = 6−2 = 4 ⎝ ⎣1 3 ⎦⎟⎠

λ1 ⋅ λ2 = 4 ⋅1 = 4 The results are the same, so verification is established. 1 d. Verify that if l is an eigenvalue of A, then is an eigenvalue of A−1. l 1 ⎤ ⎡ 3 4 − 2⎥ A−1 = ⎢ , so it has characteristic equation 1 ⎥ ⎢− 1 ⎣ 4 2 ⎦ 1 ⎤⎞ ⎛ ⎡λ − 3 4 2 ⎥⎟ det(λ I − A ) = det ⎜ ⎢ =0 ⎜⎝ ⎢ 1 λ − 1 2 ⎥⎦⎟⎠ ⎣ 4 −1

The eigenvalues of A are λ1 = 4 and λ2 = 1, so Substituting λ =

1 1 = in the characteristic equation of A−1 gives λ1 4

⎛ ⎡1 − 3 ⎛1 ⎞ 4 4 det ⎜ I − A−1 ⎟ = det ⎜ ⎢ ⎝4 ⎠ 1 ⎢ ⎜⎝ ⎣ 4

Because

practice makes perfect

1 ⎤⎞ ⎛ ⎡− 1 ⎤⎞ 2 2 ⎥⎟ 1 − 1 2 ⎥⎟ = det ⎜ ⎢ = =0 1 1 1 − 1 ⎥⎟ ⎢ ⎥⎟ 8 8 ⎜ − ⎝⎣ 4 4 ⎦⎠ 4 2 ⎦⎠ 1

1 1 = satisfies the characteristic equation of A−1, verification for λ1 4

l1 = 4 is established.

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1 1 = and 1 = 1. λ1 4 λ2

Linear Algebra

Substituting λ =

−1 1 1 = = 1 in the characteristic equation of A gives λ2 1

1 ⎤⎞ ⎛ ⎡1 − 3 ⎛ ⎡1 4 4 2 ⎥ ⎢ ⎟ ⎜ = det ⎜ ⎢ det(1 ⋅ I − A ) = det 1 1 1 ⎥⎟ ⎜⎝ ⎢ ⎜⎝ ⎢ ⎣ 4 ⎣ 4 1 − 2 ⎦⎠

1 ⎤⎞ 2 ⎥⎟ 1 − 1 = =0 1 ⎥⎟ 8 8 2 ⎦⎠

−1

Because

1 = 1 satisfies the characteristic equation of A−1, verification for λ2

λ2 = 1 is established. e. Verify that if λ is an eigenvalue of A, then λ 3 is an eigenvalue of A3 . Using a graphing calculator, ⎡ 22 42 ⎤ A3 = ⎢ ⎥, so it has characteristic equation ⎣ 21 43 ⎦ ⎛ ⎡ λ − 22 −42 ⎤⎞ det(λ I − A3 ) = det ⎜ ⎢ ⎥⎟ = 0 ⎝ ⎣ −21 λ − 43 ⎦⎠

The eigenvalues of A are λ1 = 4 and λ2 = 1, so λ13 = 64 and λ23 = 1. Substituting λ13 = 64 in the characteristic equation of A3 gives −42 ⎤⎞ ⎛ ⎡ λ − 22 −42 ⎤⎞ ⎛ ⎡64 − 22 = det ⎜ ⎢ = det(λ I − A3 ) = det ⎜ ⎢ ⎥ ⎟ 64 − 43 ⎥⎦⎟⎠ ⎝ ⎣ −21 λ − 43 ⎦⎠ ⎝ ⎣ −21 ⎛ ⎡ 42 −42 ⎤⎞ det ⎜ ⎢ ⎥ = 882 − 882 = 0 ⎝ ⎣ −21 21 ⎦⎟⎠

Because λ13 = 64 satisfies the characteristic equation of A3 , verification for λ1 = 4 is established. Substituting λ23 = 1 in the characteristic equation of A3 gives ⎛ ⎡ λ − 22 −42 ⎤⎞ ⎛ ⎡1 − 22 −42 ⎤⎞ det(λ I − A3 ) = det ⎜ ⎢ ⎥⎟ = det ⎜ ⎢ −21 1 − 43 ⎥⎟ = − λ − 21 43 ⎝⎣ ⎝⎣ ⎦⎠ ⎦⎠ ⎛ ⎡ −21 −42 ⎤⎞ det ⎜ ⎢ ⎥ = 882 − 882 = 0 ⎝ ⎣ −21 −42 ⎦⎟⎠

Because λ23 = 1 satisfies the characteristic equation of A3 , verification for λ2 = 1 is established. f. Verify that A and AT have the same eigenvalues. ⎡2 1⎤ AT = ⎢ ⎥ , so it has characteristic equation ⎣ 2 3⎦

⎛ ⎡ λ − 2 −1 ⎤⎞ 2 det(λ I − AT ) = det ⎜ ⎢ ⎥⎟ = (λ − 2)(λ − 3) − 2 = λ − 5λ + 4 = 0 ⎝ ⎣ −2 λ − 3 ⎦⎠

Because the characteristic equation of AT is the same as the characteristic equation of A, A and AT have the same eigenvalues.

Tip: One way you can use properties (a) and (b) is to verify your work when finding eigenvalues as shown in the following problem. Eigenvalues and eigenvectors

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PROBLEM

⎡1 1 −1⎤ Suppose for the matrix B = ⎢⎢ 2 2 1 ⎥⎥ you found eigenvalues ⎢⎣ 2 −1 4 ⎥⎦ l1 = 1, l2 = 3 , and l3 = 3 . a. Verify that tr ( B) is the sum of the eigenvalues of B. b. Verify that det ( B) is the product of the eigenvalues of B.

SOLUTION

a. Verify that tr ( B) is the sum of the eigenvalues of B. ⎡1 1 −1⎤ tr ⎢⎢ 2 2 1 ⎥⎥ = 1 + 2 + 4 = 7 ⎢⎣ 2 −1 4 ⎥⎦ l1 + l2 + l3 = 1 + 3 + 3 = 7

The results are the same, so verification is established. b. Verify that det ( B) is the product of the eigenvalues of B. ⎛ ⎡1 1 −1⎤⎞ Using a graphing calculator, det ⎜ ⎢⎢ 2 2 1 ⎥⎥⎟ = 9 ⎜ ⎟ ⎜⎝ ⎢ 2 −1 4 ⎥⎟⎠ ⎦ ⎣ l1 ⋅ l 2 ⋅ l 3 = 1⋅ 3 ⋅ 3 = 9

The results are the same, so verification is established. Notice from this problem that you must list an eigenvalue two times if it repeats two times. In general, an eigenvalue that repeats exactly k times must be listed k times and is said to have multiplicity k.

PROBLEM

⎡ −3 0 0⎤ ⎢ ⎥ Verify that the eigenvalues of the lower triangular matrix C = ⎢ −3 2 0⎥ ⎥ ⎢ ⎢⎣ 8 − 4 9 5 ⎥⎦ are the same as its diagonal elements.

SOLUTION

⎞ ⎛ ⎡l + 3 0 0 ⎤ ⎢ ⎥⎟ ⎜ 0 ⎥⎟ = (l + 3)(l − 2)(l − 5) det(l I − C ) = det ⎜ C = ⎢ 3 l−2 ⎥⎟ ⎢ ⎜ 4 ⎟ ⎜⎝ ⎢⎣ −8 9 l − 5 ⎥⎦⎠

Thus, the characteristic equation is (l + 3)(l − 2)(l − 5) = 0 , which yields eigenvalues l 1 = −3, l 2 = 2, and l 3 = 5. The eigenvalues are the same as the diagonal elements of C, so verification is established. PROBLEM

Verify that the eigenvalues of the diagonal matrix ⎡ −1 0 ⎢ 0 −4 ⎢ D=⎢ 0 0 ⎢ ⎢0 0 ⎢⎣ 0 0

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practice makes perfect

Linear Algebra

0 0 6 0 0

0 0⎤ 0 0⎥ ⎥ 0 0 ⎥ are the same as its diagonal elements. ⎥ 2 0⎥ 0 −5 ⎥⎦

SOLUTION

0 0 0 0 ⎤ ⎡λ + 1 ⎢ 0 λ+4 0 0 0 ⎥ ⎥ ⎢ 0 λ−6 0 0 ⎥= det(λ I − D ) = ⎢ 0 ⎥ ⎢ λ−2 0 0 0 ⎥ ⎢ 0 ⎢⎣ 0 λ + 5 ⎥⎦ 0 0 0 (λ + 1)(λ + 4 )(λ − 6)(λ − 2)(λ + 5). Thus, the characteristic equation is (λ + 1)(λ + 4 )(λ − 6)(λ − 2)(λ + 5) = 0, which yields eigenvalues λ1 = −1, λ2 = −4, λ3 = 6 , λ4 = 2, and λ5 = −5 . The eigenvalues are the same as the diagonal elements of D, so verification is established.

EXERCISE

10·2 For 1–5, fill in the blank with the best response.

1. The trace of a square matrix is the ________ of its eigenvalues. 2. The determinant of a square matrix is the ________ of its eigenvalues. 3. If λ is an eigenvalue of a nonsingular matrix A, then ________ is an eigenvalue of A−1. 4. A square matrix and its transpose have the ________ eigenvalues. 5. If A is a triangular matrix, its eigenvalues are the same as the elements on its ________.

⎡2 −1⎤ ⎥ ⎣1 4 ⎦

For 6–11, let A = ⎢

6. Find the eigenvalues of A. 7. Verify that tr ( A) is the sum of the eigenvalues of A. 8. Verify that det ( A) is the product of the eigenvalues of A.

1 is an eigenvalue of A−1. λ 10. Verify that if λ is an eigenvalue of A, then λ 3 is an eigenvalue of A3 . 9. Verify that if λ is an eigenvalue of A, then

11. Verify that A and AT have the same eigenvalues. In 12–13, for each matrix and given eigenvalues, verify that the matrix’s trace is the sum of its eigenvalues.

⎡1 5⎤ 12. ⎢ ⎥ ; λ1 = −1, λ2 = 6 ⎣2 4 ⎦ ⎡ 2 1 1⎤ 13. ⎢⎢ −2 1 3 ⎥⎥ ; λ1 = 1, λ2 = −2 , λ2 = 3 ⎢⎣ 3 1 −1⎥⎦

Eigenvalues and eigenvectors

139

In 14–15, for the matrix and given eigenvalues, verify that the matrix’s determinant is the product of its eigenvalues.

⎡1 5⎤ 14. ⎢ ⎥ ; λ1 = −1, λ2 = 6 ⎣2 4 ⎦ ⎡ 2 1 1⎤ 15. ⎢⎢ −2 1 3 ⎥⎥ ; λ1 = 1, λ2 = −2 , λ3 = 3 ⎢⎣ 3 1 −1⎥⎦ For 16–20, by inspection, find the eigenvalues of the matrix.

⎡ − 4.5 16. ⎢ ⎢⎣ 0

7⎤ ⎥ 2 ⎥⎦

⎡ −40 0 ⎤ 17. ⎢ 1 3 ⎥ ⎢⎣ 5 5 ⎥⎦ ⎡ 0.25 0 ⎤ 18. ⎢ −8 ⎥⎦ ⎣ 0 0 0 ⎤ ⎡100 ⎢ 19. −50 −30 0 ⎥ ⎢ ⎥ ⎢⎣ 60 90 −10 ⎥⎦ ⎡− 1 5 ⎤ 3 −7 0 ⎥ ⎢ 3 9 5 −10 ⎥ ⎢ 0 20. ⎢ ⎥ 0 −2 1 ⎥ ⎢ 0 ⎢⎣ 0 0 0 5 ⎥⎦

Diagonalization A square matrix A of size n is similar to a diagonal matrix D if and only if A has n linearly independent eigenvectors. If so, there exists an n × n matrix P whose columns are the n linearly independent eigenvectors of A such that P −1 AP = D . In this case, A is diagonalizable and is said to be diagonalized by P. Furthermore, the corresponding eigenvalues of A are the same as the diagonal elements of D. Remark: The diagonalizing matrix P is not unique. Rearranging the columns or multiplying one or more by a nonzero scalar will yield different matrices D. If A is diagonalizable, the process of finding D = P −1 AP for A is as follows: 1. Find the eigenvalues for A. 2. Determine the corresponding linearly independent eigenvectors.

140

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Linear Algebra

3. Construct P as a matrix whose columns are the eigenvectors. 4. Calculate P −1. 5. Calculate D = P −1 AP . PROBLEM

⎡2 2⎤ Given A = ⎢ ⎥ that has eigenvalues l 1 = 4 and l 2 = 1. Find a nonsingular ⎣1 3 ⎦ matrix P such that P −1 AP is diagonal.

SOLUTION

1. Find the eigenvalues for A. The eigenvalues are λ1 = 4 and λ2 = 1. 2. Determine the corresponding linearly independent eigenvectors. ⎡2 2⎤ ⎡ x1 ⎤ When A = ⎢ and x = ⎢ ⎥ , the equation ( l I − A ) x = 0 becomes ⎥ ⎣1 3 ⎦ ⎣ x2 ⎦ ⎡ λ − 2 −2 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢ −1 λ − 3 ⎥ ⎢ x ⎥ = ⎢0 ⎥ ⎣ ⎦⎣ 2 ⎦ ⎣ ⎦ ⎡a ⎤ Let ⎢ ⎥ be an eigenvector corresponding to λ 1 = 4, then substituting into ⎣b ⎦ ⎡2

( λ I − A) x = 0 yields ⎢ −1 ⎣

−2 ⎤ ⎡a ⎤ ⎡0 ⎤ = , which has augmented matrix 1 ⎥⎦ ⎢⎣b ⎥⎦ ⎢⎣0 ⎥⎦

⎡ 2 −2 0 ⎤ ⎡1 −1 0 ⎤ ⎢ −1 1 0 ⎥ with reduced ow-echelon form ⎢0 0 0 ⎥ . So, a = t , b = t , and ⎣ ⎦ ⎣ ⎦ ⎡ a ⎤ ⎡t ⎤ ⎡1⎤ ⎢b ⎥ = ⎢t ⎥ , where t is arbitrary. Thus, any nonzero vector that has the form t ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣1⎦ ⎡1⎤ is a eigenvector corresponding to λ1 = 4 . Thus, ⎢ ⎥ is a basis for the eigenspace ⎣1⎦ of A corresponding to λ1 = 4 . ⎡c ⎤ Let ⎢ ⎥ be an eigenvector corresponding to λ2 = 1, then substituting into ⎣d ⎦ ⎡ −1 −2 ⎤ ⎡ c ⎤ ⎡0 ⎤ = , which has augmented matrix −2 ⎥⎦ ⎢⎣d ⎥⎦ ⎢⎣0 ⎥⎦ ⎣

( λ I − A) x = 0 yields ⎢ −1

⎡ −1 −2 0 ⎤ ⎡1 2 0 ⎤ ⎢ −1 −2 0 ⎥ with reduced row-echelon form ⎢0 0 0 ⎥ . So, c = −2t , d = t , and ⎣ ⎣ ⎦ ⎦ ⎡ c ⎤ ⎡ −2t ⎤ ⎢d ⎥ = ⎢ t ⎥ where t is arbitrary. Thus, any nonzero vector that has the form ⎣ ⎦ ⎣ ⎦

Eigenvalues and eigenvectors

141

⎡ −2 ⎤ ⎡ −2 ⎤ t ⎢ ⎥ is a eigenvector corresponding to λ2 = 1 . Thus, ⎢ ⎥ is a basis for the ⎣1⎦ ⎣1⎦

eigenspace of A corresponding to λ2 = 1. 3. Construct P as a matrix whose columns are the eigenvectors, being sure to check for independence first. ⎡1⎤ ⎡ −2 ⎤ The two eigenvectors ⎢ ⎥ and ⎢ ⎥ are linearly independent because ⎣1⎦ ⎣1⎦ ⎡1 −2 ⎤ neither is a scalar multiple of the other, so P = ⎢ ⎥. ⎣1 1 ⎦

4. Calculate P −1. ⎡ 1 3 P −1 = ⎢ ⎢− 1 ⎣ 3

2 ⎤ 3⎥ 1 ⎥ 3⎦

5. Calculate D = P −1 AP . ⎡ 1 3 D = P −1 AP = ⎢ ⎢− 1 ⎣ 3

2 ⎤ 2 2 1 −2 ⎤ ⎡4 0⎤ ⎤⎡ 3⎥⎡ = ⎢ ⎥ ⎢ 1 ⎥ ⎣1 3 ⎦ ⎣1 1 ⎥⎦ ⎢⎣ 0 1 ⎥⎦ 3⎦

As expected, D is diagonal and its diagonal elements are the same as the eigenvalues of A. There are many matrices that are not diagonalizable, so it’s helpful to have an idea whether a matrix is diagonalizable. If A has n distinct eigenvalues, then A is diagonalizable. On the other hand, if the number of linearly independent eigenvectors corresponding to the eigenvalues of an n × n matrix A is less than n, then A is not diagonalizable. A useful outcome of the diagonalization of matrices is that if A is diagonalizable, then it can be factored into a product PDP −1. It follows that because A = PDP −1, then A2 = PD 2 P −1; and in general, Ak = PD k P −1 , for k ≥1. This result is helpful because raising diagonal matrices to powers is rather simple as compared to doing so with nondiagonal matrices. ⎡2 2⎤ Given A = ⎢ ⎥ , use the results from the previous problem to calculate ⎣1 3 ⎦

PROBLEM

A5 = PD 5 P −1. SOLUTION 5 1 ⎡1 −2 ⎤ ⎡ 4 0 ⎤ ⎡⎢ 3 A = PD P = ⎢ ⎥⎢ ⎥ ⎣1 1 ⎦ ⎣ 0 1 ⎦ ⎣⎢ − 1 3 5

142

practice makes perfect

5

−1

Linear Algebra

2 ⎤ 1 3 ⎥ ⎡1 −2 ⎤ ⎡1024 0 ⎤ ⎡ 3 ⎢ = 1 ⎥ ⎢⎣1 1 ⎥⎦ ⎢⎣ 0 1 ⎥⎦ ⎢ − 1 3⎦ ⎣ 3

2 ⎤ 3 ⎥ ⎡342 682 ⎤ = 1 ⎥ ⎢⎣ 341 683 ⎥⎦ 3⎦

EXERCISE

10·3 For 1–3, fill in the blank with the best response.

1. A square matrix A of size n is _________ to a diagonal matrix D if and only if A has n linearly independent eigenvectors. 2. If there exists an n × n matrix P whose columns are the n linearly independent eigenvectors of A such that P −1AP = D, then the eigenvalues of A are the same as the ________ elements of D. 3. If A has n _________ eigenvalues, then A is diagonalizable.

⎡2 4 ⎤ 4. Given A = ⎢ ⎥ that has eigenvalues λ1 = 5 and λ2 = −2 with corresponding ⎣3 1 ⎦ ⎡4 ⎤ ⎡ −1⎤ eigenvectors ⎢ 3 ⎥ and ⎢ ⎥ , respectively, find a nonsingular matrix P such that ⎣1⎦ ⎢⎣ 1 ⎥⎦ P −1AP is diagonal. ⎡2 4 ⎤ 6 6 −1 5. Given A = ⎢ ⎥ , use the results from problem 4 to calculate A = PD P . 3 1 ⎣ ⎦

Eigenvalues and eigenvectors

143

Answer key

1

Systems of linear equations and matrices

1·1

1.

x+ y=3 −2 x + y = −3 Solve the first equation for y: y = 3− x Substitute into the second equation and solve for x: −2 x + (3 − x ) = −3 x=2 Then, y = 3 − x = 3 − 2 = 1

2.

5x − 2 y = 3 x + 6y =1 Solve the second equation for x: x =1− 6y Substitute into the first equation and solve for y: 5(1 − 6 y ) − 2 y = 3 1 y= 16 ⎛ 1 ⎞ 10 5 Thus, x = 1 − 6 y = 1 − 6 ⎜ ⎟ = = ⎝ 16 ⎠ 16 8

3.

x − 2y = 7 3x + 2 y = 5 Eliminate the x variable and solve for y: −3x + 6 y = −21 3x + 2 y = 5 8 y = −16 , y = −2 Substitute into the second equation and solve for x: 3x + 2(−2) = 5, x = 3

144

4.

2x − 3 y = 16 5x − 2y = −4 Eliminate the y variable and solve for x: 4x − 6 y = 32 −15x + 6 y = 12 −11x = 44 x = −4 Substitute into the first equation and solve for y: 2(−4 ) − 3 y = 16, y = −8

5.

3x − 2 y = 3 −6 x + 4 y = −8 Eliminate the x variable and solve for y. 6x − 4 y = 6 −6 x + 4 y = −8 0 = −2, but this is obviously false. There is no solution.

1·2

1. 3x − 5 y = 8 Since there is only one equation, solve for one of the variables in terms of the other. x=

8 + 5y 8 5 = + y . There are infinitely many solutions. For convenience, let y = 3t . The parametric form 3 3 3

8 is then x = + 5t , y = 3t, t arbitrary. 3 2.

2x − 3 y = 4 x − 2y = 6 Eliminate the x variable and solve for y: 2x − 3 y = 4 −2 x + 4 y = −12 y = −8. Substitute into the second equation and solve for x: x − 2(−8) = 6, x = −10

3. 2 x + 3 y − z = 5 Solve for one variable in terms of the other two: z = 2x + 3 y − 5 There are infinitely many solutions with two free variables. 4. x + 2 y + 3z + 4w = 0 Solve for one variable in terms of the other three: x = −2 y − 3z − 4w There are infinitely many solutions with three free variables. 3x − 2 y = 0 5. 6x − 4 y = 1 Eliminating either variable eliminates the other and leads to the absurd result that 0 = 1. Hence, there is no solution.

Answer key

145

1·3

1. Exhibit the 3 × 1 column vector whose elements are 1, 5, and 7, in this order. ⎡1 ⎤ ⎢5 ⎥ ⎢ ⎥ ⎢⎣7 ⎥⎦ 2. Exhibit the I 3×3 matrix. ⎡1 0 0 ⎤ ⎢0 1 0 ⎥ ⎢ ⎥ ⎢⎣0 0 1 ⎥⎦ ⎡1 2 ⎤ 3. What is the size of the matrix ⎢⎢ 0 0 ⎥⎥ ? ⎢⎣ 4 5 ⎥⎦ 3× 2 4. Exhibit the 04 × 4 matrix. ⎡0 ⎢0 ⎢ ⎢0 ⎢ ⎣0

0 0 0 0

0 0 0 0

0⎤ 0⎥ ⎥ 0⎥ ⎥ 0⎦

5. Exhibit the 1 × 3 row vector whose elements are 2, 0, and 9, in this order. [2 0 9]

1·4

1.

3x + 2 y = 4 x − 2y = 3 ⎡3 2 4 ⎤ R ↔ R ⎡1 −2 3 ⎤ −3R + R ⎡1 −2 3 ⎤ 1 R 1 1 2 ⎢3 2 4 ⎥ 2 ⎢0 8 −5 ⎥ 8 2 ⎢1 −2 3 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 7 5 x = ,y=− 4 8

2.

7 ⎤ 4⎥ ⎡1 −2 3 ⎤ 2 R + R ⎡1 0 2 1 ⎢0 1 −5 ⎥ ⎢0 1 −5 ⎥ ⎣ ⎣ 8 ⎥⎦ 8⎦

5x − y = 3 2x + 2 y = 2 ⎡5 −1 3 ⎤ 1 R ⎢ 2 2 2 ⎥ 22 ⎣ ⎦

⎡5 −1 3 ⎤ −5 R + R 2 1 ⎢1 1 1 ⎥ ⎣ ⎦

⎡0 −6 −2 ⎤ R ↔ R 1 2 ⎢1 1 1 ⎥ ⎣ ⎦

⎡1 1 1 ⎤ − 1 R 2 ⎢0 −6 −2 ⎥ ⎣ ⎦ 6

⎡1 1 1 ⎤ ⎢0 1 1 ⎥ ⎣ 3 ⎥⎦

2 ⎡1 0 3 ⎤⎥ R R − 1 + 2 1 ⎢ 0 1 1 ⎥ ⎣ 3⎦ 2 1 x= , y= 3 3 3.

3x + 2 y = 4 x + 2y = 3 1 ⎡3 2 4 ⎤ ⎡0 −4 −5 ⎤ ⎡1 2 3 ⎤ 1 ⎡1 2 3 ⎤ −2 R + R ⎡1 0 2 ⎤⎥ 3R2 + R1 ⎢ R − R2 ⎢ 1 ↔ R 2 ⎢ 2 1 ⎢ ⎥ ⎢1 2 3 ⎥ − ⎥ ⎥ 4 ⎣0 1 5 ⎥ ⎣ ⎦ ⎣1 2 3 ⎦ ⎣0 −4 −5 ⎦ ⎣0 1 5 4 ⎥⎦ 4⎦ 1 5 x= , y= 2 4

146

Answer key

x − 3y + z = 2 4. 2 x − y − 2z = 1 3x + 2 y − z = 5 ⎡ 1 −3 1 ⎢ 2 −1 −2 ⎢ ⎢⎣ 3 2 −1 1 R3 2 R 2 ↔ R 3

2⎤ ⎡1 −3 1 −2 R1 + R2 ⎢ ⎥ 1⎥ 0 5 −4 −3R1 + R3 ⎢ 5 ⎥⎦ ⎢⎣0 11 −4

⎡1 −3 1 ⎢0 3 0 ⎢ ⎢⎣0 5 −4

⎡1 0 1 1 ⎢ R2 0 1 0 3 ⎢⎢ ⎣0 0 −4 x=

3 1 3 − 14

2⎤ 5 1 ⎥⎥ − R2 + R3 3 −3 ⎥⎦ ⎤ ⎥ ⎥ − 1 R3 ⎥ 4 ⎥ 3⎦

⎡1 −3 1 2⎤ ⎥ 1R2 + R3 ⎢⎢0 5 −4 −3 ⎥ − ⎢⎣0 6 0 −1⎥⎦

⎡1 −3 1 ⎢0 3 0 ⎢ ⎢⎣0 0 −4

⎡1 0 1 ⎢0 1 0 ⎢ ⎢⎣0 0 1

2 1 14 −

2⎤ −3 ⎥⎥ 2 ⎥⎦

⎤ ⎡1 0 1 ⎥ ⎢ ⎥ 1 R2 + R1 ⎢0 3 0 ⎥ ⎢⎣0 0 −4 3 ⎥⎦

3 ⎤ ⎡1 0 0 ⎥ 1 ⎥ −1R + R ⎢0 1 0 3 1 ⎢ 3 ⎥ ⎢⎣0 0 1 ⎥ 7 6⎦

3 1 14 −

⎤ ⎥ ⎥ ⎥ 3 ⎥⎦

11 ⎤ 6⎥ 1 ⎥ 3⎥ 7 ⎥ 6⎦

11 1 7 , y= , z= 6 3 6

2x − y = 0 5. 2y − z = 0 x + 2y − z = 3 ⎡0 −1 0 ⎡ 2 −1 0 0 ⎤ ⎡ 2 −1 0 0 ⎤ ⎢ 0 2 −1 0 ⎥ −1R + R ⎢ 0 2 −1 0 ⎥ −2 R + R ⎢0 2 −1 2 3 3 ⎢ 1 ⎢ ⎢ ⎥ ⎥ ⎢⎣1 0 0 ⎢⎣ 1 2 −1 3 ⎥⎦ ⎢⎣ 1 0 0 3 ⎥⎦ ⎡1 0 0 ⎢ ↔ R R 1 3 ⎢0 0 −1 ⎢⎣0 −1 0

3 ⎤ −12 ⎥⎥ R 2 ↔ R 3 ⎥ −6 ⎦

⎡1 0 0 ⎢0 −1 0 ⎢ ⎢⎣0 0 −1

−6 ⎤ 0 ⎥⎥ 2 R1 + R2 3 ⎥⎦

3 ⎤ 1R2 and − 1R3 −6 ⎥⎥ − ⎥ −12 ⎦

⎡0 −1 0 ⎢0 0 −1 ⎢ ⎢⎣1 0 0

−6 ⎤ −12 ⎥⎥ 3 ⎥⎦

⎡1 0 0 3 ⎤ ⎢0 1 0 6 ⎥ ⎢ ⎥ ⎢⎣0 0 1 12 ⎥⎦

x = 3, y = 6, z = 12

1·5

1.

x+ y=4 2 x + 3 y = 10 ⎡1 1 4 ⎤ ⎡1 1 4 ⎤ 2 R1 + R2 ⎢ 1R2 + R1 ⎡1 0 2 ⎤ ⎢ 2 3 10 ⎥ − ⎥− ⎢0 1 2 ⎥ ⎣ ⎦ ⎣0 1 2 ⎦ ⎣ ⎦ x= y=2

2.

2x − 4 y = 3 x + 2y = 4 ⎡ 2 −4 3 ⎤ ⎡0 −8 −5 ⎤ 1 2 R2 + R1 ⎢ ⎢1 2 4 ⎥ − ⎥ − 8 R1 ⎣ ⎦ ⎣1 2 4 ⎦ x=

5⎤ ⎡0 1 ⎥ 2 R1 + R2 ⎢1 2 8 ⎥ − ⎣ 4⎦

11 ⎤ 5⎤ 1 0 ⎥ 0 1 ⎡ ⎡ 4⎥ 8 ⎥ 1 ↔ R 2 ⎢ ⎢1 0 11 ⎥ R ⎣0 1 5 ⎥ ⎣ ⎥ 8 ⎥⎦ 4 ⎥⎦

11 5 ,y= 4 8

Answer key

147

x + y + 2z = 9 3. 2 x + 4 y − 3z = 1 3x + 6 y − 5z = 0 ⎡1 1 2 ⎢ 2 4 −3 ⎢ ⎢⎣ 3 6 −5

9⎤ ⎡1 1 2 1 ⎥⎥ −2 R1 + R2 ⎢⎢0 2 −7 3R1 + R3 ⎢0 3 −11 0 ⎥⎦ − ⎣

⎡1 1 0 −2 R2 + R1 ⎢ 0 0 1 R2 + R3 ⎢ 4 ⎢⎣0 1 0

9 ⎤ ⎡1 1 2 ⎢ 1 R + R −17 ⎥⎥ − 2 3 ⎢0 2 −7 ⎢⎣0 1 −4 −27 ⎥⎦

3⎤ ⎡1 0 0 ⎢ 1 R + R 3 ⎥⎥ − 3 1 ⎢0 0 1 ⎢⎣0 1 0 2 ⎥⎦

1⎤ 3 ⎥⎥ R 2 ↔ R 3 2 ⎥⎦

⎡1 0 0 ⎢0 1 0 ⎢ ⎢⎣0 0 1

9 ⎤ 2 R3 + R2 −17 ⎥⎥ − −10 ⎥⎦

⎡1 1 2 ⎢0 0 1 ⎢ ⎢⎣0 1 −4

9 ⎤ 3 ⎥⎥ −10 ⎥⎦

1⎤ 2 ⎥⎥ 3 ⎥⎦

x = 1, y = 2, z = 3 4.

5 x − 2 y + 6z = 0 −2 x + y + 3z = 1 ⎡ 5 −2 6 ⎢ −2 1 3 ⎣

0⎤ R2 + R1 2 1 ⎥⎦

⎡ 1 0 12 ⎢ −2 1 3 ⎣

2⎤ R1 + R2 2 1 ⎥⎦

⎡1 0 12 ⎢0 1 27 ⎣

2⎤ 5 ⎥⎦

x = 2 − 12z y = 5 − 27 z , and z is a free variable

5.

6a + 6b + 3c = 5 −2b + 3c = 1 3a + 6b − 3c = −2 ⎡6 6 3 ⎢0 −2 3 ⎢ ⎢⎣ 3 6 −3

⎡0 −6 9 5⎤ ⎢0 −2 3 2 + R R 1 ⎥⎥ − 3 1 ⎢ ⎢⎣ 3 6 −3 −2 ⎥⎦

9⎤ 1 1 ⎥⎥ − R1 3 −2 ⎥⎦

⎡0 2 −3 ⎢0 −2 3 ⎢ ⎢⎣ 3 6 −3

−3 ⎤ ⎡0 0 0 1 ⎥⎥ 1 R2 + R1 ⎢⎢0 −2 3 ⎢⎣ 3 6 −3 −2 ⎥⎦

No solution. 2x + 2 y + 4z = 2 − 3z = −9 6. −3x 6 x + 3 y + 9z = 15

⎡2 2 4 ⎢ −3 0 −3 ⎢ ⎢⎣ 6 3 9

1 R1 2⎤2 ⎡1 1 2 1 −9 ⎥⎥ − R2 ⎢ 1 0 1 3 ⎢ ⎢⎣ 2 1 3 15 ⎥⎦ 1 R3 3

No solution. ⎡1 0 0 7. ⎢⎢0 1 0 ⎢⎣0 0 1

−3 ⎤ 0 ⎥⎥ 7 ⎥⎦

x = −3, y = 0, z = 7

148

Answer key

1⎤ ⎡1 1 2 1 ⎤ ⎡1 1 2 −1R1 + R2 ⎢ ⎥ 0 − 1 − 1 2 3 ⎥⎥ − 1R2 + R3 ⎢⎢0 −1 −1 ⎢ ⎥ −2 R1 + R3 ⎢ ⎢⎣0 0 0 5 ⎥⎦ ⎣0 −1 −1 3 ⎥⎦

1⎤ 2 ⎥⎥ 1 ⎥⎦

−2 ⎤ 1 ⎥⎥ −2 ⎥⎦

⎡1 −3 0 8. ⎢⎢0 0 1 ⎢⎣0 0 0

0⎤ 0 ⎥⎥ 1 ⎥⎦

No solution. ⎡1 0 0 −7 8 ⎤ ⎢ 9. ⎢0 1 0 3 2 ⎥⎥ ⎢0 0 1 1 −5 ⎥⎦ ⎣ x = 8 + 7t y = 2 − 3t z = −5 − t , and t is a free variable. ⎡1 −6 ⎢0 0 10. ⎢ ⎢0 0 ⎢ ⎢⎣0 0

0 1 0 0

0 0 1 0

3 4 5 0

−2 ⎤ 7⎥ ⎥ 8⎥ ⎥ 0⎦

t = 8 − 5s z = 7 − 4s x = −2 + 6 y − 3s , and y and s are free variables.

1·6

x + 2 y − 3z = 0 1. 2 x + 5 y + 2z = 0 3x − y − 4 z = 0 ⎡ 1 2 −3 ⎤ ⎢ 2 5 2 ⎥ −2 R1 + R2 ⎢ ⎥ −3R + R 1 3 ⎢⎣ 3 −1 −4 ⎥⎦

⎡1 2 −3 ⎤ ⎢0 1 8 ⎥ R2 + R3 ⎢ ⎥ 7 ⎢⎣0 −7 5 ⎥⎦

⎡1 2 −3 ⎤ ⎢0 1 8 ⎥ ⎢ ⎥ ⎢⎣0 0 61 ⎥⎦

There is no nontrivial solution. x − 3 y − 4z + w = 0 2. 2 x − 5 y − 2z + 3w = 0 x − y − z − 2w = 0 There are more unknowns than equations, so there is a nontrivial solution, in fact, infinitely many. 2 x + 4 y − 2z = 0 2 x + 5 y + 2z = 0 3. − x − 4 y − 7z = 0 3x + 9 y + 9z = 0 ⎡1 2 −1 ⎤ ⎡ 2 4 −2 ⎤ 1 ⎡ 1 2 −1 ⎤ −2 R1 + R2 ⎢ R1 ⎢ ⎥ ⎥ ⎢2 5 0 1 4 ⎥ 2 R2 + R3 2 5 2 2 2 ⎥ ⎢ ⎥ 1R1 + R3 ⎢ ⎢ ⎥ ⎢0 −2 −8 ⎥ − ⎢ −1 −4 −7 ⎥ 1 ⎢ −1 −4 −7 ⎥ 1R2 + R4 1R1 + R4 ⎢ ⎥ ⎥− ⎢ ⎥ 3 R4 ⎢ ⎣0 1 4 ⎦ ⎣ 3 9 9 ⎦ ⎣ 1 3 3 ⎦

⎡1 ⎢0 ⎢ ⎢0 ⎢ ⎣0

2 −1⎤ ⎡1 ⎢ 1 4⎥ ⎥ −2 R2 + R1 ⎢0 0 0⎥ ⎢0 ⎥ ⎢ 0 0⎦ ⎣0

0 −9 ⎤ 1 4⎥ ⎥ 0 0⎥ ⎥ 0 0⎦

There are infinitely many nontrivial solutions.

Answer key

149

4.

3x + 2 y = 0 5x − 4 y = 0 ⎡3 2 ⎤ R1 + R2 ⎢5 −4 ⎥ 2 ⎣ ⎦

⎡ 3 2⎤ 1 ⎢11 0 ⎥ 11 R2 ⎣ ⎦

⎡3 2 ⎤ −3R + R ⎡0 2 ⎤ 2 1 ⎢ ⎢1 0 ⎥ ⎥ ⎣ ⎣1 0 ⎦ ⎦

There is no nontrivial solution.

5. Solve the system:

x + y + 2z =0 y − 3z + w = 0 3x + y + z + 2w = 0 x + 3 y − 2z − 2w = 0

⎡1 ⎢0 ⎢ ⎢3 ⎢ ⎣1

1 2 0⎤ 1 −3 1 ⎥ −3R1 + R3 ⎥ 1R1 + R4 1 1 2 ⎥− ⎥ 3 −2 −2 ⎦

1 1 0 0

2 0 ⎤ 0 −5 ⎥ 1 ⎥− R 3 0 −18 ⎥ 18 ⎥ 1 −2 ⎦

⎡1 ⎢0 ⎢ ⎢0 ⎢ ⎣0

⎡1 ⎢0 ⎢ ⎢0 ⎢ ⎣0

1 1 0 0

2 0⎤ ⎡1 ⎡1 1 ⎢0 1 −3 1 ⎥ 2 R + R ⎢0 3 ⎢ ⎢ ⎥ 2 R + R 2 ⎢0 −2 −5 2 ⎥ − 2 4 ⎢0 ⎢ ⎢ ⎥ ⎣0 ⎣0 2 −4 −2 ⎦ 2 0⎤ ⎡1 0 −5 ⎥ 2 R3 + R4 ⎢0 ⎥ ⎢ R3 + R2 ⎢0 0 1 ⎥ 5 ⎥ ⎢ 1 −2 ⎦ ⎣0

1 1 0 0

2 0 0 1

1 2 0⎤ ⎡1 ⎥ 1 −3 1 1 ⎢ 0 ⎥ R ⎢ 0 −11 4 ⎥ 24 ⎢ 0 ⎥ ⎢ 0 2 −4 ⎦ ⎣0

0⎤ ⎡1 0 ⎥ −1R2 + R1 ⎢0 ⎥ ⎢ 2 R4 + R1 ⎢0 1⎥ − ⎥ ⎢ 0⎦ ⎣0

0 1 0 0

The only solution is the trivial one, x = y = z = w = 0.

2

Matrix algebra

2·1

⎡ 2 8 ⎤ ⎡ 5 4 ⎤ ⎡ 2 8 ⎤ ⎡ 15 12 ⎤ ⎡17 20 ⎤ 1. ⎢ ⎥=⎢ ⎥ + 3⎢ ⎥ ⎥=⎢ ⎥+⎢ ⎣ −3 0 ⎦ ⎣ −2 3 ⎦ ⎣ −3 0 ⎦ ⎣ −6 9 ⎦ ⎣ −9 9 ⎦ ⎡ 3 6 0 ⎤ ⎡ −1 6 4 ⎤ ⎡ 3 6 0 ⎤ ⎡ 1 −6 −4 ⎤ ⎡ 4 0 −4 ⎤ ⎥ ⎢ ⎥ ⎢ 2. ⎢⎢ 4 −2 −5 ⎥⎥ − ⎢⎢ −4 3 −1⎥⎥ = ⎢ 4 −2 −5 ⎥ + ⎢ 4 −3 1 ⎥ = ⎢⎢ 8 −5 −4 ⎥⎥ ⎢⎣ 6 0 −1 ⎥⎦ ⎢⎣ 5 4 2 ⎥⎦ ⎢⎣ 6 0 −1 ⎥⎦ ⎢⎣ −5 −4 −2 ⎥⎦ ⎢⎣ 1 −4 −3 ⎥⎦ ⎡ 2 4 ⎤ 2 − 18 + 15 4 + 6 ⎤ ⎡ −1 10 ⎤ ⎡1 6 3 ⎤ ⎢ ⎥=⎡ = 3. ⎢ − 3 1 ⎢ ⎥ ⎢ ⎥ ⎣ −15 + 20 5 ⎥⎦ ⎢⎣ 5 5 ⎥⎦ ⎣0 5 4 ⎦ ⎢ ⎥ 5 0 ⎣ ⎦ ⎡1 0 ⎤ ⎡ 3 6 ⎤ ⎡ 3 6 ⎤ 4. ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣0 1 ⎦ ⎣ −4 5 ⎦ ⎣ −4 5 ⎦ ⎡0 0 ⎤ ⎡ 5 6 ⎤ ⎡0 0 ⎤ 5. ⎢ ⎥ ⎥⎢ ⎥=⎢ ⎣0 0 ⎦ ⎣ 4 9 ⎦ ⎣0 0 ⎦ ⎡0 1 ⎤ ⎡3 7 ⎤ ⎡1 5 ⎤ 6. If A = ⎢ ,B = ⎢ , and C = ⎢ ⎥ ⎥ ⎥ , show that A( BC ) = ( AB)C . ⎣0 2 ⎦ ⎣1 0 ⎦ ⎣2 8⎦ ⎡0 1 ⎤ ⎛ ⎡3 7 ⎤ ⎡ 1 5 ⎤⎞ ⎡0 1 ⎤ ⎡17 71⎤ ⎡ 1 5 ⎤ A( BC ) = ⎢0 2 ⎥ ⎜ ⎢1 0 ⎥ ⎢ 2 8 ⎥⎟ = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣ ⎦⎝ ⎣ ⎦⎣ ⎦⎠ ⎣0 2 ⎦ ⎣ 1 5 ⎦ ⎣ 2 10 ⎦

150

Answer key

0 0 0 1

1 2 0⎤ 1 −3 1 ⎥ 11R4 + R3 ⎥ R4 + R2 0 −11 4 ⎥ 3 ⎥ 0 1 −2 ⎦ 0⎤ 0⎥ ⎥R ↔R 3 4 1 ⎥ ⎥ 0⎦

⎡1 ⎢0 ⎢ ⎢0 ⎢ ⎣0

0 1 0 0

0 0 1 0

0⎤ 0⎥ ⎥ 0⎥ ⎥ 1⎦

⎛ ⎡0 1 ⎤ ⎡3 7 ⎤⎞ ⎡1 5 ⎤ ⎡1 0 ⎤ ⎡1 5 ⎤ ⎡1 5 ⎤ ( AB)C = ⎜ ⎢ = ⎥⎢ ⎥ ⎢ ⎥= ⎝ ⎣0 2 ⎦ ⎣1 0 ⎦⎟⎠ ⎣ 2 8 ⎦ ⎢⎣ 2 0 ⎥⎦ ⎢⎣ 2 8 ⎥⎦ ⎢⎣ 2 10 ⎥⎦ Thus, A( BC ) = ( AB)C . ⎡1 5 ⎤ ⎡ 2 −3 ⎤ 7. If A = ⎢ ⎥ and B = ⎢5 −4 ⎥ , show that tr( A + B) = tr( A) + tr( B) . 2 8 ⎣ ⎦ ⎣ ⎦ ⎡3 2 ⎤ ⎡1 5 ⎤ ⎡ 2 −3 ⎤ A+ B= ⎢ + tr ⎢ . Thus, tr( A + B) = 7 and tr(A) + tr( B) = tr ⎢ ⎥ ⎥ ⎥ = 9 − 2 = 7. ⎣7 4 ⎦ ⎣2 8⎦ ⎣5 −4 ⎦ Hence, tr( A + B) = tr( A) + tr( B) = 7. ⎡1 5 ⎤ 8. For A = ⎢ ⎥ , compute (3x − y )A. ⎣2 8⎦ ⎡ 1 5 ⎤ ⎡ 3x − y 15 x − 5 y ⎤ (3 x − y ) A = (3 x − y ) ⎢ ⎥=⎢ ⎥ ⎣ 2 8 ⎦ ⎣6 x − 2 y 24 x − 8y ⎦ ⎡ 2 1⎤ ⎡1 −2 ⎤ and B = ⎢ 9. For A = ⎢ ⎥ , compute 3 A + 3B . ⎥ ⎣ −2 3 ⎦ ⎣3 4 ⎦ ⎡1 −2 ⎤ ⎡ 2 1 ⎤ ⎡ 3 −6 ⎤ ⎡ 6 3 ⎤ ⎡9 −3 ⎤ 3 A + 3B = 3 ⎢ ⎥=⎢ ⎥ + 3⎢ ⎥=⎢ ⎥ ⎥+⎢ ⎣3 4 ⎦ ⎣ −2 3 ⎦ ⎣9 12 ⎦ ⎣ −6 9 ⎦ ⎣ 3 21 ⎦ ⎡ 2 1⎤ ⎡1 −2 ⎤ 10. If A = ⎢ and B = ⎢ ⎥ , prove that A + B = B + A. ⎥ ⎣ −2 3 ⎦ ⎣3 4 ⎦ −2 + 1⎤ ⎡3 −1⎤ ⎡1 −2 ⎤ ⎡ 2 1 ⎤ ⎡ 1 + 2 A+ B = ⎢ ⎥ + ⎢ −2 3 ⎥ = ⎢3 + (−2) 4 + 3 ⎥ = ⎢1 7 ⎥ 3 4 ⎦ ⎣ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ 2 1 ⎤ ⎡1 −2 ⎤ ⎡ 2 + 1 1 + (−2)⎤ ⎡3 −1⎤ B+ A = ⎢ ⎥+⎢ ⎥=⎢ ⎥ ⎥=⎢ ⎣ −2 3 ⎦ ⎣3 4 ⎦ ⎣ −2 + 3 3 + 4 ⎦ ⎣1 7 ⎦ ⎡3 −1⎤ Thus, A + B = B + A = ⎢ ⎥. ⎣1 7 ⎦

2·2

1 ⎤ ⎡ 2 −1 ⎢ − − − 2 − 1 2 1 4 1 ⎡ ⎤ ⎡ ⎤ 1 1 ⎡ ⎤ 5 5 ⎥ =⎢ = ⎢ 1. A−1 = ⎢ = ⎥ ⎥ ⎢ ⎥ ⎥ 4 −8 + 3 ⎣ 3 4 ⎦ −5 ⎣ 3 4 ⎦ ⎢ 3 ⎣ −3 −2 ⎦ − − ⎥ ⎢⎣ 5 5 ⎥⎦ ⎡ −1 ⎢ −2 − 4 − 1 4 1 1 1 − ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 1 2. B −1 = ⎢ = =⎢ = ⎢ ⎥ ⎢ ⎥ ⎥ −4 + 2 ⎣ 2 −1⎦ −2 ⎣ 2 −1⎦ ⎢ ⎣ −2 4 ⎦ −1 ⎢⎣

1⎤ 2⎥ ⎥ 1⎥ 2 ⎥⎦

4⎤ ⎡ 11 −1 −1 − − ⎥ ⎛ ⎡ 4 1 ⎤ ⎡ −1 1 ⎤⎞ ⎛ ⎡ −6 8 ⎤⎞ 1 ⎡ −11 −8 ⎤ ⎢ 10 5 3. ( AB) = ⎜ ⎢ ⎥ ⎥⎢ ⎥ = ⎜ ⎢ 7 −11⎥⎟ = 10 ⎢ −7 −6 ⎥ = ⎢ 7 3 ⎝ ⎣ −3 −2 ⎦ ⎣ −2 4 ⎦⎟⎠ ⎝⎣ ⎦⎠ ⎣ ⎦ ⎢− − ⎥ ⎢⎣ 10 5 ⎥⎦ −1

1 ⎤⎡ ⎡ 2 −2 ⎢ 5 5 ⎥⎢ 4. A B = ⎢ ⎥⎢ ⎢ − 3 − 4 ⎥ ⎢ −1 ⎢⎣ 5 5 ⎥⎦ ⎢⎣ −1

−1

3 ⎤ 1⎤ ⎡ −1 10 ⎥ 2⎥ ⎢ ⎥ ⎥=⎢ 1⎥ ⎢ 7 2 − ⎥ 2 ⎥⎦ ⎢⎣ 10 ⎥⎦ Answer key

151

⎡ ⎢ −2 5. B A = ⎢ ⎢ −1 ⎢⎣ −1

−1

1 ⎤ ⎡ 11 4⎤ 1⎤⎡ 2 − − ⎥ 5 ⎥ ⎢ 10 5 2⎥⎢ 5 ⎥=⎢ ⎥ ⎥⎢ 1⎥⎢ 3 4 7 3 − − ⎥ ⎢− − ⎥ 2 ⎥⎦ ⎢⎣ 5 5 ⎥⎦ ⎢⎣ 10 5 ⎥⎦

⎛ ⎡ 4 1 ⎜ ⎢ −2 ⎡ ⎤ 6. A(− B −1 ) = ⎢ ⎜−⎢ ⎥ ⎣ −3 −2 ⎦ ⎜ ⎢ −1 ⎝ ⎢⎣

1⎤ 5 1 ⎤⎞ ⎡ 2 − ⎥ ⎡9 − ⎤ ⎢ 4 1 ⎥ ⎢ ⎟ ⎡ ⎤ 2 2⎥ 2 ⎥=⎢ ⎥⎟ = ⎢ ⎥ ⎥⎢ 1⎥ ⎢ 5 ⎥ 1 ⎥⎟ ⎣ −3 −2 ⎦ ⎢ 1 − − 8 ⎢⎣ 2 ⎥⎦ ⎢⎣ 2 ⎥⎦ 2 ⎥⎦⎠

⎛⎡ 4 1 ⎜ ⎢ −2 ⎡ ⎤ −1 7. − A( B ) = − ⎢ ⎜⎢ ⎥ ⎣ −3 −2 ⎦ ⎜ ⎢ −1 ⎝ ⎢⎣

1 ⎤⎞ ⎡ −2 2 ⎥⎟ ⎡ −4 −1⎤ ⎢ ⎥⎟ = ⎢ ⎥⎢ 1 ⎥⎟ ⎣ 3 2 ⎦ ⎢ −1 ⎢⎣ 2 ⎥⎦⎠

1⎤ ⎡ 5⎤ 9 − ⎥ 2⎥ ⎢ 2 ⎥= ⎥ 1⎥ ⎢ ⎢ −8 5 ⎥ 2 ⎥⎦ ⎢⎣ 2 ⎥⎦

8. ( AB)−1 = B −1 A−1 9. A(− B −1 ) = − A( B −1 ) 10. ( AB)−1 ≠ A−1 B −1

2·3

−1 ⎡1 5 ⎤ ⎛1 ⎞ 1. For A = ⎢ , find ⎜ A⎟ . ⎥ ⎝5 ⎠ ⎣2 8⎦ −1 ⎛ 1 ⎡ 8 −5 ⎤⎞ 5 ⎡ 8 −5 ⎤ ⎛1 ⎞ −1 ⎜⎝ A⎟⎠ = 5 A = 5 ⎜ − ⎢ ⎥ = − 2 ⎢ −2 1 ⎥ 5 ⎝ 2 ⎣ −2 1 ⎦⎟⎠ ⎣ ⎦

⎡1 5 ⎤ −2 2. For A = ⎢ ⎥ , compute A . 2 8 ⎣ ⎦ ⎛ 1 ⎡ 8 −5 ⎤⎞ ⎛ 1 ⎡ 8 −5 ⎤⎞ 1 ⎡ 74 −45 ⎤ A−2 = ⎜ − ⎢ ⎥ − ⎢ ⎥ = ⎢ ⎥ ⎝ 2 ⎣ −2 1 ⎦⎟⎠ ⎜⎝ 2 ⎣ −2 1 ⎦⎟⎠ 4 ⎣ −18 11 ⎦ ⎡x ⎤ ⎡1 5 ⎤ ⎡ x ⎤ ⎡ 2 ⎤ 3. For ⎢ = ⎢ ⎥ , use the inverse matrix to solve for ⎢ ⎥ ⎥ ⎢ ⎥ ⎣y⎦ ⎣2 8⎦ ⎣ y ⎦ ⎣3⎦ 1⎡ 1 ⎤ ⎡x ⎤ 1 ⎡ 8 −5 ⎤ ⎡ 2 ⎤ ⎢ y ⎥ = − 2 ⎢ −2 1 ⎥ ⎢ 3 ⎥ = − 2 ⎢ −1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ x=−

1 1 and y = . 2 2

⎡ 2 1⎤ ⎡1 −2 ⎤ −1 −1 −1 4. For A = ⎢ ⎥ and B = ⎢ −2 3 ⎥ , show that ( AB) = B A . 3 4 ⎣ ⎦ ⎣ ⎦ −1

−1

⎛ ⎡1 −2 ⎤ ⎡ 2 1 ⎤⎞ ⎛ ⎡ 6 −5 ⎤⎞ 1 ⎡15 5 ⎤ ( AB) = ⎜ ⎢ ⎥ ⎢ −2 3 ⎥⎟ = ⎜ ⎢ −2 15 ⎥⎟ = 80 ⎢ 2 6 ⎥ 3 4 ⎝⎣ ⎝⎣ ⎣ ⎦ ⎦⎣ ⎦⎠ ⎦⎠ −1

⎛ 1 ⎡ 3 −1⎤⎞ ⎛ 1 ⎡ 4 2 ⎤⎞ 1 ⎡15 5 ⎤ B −1 A−1 = ⎜ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎝ 8 ⎣ 2 2 ⎦⎟⎠ ⎜⎝ 10 ⎣ −3 1 ⎦⎟⎠ 80 ⎣ 2 6 ⎦ Thus, ( AB)−1 = B−1 A−1 . ⎡3 2 ⎤ −1 −1 5. For A = ⎢ ⎥ , show that ( A ) = A. 5 4 ⎣ ⎦ −1

⎛ 1 ⎡3 2 ⎤⎞ ⎡3 2 ⎤ ⎛ 1 ⎡ 4 −2 ⎤⎞ (A ) = ⎜ ⎢ ⎥ ⎟ = 2 ⎜ 2 ⎢5 4 ⎥ ⎟ = ⎢5 4 ⎥ = A 5 3 − 2 ⎝ ⎣ ⎝ ⎣ ⎦⎠ ⎣ ⎦ ⎦⎠ −1 −1

152

Answer key

⎡1 5 ⎤ −2 −1 6. For A = ⎢ ⎥ , show that A A = A (see Prob. 2 above). 2 8 ⎣ ⎦ ⎛ 1 ⎡ 74 −45 ⎤⎞ ⎡ 1 5 ⎤ 1 ⎡ 74 − 90 370 − 360 ⎤ 1 ⎡ −16 10 ⎤ 1 ⎡ 8 −5 ⎤ = ⎢ =− ⎢ = ⎢ A−2 A = ⎜ ⎢ ⎥ ⎥ ⎥ ⎢ ⎥ ⎟ 2 ⎣ −2 1 ⎥⎦ ⎝ 4 ⎣ −18 11 ⎦⎠ ⎣ 2 8 ⎦ 4 ⎣ −18 + 22 −90 + 88 ⎦ 4 ⎣ 4 −2 ⎦ = A−1 ⎡3 2 ⎤ −1 7. For A = ⎢ ⎥ , show that AA = I . 5 4 ⎣ ⎦ ⎡3 2 ⎤ ⎛ 1 ⎡ 4 −2 ⎤⎞ 1 ⎡ 2 0 ⎤ ⎡1 0 ⎤ AA−1 = ⎢ ⎥=⎢ ⎥⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎣5 4 ⎦ ⎝ 2 ⎣ −5 3 ⎦⎠ 2 ⎣ 0 2 ⎦ ⎣0 1 ⎦ ⎡2 0⎤ 2 8. For A = ⎢ ⎥ , compute A − 2 A + I . 4 1 ⎣ ⎦ ⎡ 2 0 ⎤ ⎡ 2 0 ⎤ ⎡ 2 0 ⎤ ⎡1 0 ⎤ ⎡ 4 0 ⎤ ⎡ 4 0 ⎤ ⎡1 0 ⎤ ⎡ 1 0 ⎤ A2 − 2 A + I = ⎢ ⎥⎢ ⎥ − 2⎢ ⎥+⎢ ⎥=⎢ ⎥−⎢ ⎥+⎢ ⎥=⎢ ⎥ ⎣ 4 1 ⎦ ⎣ 4 1 ⎦ ⎣ 4 1 ⎦ ⎣0 1 ⎦ ⎣12 1 ⎦ ⎣ 8 2 ⎦ ⎣0 1 ⎦ ⎣ 4 0 ⎦ ⎡ cosθ 9. Find the inverse of ⎢ ⎣ − sinθ ⎛ ⎡ cosθ ⎜⎝ ⎢ − sinθ ⎣

sinθ ⎤ . cosθ ⎥⎦

−1

sinθ ⎤⎞ ⎡cosθ 1 = 2 2 ⎥ ⎟ cos θ + sin θ ⎢⎣ sinθ cosθ ⎦⎠

− sinθ ⎤ ⎡cosθ = cosθ ⎥⎦ ⎢⎣ sinθ

− sinθ ⎤ cosθ ⎥⎦

10. Using matrix properties, prove that if ABA −1 = C , then B = A−1CA. Proof: Given ABA −1 = C Multiply the expression on each side of the equation on the left by A−1 and the right by A. A−1 ( ABA −1 )A = A−1 (C )A Using the associative law for multiplication, ( A −1 A)B( A −1 A) = A−1CA Using the property of inverses, IBI = A−1CA Using the identity property, B = A−1CA Thus, if ABA−1 = C , then B = A−1CA.

2·4

2 3⎤ 1. ⎡⎢ ⎥ ⎣ −1 4 ⎦

⎡ 2 3 ⎤ ⎡a b ⎤ ⎡ 2a + 3c 2b + 3d ⎤ ⎡1 0 ⎤ Using Method 1, AA−1 = ⎢ ⎥ ⎥⎢ ⎥=⎢ ⎥=⎢ ⎣ −1 4 ⎦ ⎣ c d ⎦ ⎣ −a + 4c −b + 4d ⎦ ⎣0 1 ⎦

2a + 3c = 1 2b + 3d = 0 4 3 1 2 and . Solving the equations gives a = , b = − , c = , and d = 11 11 11 11 − a + 4c = 0 −b + 4 d = 1 Thus, A−1 =

1 ⎡ 4 −3 ⎤ . 11 ⎢⎣ 1 2 ⎥⎦

Answer key

153

⎡1 1 ⎤ 2. ⎢ ⎥ ⎣1 2 ⎦ Using Method 2. 1 ⎡ 2 −1⎤ ⎡ 2 −1⎤ A−1 = ⎢ = 1 ⎣ −1 1 ⎥⎦ ⎢⎣ −1 1 ⎥⎦ ⎡ 2 −2 ⎤ 3. ⎢ ⎥ ⎣ 1 −1 ⎦ Using Method 2, ad − bc = ( 2 ⋅−1) − (1 ⋅−2 ) = −2 + 2 = 0, so the matrix does not have an inverse. 4.

3x + 2 y = 4 x − 2y = 3 ⎡3 2 ⎤ ⎡ x ⎤ ⎡ 4 ⎤ ⎢1 −2 ⎥ ⎢ y ⎥ = ⎢ 3 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ −1

⎡ x ⎤ ⎛ ⎡3 2 ⎤⎞ ⎡ 4 ⎤ ⎛ 1 ⎡ −2 −2 ⎤⎞ ⎡ 4 ⎤ 1 ⎡ −14 ⎤ ⎢ y ⎥ = ⎜ ⎢1 −2 ⎥⎟ ⎢ 3 ⎥ = ⎜ − 8 ⎢ −1 3 ⎥⎟ ⎢ 3 ⎥ = − 8 ⎢ 5 ⎥ ⎣ ⎦⎠ ⎣ ⎦ ⎦ ⎣ ⎦⎠ ⎣ ⎦ ⎝ ⎣ ⎦ ⎝⎣ x=

5.

7 5 and y = − 4 8

5x − y = 3 2x + 2 y = 2 −1

⎡ x ⎤ ⎛ ⎡5 −1⎤⎞ ⎡ 3 ⎤ ⎛ 1 ⎡ 2 1 ⎤⎞ ⎡ 3 ⎤ 1 ⎡ 8 ⎤ ⎢ y ⎥ = ⎜ ⎢ 2 2 ⎥⎟ ⎢ 2 ⎥ = ⎜⎝ 12 ⎢ −2 5 ⎥⎟⎠ ⎢ 2 ⎥ = 12 ⎢ 4 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎝⎣ ⎦⎠ ⎣ ⎦ x=

2·5

2 1 and y = 3 3

⎡0 4 ⎤ ⎡3 2 ⎤ 1. Show that ( AB)T = BT AT for A = ⎢ ⎥ and B = ⎢ 2 −3 ⎥ . 0 5 ⎣ ⎦ ⎣ ⎦ T

T

⎛ ⎡ 3 2 ⎤ ⎡ 0 4 ⎤⎞ 6 ⎤⎞ ⎛⎡ 4 ⎡ 4 10 ⎤ ( AB) = ⎜ ⎢ ⎥ ⎢ 2 −3 ⎥⎟ = ⎜ ⎢10 −15 ⎥⎟ = ⎢ 6 −15 ⎥ 0 5 ⎝⎣ ⎝⎣ ⎦⎣ ⎦⎠ ⎣ ⎦ ⎦⎠ T

T T ⎡ 0 2 ⎤ ⎡ 3 0 ⎤ ⎡ 4 10 ⎤ ⎛ ⎡ 0 4 ⎤⎞ ⎛ ⎡ 3 2 ⎤⎞ =⎢ BT AT = ⎜ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎥ ⎢ ⎥ ⎟ ⎜ ⎟ ⎣ 4 −3 ⎦ ⎣ 2 5 ⎦ ⎣ 6 −15 ⎦ ⎝ ⎣ 2 −3 ⎦⎠ ⎝ ⎣0 5 ⎦⎠

Thus, ( AB)T = BT AT . ⎡3 2 ⎤ ⎡0 4 ⎤ 2. Show that ( A + B)T = AT + BT for A = ⎢ ⎥ and B = ⎢ 2 −3 ⎥ . 0 5 ⎣ ⎦ ⎣ ⎦ T

T

⎛ ⎡ 3 2 ⎤ ⎡ 0 4 ⎤⎞ ⎛ ⎡ 3 6 ⎤⎞ ⎡3 2 ⎤ ( A + B) = ⎜ ⎢ ⎥ + ⎢ 2 −3 ⎥⎟ = ⎜ ⎢ 2 2 ⎥⎟ = ⎢6 2 ⎥ 0 5 ⎝⎣ ⎝⎣ ⎦⎠ ⎣ ⎦ ⎦ ⎣ ⎦⎠ T

154

Answer key

T

T

⎡ 3 0 ⎤ ⎡ 0 2 ⎤ ⎡3 2 ⎤ ⎛ ⎡ 3 2 ⎤⎞ ⎛ ⎡ 0 4 ⎤⎞ AT + BT = ⎜ ⎢ +⎜⎢ =⎢ ⎥=⎢ ⎥+⎢ ⎥ ⎥ ⎥ ⎟ ⎟ ⎝ ⎣0 5 ⎦⎠ ⎝ ⎣ 2 −3 ⎦⎠ ⎣ 2 5 ⎦ ⎣ 4 −3 ⎦ ⎣6 2 ⎦ Thus, ( A + B)T = AT + BT . ⎡3 2 ⎤ 3. Show that (5 A)T = 5 AT for A = ⎢ ⎥. ⎣0 5 ⎦ T

T

⎛ ⎡ 3 2 ⎤⎞ ⎡3 0⎤ ⎡15 0 ⎤ ⎛ ⎡15 10 ⎤⎞ T =⎢ (5 A) = ⎜ 5 ⎢ ⎥ = 5 ⎢ 2 5 ⎥ = 5A ⎥⎟ = ⎜ ⎢ ⎥ ⎟ 10 25 0 5 ⎝ ⎣ ⎝ ⎣ 0 25 ⎦⎠ ⎣ ⎦ ⎣ ⎦ ⎦⎠ T

⎡ 6 −2 ⎤ 4. Show that (C T )−1 = (C −1 )T for C = ⎢ ⎥. ⎣4 2 ⎦ −1

−1 ⎛ ⎛ ⎡ 6 −2 ⎤⎞ T ⎞ ⎛ ⎡ 6 4 ⎤⎞ 1 ⎡ 2 −4 ⎤ (C ) = ⎜ ⎜ ⎢ ⎟ =⎜⎢ ⎥⎟ = 20 ⎢ 2 6 ⎥ and ⎜⎝ ⎝ ⎣ 4 2 ⎥⎦⎟⎠ ⎟⎠ − 2 2 ⎝⎣ ⎣ ⎦ ⎦⎠ T −1

T

T ⎛ ⎛ ⎡ 6 −2 ⎤⎞ −1 ⎞ ⎛ 1 ⎡ 2 2 ⎤⎞ 1 ⎡ 2 −4 ⎤ . = = ⎢ (C ) = ⎜ ⎜ ⎢ ⎟ ⎢ ⎥ ⎥ ⎜ ⎟ ⎟ 20 ⎣ 2 6 ⎥⎦ ⎜⎝ ⎝ ⎣ 4 2 ⎦⎠ ⎟⎠ ⎝ 20 ⎣ −4 6 ⎦⎠ −1 T

Thus, (C T )−1 = (C −1 )T . ⎡ 6 −2 ⎤ T −1 5. Given C = ⎢ ⎥ , find (CC ) . ⎣4 2 ⎦ −1

−1 −1 ⎛ ⎡ 6 −2 ⎤ ⎡ 6 −2 ⎤T ⎞ ⎛ ⎡ 6 −2 ⎤ ⎡ 6 4 ⎤⎞ 1 ⎡ 20 −20 ⎤ ⎡ 40 20 ⎤ = = = (CC ) = ⎜ ⎢ ⎟ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎜ ⎟ 400 ⎢⎣ −20 40 ⎥⎦ ⎝ ⎣ 4 2 ⎦ ⎣ −2 2 ⎦⎠ ⎣ 20 20 ⎦ ⎝ ⎣4 2 ⎦ ⎣4 2 ⎦ ⎠ T −1

⎡ 6 −2 ⎤ T −1 6. Given C = ⎢ ⎥ , find (C C ) . ⎣4 2 ⎦ −1

−1 −1 ⎛ ⎡ 6 −2 ⎤T ⎡ 6 −2 ⎤⎞ ⎛ ⎡ 6 4 ⎤ ⎡ 6 −2 ⎤⎞ 1 ⎡8 4 ⎤ ⎡ 52 −4 ⎤ = = = (C T C )−1 = ⎜ ⎢ ⎟ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎜ ⎟ 400 ⎢⎣ 4 52 ⎥⎦ ⎝ ⎣ −2 2 ⎦ ⎣ 4 2 ⎦⎠ ⎣ −4 8 ⎦ ⎝ ⎣ 4 2 ⎦ ⎣ 4 2 ⎦⎠ T

⎡4⎤ ⎢ ⎥ 7. Find ⎢ −1⎥ . ⎢⎣ 1 ⎥⎦ T

⎡4⎤ ⎢ −1⎥ = [4 −1 1] ⎢ ⎥ ⎢⎣ 1 ⎥⎦ 8. Find [2 −1 5 7]T . ⎡2⎤ ⎢ −1⎥ [2 −1 5 7]T = ⎢ ⎥ ⎢5⎥ ⎢ ⎥ ⎣7⎦

Answer key

155

⎡ 2 −1 3 7 ⎤ T T 9. Given D = ⎢ ⎥ , show that (D ) = D . ⎣ 4 −5 0 8 ⎦ T

⎡2 4⎤ ⎢ −1 −5 ⎥ ⎛ ⎞ ⎡ 2 −1 3 7 ⎤ ⎥ = ⎡ 2 −1 3 7 ⎤ = D (DT )T = ⎜ ⎢ ⎟ = ⎢⎢ ⎥ ⎥ ⎢⎣ 4 −5 0 8 ⎥⎦ 3 0 4 − 5 0 8 ⎦ ⎠ ⎝⎣ ⎢ ⎥ ⎣7 8⎦ T

T

⎡1 5 2 ⎤ 10. Given E = ⎢⎢5 −1 −3 ⎥⎥ , show that E T = E . ⎢⎣ 2 −3 −4 ⎥⎦ T

⎡1 5 2 ⎤ ⎡1 5 2 ⎤ E = ⎢⎢5 −1 −3 ⎥⎥ = ⎢⎢5 −1 −3 ⎥⎥ = E ⎢⎣ 2 −3 −4 ⎥⎦ ⎢⎣ 2 −3 −4 ⎥⎦ T

3

Graphing calculators and matrices Note: For the problems in this exercise, follow the guidelines given in Chapter 3. Screen shorts are provided for verification of your results.

3·1

1. name 2. square 3. scalar 4. rows 5. rref(

3·2

1.

2.

156

Answer key

3.

4.

5.

8.

6.

9.

7.

10.

4. Undefined

3·3

1.

2. 5.

6. Undefined 3.

Answer key

157

7.

8. Undefined, because G does not have an inverse. It is a singular matrix.

9.

10.

11. Note: Be sure to use the negation key, not the minus key in this problem.

12.

158

Answer key

13.

14.

15.

3·4

1.

5.

2.

6.

3.

7.

4.

8.

Answer key

159

9. Undefined

13.

10.

11.

14.

15.

12.

3·5

160

1.

4.

2.

5.

3.

6.

Answer key

7.

12.

8.

13.

9.

14.

10.

15.

11.

3·6

1.

2x + 4y = 2 3x − 2y = 1

1 1 x = 0.5, y = 0.25 or x = , y = 2 4

Answer key

161

4 x + 2 y − 6z = 2 2. 3x − y − 4 z = 7 5 x + 2 y − 6z = 5

x = 3, y = −2, z = 1 3.

x − 2y =1 3x − 6 y = 3

x = 2t + 1, y = t , t arbitrary 4 x + 2 y − 6z = 2 4. 3x − y − 4 z = 7 2 x + y − 3z = 2

No solution 2 x + 5 y − 8z = 4 5. 2 x + 4 y − 6z = 2 3x + 8 y − 13z = 7

x = −t − 3, y = 2t + 2, z = t , t arbitrary

162

Answer key

6.

x + 2y = 0 2x − 3 y = 0

x = 0, y = 0 4 x + 2 y − 6z = 0 7. 3x − y − 4 z = 0 5 x + 2 y − 6z = 0

x = 0, y = 0, z = 0 4 x + 2 y − 6z = 0 8. −2 x − y + 3z = 0 5 x + 2 y − 6z = 0

x = 0, y = 3t , z = t , t arbitrary 4 x + 2 y − 6z = 3 9. −2 x − y + 3z = 5 5 x + 2 y − 6z = 0

No solution.

Answer key

163

10.

4 x + 2 y − 6z = 0 5 x + 2 y − 6z = 0

x = 0, y = 3t , z = t , t arbitrary 11.

x+ y=4 2 x + 3 y = 10

x = 2, y = 2 2x − 4 y = 3

12.

x + 2y = 4

x = 2.75, y = 0.625 or x = x + y + 2z = 9 13. 2 x + 4 y − 3z = 1 3x + 6 y − 5z = 0

x = 1, y = 2, z = 3

164

Answer key

11 5 , y= 4 8

14.

5 x − 2 y + 6z = 0 −2 x + y + 3z = 1

x = −12t + 2, y = −27t + 5, z = t , t arbitrary 6a + 6b + 3c = 5 15.

−2b + 3c = 1 3a + 6b − 3c = −2

No solution.

3·7

1.

2 x1 + 4 x 2 = 2 3 x1 − 2 x 2 = 1

1 1 x1 = 0.5, x 2 = 0.25 or x1 = , x 2 = 2 4 4 x1 + 2 x 2 − 6 x 3 = 2 2. 3x1 − x 2 − 4 x 3 = 7 5 x1 + 2 x 2 − 6 x 3 = 5

x1 = 3, x 2 = −2, x 3 = 1

Answer key

165

3.

x1 − 2 x 2 = 1 3 x1 − 6 x 2 = 3

A is singular, so X = A−1C does not apply. 4 x1 + 2 x 2 − 6 x 3 = 2 4. 3x1 − x 2 − 4 x 3 = 7 2 x1 + x 2 − 3 x 3 = 5

A is singular, so X = A−1C does not apply. 2 x1 + 5 x 2 − 8 x 3 = 4 5. 2 x1 + 4 x 2 − 6 x 3 = 2 3x1 + 8 x 2 − 13x 3 = 7

A is singular, so X = A−1C does not apply. 3x1 + 2 x 2 = 12 3 x1 − 2 x 2 = 0

6.

x1 = 2 , x 2 = 3

166

Answer key

4 x1 + 2 x 2 − 6 x 3 = 0 7. 3x1 − x 2 − 4 x 3 = 0 5 x1 + 2 x 2 − 6 x 3 = 0

x1 = 0 , x 2 = 0 , x 3 = 0 2 x1 − x 2 + x 3 = 6 8. x1 + x 2 + x 3 = 0 4 x1 + 5 x 2 − x 3 = 2

5 7 3 x1 = 2.5, x 2 = −1.75, x 3 = −0.75 or x1 = , x 2 = − , x 3 = − 2 4 4 11x1 + x 2 + 2 x 3 = 2 9. x1 + 5 x 2 + 2 x 3 = 2 =0 4 x1 + 5 x 2

x1 = 0 , x 2 = 0 , x 3 = 1

Answer key

167

7 x1 + 3x 2 − 5 x 3 + 8 x 4 = −3 x1 + x 2 + x 3 + 2 x 4 = 3 10. 4 x1 + x 2 + x 3 + x 4 = 6 3 x1 + 7 x 2 − x 3 + x 4 = 1

x1 = 1, x 2 = 0, x 3 = 2, x 4 = 0

4

Special types of square matrices

4·1

1. Show that A and B are inverses.

2. Show that (A−1) −1 = A.

3. Show that (B −1)T = (BT ) −1.

4. Show that (10A) −1 =

168

Answer key

1 −1 A . 10

1 5. Show that (A + B) −1 = I. 5

6. Show that A−1 + B −1 = 5I.

7. Show that (A2) −1 = (A−1)2.

8. Find C −1.

9. Find D −1.

10. Find E −1.

Answer key

169

11. Show that (C + D) −1 ≠ C −1 + D −1.

12. Show that (CDE) −1 = E −1D −1C −1.

13.

Nonsingular, because det( F ) = −1 ≠ 0.

14.

Singular, because det(G ) = 0.

15.

Nonsingular, because det( H ) = 27 ≠ 0.

170

Answer key

4·2

1. L 2. D 3. U 4. D 5. L 6. Show that A + B is lower triangular.

7. Show that AB is lower triangular.

8. Show that 5B is lower triangular.

9. Show that AT is upper triangular.

10. Show that A−1 is lower triangular.

11. Compute A + B.

Answer key

171

12. Compute AB.

13. Compute |5B|.

14. Find A−1.

15. Find A4.

4·3

1. involutory 2 2. A

3. identity 4. zero 5. singular 2⎤ ⎡ 5 6. Show that A = ⎢ ⎥ is involutory. ⎣ −12 −5 ⎦

172

Answer key

⎡ 4 −1 −4 ⎤ 7. Show that B = ⎢⎢ 3 0 −4 ⎥⎥ is involutory. ⎢⎣ 3 −1 −3 ⎥⎦

⎡3 1⎤ 8. Show that C = ⎢ ⎥ is idempotent. ⎣ −6 −2 ⎦

⎡ −1.5 0.5 2 ⎤ ⎢ ⎥ 9. Show that D = ⎢ −1.5 0.5 2 ⎥ is idempotent. ⎢⎣ −1.5 0.5 2 ⎥⎦

⎡1 5 −2 ⎤ 10. Show that A = ⎢⎢1 2 −1 ⎥⎥ is nilpotent of index 3. ⎢⎣3 6 −3 ⎥⎦

4·4

⎡ 0 4 −3 ⎤ 1. ⎢⎢ −4 0 8 ⎥⎥ , skew-symmetric ⎢⎣ 3 −8 0 ⎥⎦ ⎡2 4 3 ⎤ 2. ⎢⎢ 4 6 −8 ⎥⎥ , symmetric ⎢⎣ 3 −8 0 ⎥⎦

Answer key

173

⎡ −7 4 3 ⎤ 3. ⎢⎢ 4 5 −2 ⎥⎥ , symmetric ⎢⎣ 3 −2 9 ⎥⎦ ⎡ 0 10 7 −9 ⎤ ⎢ −10 0 −5 2 ⎥ ⎥ , skew-symmetric 4. ⎢ ⎢ −7 5 0 −15 ⎥ ⎥ ⎢ 0 ⎦ ⎣ 9 −2 15 ⎡ 0 5. ⎢ ⎢⎣ − 2

2⎤ ⎥ , skew-symmetric 0 ⎥⎦

6. skew-symmetric 7. symmetric 8. skew-symmetric 9. symmetric 10. neither 11. symmetric 12. Show that A + B is symmetric.

13. Show that A−1 is symmetric.

14. Show that B5 is symmetric.

15. Show that C T is skew-symmetric.

174

Answer key

4·5

1. A−1 2. I ⎡0 0 1 ⎤ ⎥ ⎢ 3. Given A = ⎢1 0 0 ⎥ , find det (A). ⎢⎣0 1 0 ⎥⎦

⎡0 −1 0 ⎤ 4. Given B = ⎢⎢1 0 0 ⎥⎥ , show that B is orthogonal. ⎢⎣0 0 1 ⎥⎦

⎡1 2 5. Given C = ⎢⎢ 1 ⎢⎣ 2

4·6

⎤ 2⎥ ⎥ , show that C is orthogonal. 1 2 ⎥⎦

− 1

1. A 2. A ∗ 3. real 4. −A 5. pure imaginary 6. Hermitian 7. skew-Hermitian 8. neither 9. Hermitian 10. skew-Hermitian

Answer key

175

5 5·1

Determinants ⎡2 5 ⎤ 1. A = ⎢ ⎣1 − 4⎥⎦ det( A) = − 8 − 5 = − 13 ⎡ 3 − 6⎤ 2. A = ⎢ ⎣− 1 2 ⎥⎦ det( A) = 6 − 6 = 0 ⎡3 4 2⎤ 3. A = ⎢ 0 7 − 3⎥ ⎢ ⎥ ⎣− 2 1 5 ⎦ 7 −3 4 2 4 2 −0 + ( −2) = 166 1 5 1 5 7 −3

det( A) = 3

⎡1 0 0⎤ 4. A = ⎢0 1 0⎥ ⎢ ⎥ ⎣0 0 1⎦ det( A) = 1

1 0 =1 0 1

⎡ 2 4⎤ 5. A = ⎢− 3 1⎥ ⎢ ⎥ ⎣ 5 0⎦ Undefined because the matrix is not a square matrix and thus has no determinant. 6. The element 7 +

3 2 = 19 −2 5

7. The element 0 −

4 2 = − 18 1 5

8. The element 1 −

3 2 =9 0 −3

⎡1 2 3⎤ 9. If A = ⎢0 0 5⎥ , find det( A) by expanding by cofactors along the “smart” row. ⎢ ⎥ ⎣6 − 2 1⎦ det( A) = − 5

1 2 = 70 6 −2

Why do you think the word smart was used? The calculation was simplified due to the 0s in the second (smart) row. ⎡2 6 − 3⎤ 10. A = ⎢4 − 1 2 ⎥ ⎢ ⎥ 3⎦ ⎣4 0 2 6 −3 2 6 4 − 1 2 4 − 1 = 2(− 1)3 + 6(2)4 + (− 3)(4 )0 − 4(− 1)(− 3) − 0(2)2 − 3(4 )6 = −42 4 0 3 4 0

176

Answer key

⎡3 4 2⎤ 11. A = ⎢ 0 7 − 3⎥ ⎢ ⎥ ⎣− 2 1 5 ⎦ 3 4 2 3 4 0 7 − 3 0 7 = 105 + 24 + 28 + 9 = 166 −2 1 5 −2 1 ⎡1 2 3⎤ 12. A = ⎢0 0 5⎥ ⎢ ⎥ ⎣6 − 2 1⎦ det( A) = 70 ⎡4 0 3⎤ 13. A = ⎢4 − 1 2⎥ ⎢ ⎥ ⎣2 6 3⎦ det( A) = 18 ⎡3 0 − 2⎤ 1⎥ 14. A = ⎢4 7 ⎥ ⎢ ⎣2 − 3 5 ⎦ det( A) = 166 ⎡1 ⎢3 15. A = ⎢ ⎢2 ⎢⎣1

0 1 1 0

6 0 2 4

7⎤ 3⎥ 3⎥⎥ 4⎥⎦

det( A) = −10

5·2

1.

2 x − 3 y = 16 5x − 2y = −4 D=

2 16 2 −3 16 − 3 = − 88 = 11, X = = − 44 , and Y = −4 −2 5 −4 5 −2

Solution: x = 2.

− 44 − 88 = −4 , y = = −8 11 11

x − 2y = 7 3x + 2 y = 5 D=

1 −2 7 −2 1 7 = 8, X = = 24 , and Y = = − 16 3 2 5 2 3 5

Solution: x =

24 − 16 = 3, y = = −2 8 8

x − 2 y − 3z = − 20 3. 2 x + 4 y − 5z = 11 3x + 7 y − 4 z = 33 1 −2 −3 − 20 − 2 − 3 1 − 20 − 3 1 − 2 − 20 D = 2 4 − 5 = 27 , X = 11 4 − 5 = 27 , Y = 2 11 − 5 = 162, and Z = 2 4 11 = 81 3 7 −4 33 7 −4 3 33 − 4 3 7 33 Solution: x =

27 162 81 = 1, y = = 6, z = =3 27 27 27 Answer key

177

x + 2y − z = 7 4. 4 x + 3 y + 2z = 1 9 x + 8 y + 3z = 4 1 2 −1 7 2 −1 D = 4 3 2 = 0, X = 1 3 2 = − 35 9 8 3 4 8 3 Thus, there is no solution. 2x − 4 y + 7z = 5 5. 3x + 2 y −

z=2

x − 10 y + 15z = 8 2 −4 7 5 −4 7 2 5 7 2 −4 5 D = 3 2 − 1 = 0, X = 2 2 − 1 = 0, Y = 3 2 − 1 = 0, and Z = 3 2 2 = 0 1 − 10 15 8 − 10 15 1 8 15 1 − 10 8 There are infinitely many solutions. Now, try to solve two equations such that the coefficient determinant involving just two of the variables is not 0. Use the first two equations. 2x − 4 y = 5 − 7z 3x + 2 y = 2 + z Eliminate y. Multiply the second equation by 2 and add the equations. 8 x = 9 − 5z 9 − 5z 23z − 11 and y = . The free variable is z. For convenience, let z = 16t . The parametric form of 8 16 9 11 the solutions is x = − 10t , y = 23t − , z = 16t , with t arbitrary. 8 16 x=

6.

3x − 2 y = − 16 2x − 5 y = −4 ⎡x ⎤ ⎡3 − 2⎤ ⎡− 16⎤ X = ⎢ ⎥, A = ⎢ , and C = ⎢ ⎥ y − 2 5 ⎣ ⎦ ⎣ ⎦ ⎣ − 4 ⎥⎦ ⎡ 72 ⎤ ⎢− ⎥ ⎡x ⎤ −1 X = ⎢ ⎥ = A C = ⎢ 11 ⎥ 20 y ⎣ ⎦ ⎢⎣− 11 ⎥⎦ x −z =1

7. 9 x − y 8x + 9 y −

+ 4z = 0 z=2

⎡x ⎤ ⎡1 0 − 1⎤ ⎡1⎤ X = ⎢ y⎥ , A = ⎢9 − 1 4 ⎥ and C = ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎣z ⎦ ⎣8 9 − 1⎦ ⎣2⎦ ⎡ 37 ⎤ ⎢ 124 ⎥ ⎡x ⎤ ⎢ 15 ⎥ −1 ⎢ ⎥ X = y = A C = ⎢− 124 ⎥ ⎢ ⎥ ⎢ 87 ⎥ ⎣z ⎦ ⎢− ⎥ ⎢⎣ 124 ⎥⎦

178

Answer key

x − y + 2z = 4 8. 3x + 2 y + z = 5 3x − 3 y + 6z = 12

9.

The third equation is a multiple of the first so the determinant is 0. In this case there are infinitely many 13 7 solutions: x = − z , y = z − , z = z (free variable) 5 5 x − 5y = 3 2 x − 10 y = 6 The determinant of the coefficient matrix is 0. Thus, there are infinitely many solutions because the lines t −3 , t arbitrary are the same line. Solutions: x = t, y = 5 0 .7 x − 2 .1 y +

10.

z = 3 .7

4 x − 0 .4 y + 0.23z = 8 .2 2x −

3 y + 3 . 2z = 4

(This is where you give thanks for the technology!) 1 ⎤ ⎡x ⎤ ⎡0 .7 − 2 .1 ⎡3 .7⎤ X = ⎢ y⎥ , A = ⎢ 4 − 0 .4 0 .23⎥ and C = ⎢8 .2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ −3 3 .2 ⎦ ⎣z ⎦ ⎣2 ⎣4⎦ ⎡x ⎤ ⎡ 1 .96 ⎤ X = ⎢ y⎥ = A−1 C ≈ ⎢− 1 .98⎥ ⎢ ⎥ ⎢ ⎥ ⎣z ⎦ ⎣− 1 .83⎦

5·3

⎡8 0 0 .03⎤ 1. If A = ⎢0 5 5 .71⎥ , then det( A) = (8)(5)(3) = 120 ⎢ ⎥ 3 ⎦ ⎣0 0 ⎛ ⎡3 1 − 2⎤ ⎞ 2. det ⎜ ⎢0 1 3 ⎥ ⎟ = 3(4 − 3) = 3(1) = 3 ⎥⎟ ⎜⎢ ⎝ ⎣0 1 4 ⎦ ⎠ ⎛ ⎡8 0 0 .03⎤ ⎞ 3. det ⎜ 4 ⎢0 5 5 .71⎥ ⎟ = 43 (120) = 7680 (See Prob. 1.) ⎥⎟ ⎜ ⎢ 3 ⎦⎠ ⎝ ⎣0 0 1 1 1 ⎡2 3⎤ = =− 4. If A = ⎢ , then det( A−1 ) = det( A) 2 − 12 10 ⎣4 1⎥⎦ ⎛ ⎡a b c ⎤ ⎞ ⎛ ⎡a b c ⎤ ⎞ 5. If det ⎜ ⎢3 11 0 ⎥ ⎟ = 12 , then det ⎜ ⎢2 5 − 1⎥ ⎟ = −12 ⎥⎟ ⎥⎟ ⎜⎢ ⎜⎢ ⎝ ⎣2 5 − 1⎦ ⎠ ⎝ ⎣3 11 0 ⎦ ⎠ ⎡1 7⎤ ⎡3 − 1⎤ 6. If A = ⎢ and B = ⎢ show that det( AB) = det( A)det( B) . ⎥ 0 2 ⎣ ⎦ ⎣2 − 1⎥⎦ ⎡1 7⎤ ⎡3 − 1⎤ ⎡17 − 8⎤ = − 34 + 32 = − 2 det( AB) = ⎢ = ⎣0 2⎥⎦ ⎢⎣2 − 1⎥⎦ ⎢⎣ 4 − 2⎦⎥ det( A)det( B) =

1 7 3 −1 = (2)(− 1) = − 2 0 2 2 −1

Thus, det( AB) = det( A)det( B) = −2. Answer key

179

⎡1 7⎤ ⎡3 − 1⎤ 7. If A = ⎢ and B = ⎢ , show that det( A + B) ≠ det( A) + det( B) . ⎣0 2⎥⎦ ⎣2 − 1⎥⎦ ⎛ ⎡4 6⎤ ⎞ ⎛ ⎡1 7⎤ ⎡3 − 1⎤ ⎞ = −8 det( A + B) = det ⎜ ⎢ = det ⎜ ⎢ +⎢ ⎟ ⎥ ⎥ 0 2 2 1 − ⎝ ⎣2 1⎥⎦ ⎟⎠ ⎝⎣ ⎦ ⎣ ⎦⎠ det( A) + det( B) =

1 7 3 −1 + = 2 −1 = 1 0 2 2 −1

Thus, det( A + B) ≠ det( A) + det( B) . ⎡2 3 2⎤ 8. A = ⎢0 7 6⎥ ⎢ ⎥ ⎣5 5 4⎦ det( A) = 16 9 ⎡7 8 ⎢6 5 4 9. A = ⎢ − − 1 9 3 ⎢ 1 ⎢⎣5 0

0⎤ 8⎥ 7⎥⎥ 2⎥⎦

det( A) = 1842 ⎡2 3 2⎤ 10. Is A = ⎢0 7 6⎥ an invertible matrix? Explain. ⎢ ⎥ ⎣5 5 4⎦ Yes. The determinant is not 0. See Prob. 8 above.

6

Vectors in Rn

6·1

1. Sketch the vectors u = (5, 2), v = (− 3, 2) , and u + v on the coordinate plane. (2, 4)

4

v

3 (–3, 2) v

–5

–4

–3

–2

u

1

–1

(5, 2)

u+v

2

1

2

3

4

5

–1

2. If u = (9, 0) and v = (5, − 4 ), compute u − v , 4 v , v − u , and 3u − 4 v . u − v = (4 , 4 ) 4v = (20, −16) v − u = (− 4 , − 4 ) 3u − 4 v = (27 , 0) − (20, − 16) = (7 ,16) 3. Write the vector (5, 8) in a i + b j form. (5, 8) = 5i + 8 j 4. Find the norm of v = 5i − 3j. v = 5i − 3j = 25 + 9 = 34

180

Answer key

5. If u = 3i + 6 j and v = − 4 i + 2 j , compute u + v and write the answer in a i + b j form. u + v = (3i + 6 j) + (− 4 i + 2 j) = − i + 8 j 6. Write the vector (3, 5) as a column vector. ⎡3⎤ (3, 5) = ⎢ ⎥ ⎣5⎦ ⎡2⎤ 7. Multiply ⎡⎣2 3⎤⎦ and ⎢ ⎥ as matrices. ⎣3⎦ ⎡2⎤ [2 3] ⎢ ⎥ = [13] ⎣3⎦ 2

8. Calculate (2, 3) . 2

(2, 3) = ( 4 + 9 )2 = 13 9. Compare the results of problems 7 and 8. What do you notice? They both give the same one dimensional number. In problem 7, the answer is a 1 × 1 matrix, and in problem 8, it is a real number. (This connection is discussed after exercise 6-3.) 10. If a i + b j = 0 , then what must be true of a and b? a=b=0

6·2

1. u = 5i + 2 j and v = − 3i − 3j cosq =

uu v − 15 − 6 − 21 ≈ − 0 .919145 = = u ⋅ v ( 25 + 4 )( 9 + 9 ) ( 29 )( 18 )

Thus, q = cos−1 (− 0 .919145) ≈ 157 °. 2. u = (− 3, 6) and v = (2, − 5) cosθ =

uu v − 6 − 30 = = u ⋅ v ( 9 + 36 )( 4 + 25 )

− 36 ≈ − 0 .99655 45 29

Thus, θ = cos−1 (− .99655) ≈ 175 °. 3. u = − 4 i + 3j and v = − 6i − 8 j cosθ =

uu v 24 − 24 = =0 u ⋅ v ( 25 )( 100 )

θ = 90 ° 4. Prove that i and j are orthogonal. i u j = (1, 0) u (0,1) = 1 ⋅ 0 + 0 ⋅1 = 0 Thus i and j are orthogonal 5. Find all vectors, u, such that u = 3 and u is perpendicular to v = 2i + 2 j. If u = ai + b j and v are perpendicular then u u v = 2a + 2b = 0 . Also, if u = 3 then a2 + b2 = 9. 3 3 2 =± Solving these simultaneously, b = −a and a2 + (−a)2 = 9. Thus, a = ± . There are two solution 2 2 vectors: u1 =

3 2 3 2 3 2 3 2 i− j and u 2 = − i+ j 2 2 2 2

6. For u = − 5i + 3j and v = − 6i − 2 j , find u u v . u u v = 30 − 6 = 24 Answer key

181

7. Find the unit vector associated with u = 5i + 2 j . 5i + 2 j 5i + 2 j 5 2 u = = i+ j = u 25 + 4 29 29 29

v=

8. Express (5, 7 ) as a linear combination of i and j. (5, 7 ) = 5i + 7 j 9. Show that (1, 2) and (2, − 1) are orthogonal. (1, 2) u (2, − 1) = 2 − 2 = 0 10. Show that (3,1) can be written as a linear combination of (1, 2) and (2, − 1). If (3,1) = a(1, 2) + b(2, − 1), then a + 2b = 3 and 2a − b = 1. Solving these simultaneously, a = 1 and b = 1. Thus (3,1) = 1(1, 2) + 1(2, − 1).

6·3

1. Find 6(2u − v ). 6(2u − v ) = 6(2(4 , 7 , − 3, 2) − (5, − 2, 8,1)) = 6((8,14 , − 6, 4 ) − (5, − 2, 8,1)) = 6(3,16, − 14 , 3) = (18, 96, − 84 ,18) 2. Find u + v . u + v = ((4 , 7 , − 3, 2)) + (5, − 2, 8,1) = (9, 5, 5, 3) = 81 + 25 + 25 + 9 = 140 3. Find u u v . u u v = (4 , 7 , − 3, 2) u (5, − 2, 8,1) = 20 − 14 − 24 + 2 = −16 4. Find

1 v. v

1 1 (5, − 2, 8,1) (5, − 2, 8,1) . v= = v 94 25 + 4 + 64 + 1 5. Find

1 v . v

1 v =1, because this is the magnitude of a unit vector. v 6. Show that u + v ≤ u + v . u + v = 140 ≈ 11 .83 . (See Prob. 2) u + v = (4 , 7 , − 3, 2) + (5, − 2, 8,1) = 16 + 49 + 9 + 4 + 25 + 4 + 64 + 1 = 78 + 94 ≈ 18 .53 Thus, u + v ≤ u + v 7. Show that u + v u+v 2

2

2

2

2

≠ u + v .

= ( 140 )2 = 140

u + v

2

= ( 78 )2 + ( 94 )2 = 172

Thus, u + v

2

2

2

≠ u + v .

8. Find a vector x such that 5 x − 2 v = 2(u − 5 x ). If 5 x − 2 v = 2(u − 5 x ), then 5 x − 2 v = 2u − 10 x . 2u + 2 v 2 2 ⎛6 2 2 = (u + v ) = Solving for x gives x = (9, 5, 5, 3) = ⎜ , , , 15 15 15 ⎝5 3 3

182

Answer key

2⎞ 5 ⎟⎠

9. Find (6u ) u v . (6u ) u v = 6(u u v ) = 6(− 16) = − 96 (See Prob. 3) 10. Show that (1, 0, 0) , (0,1, 0) , and (0, 0,1) are mutually orthogonal. (1, 0, 0) u (0,1, 0) = 0, (1, 0, 0) u (0, 0,1) = 0, and (0,1, 0) u (0, 0,1) = 0

6·4

1. If u = (2, 2, 2) and v = (0, 4 , − 2) , find u − v . u − v = (2, 2, 2) − (0, 4 , − 2) = (2, − 2, 4 ) = 4 + 4 + 16 = 24 2. Let v = (− 2, 3, 0, 6). Find all scalars k such that kv = 5. If kv = 5, then k(− 2, 3, 0, 6) = 5 k 4 + 9 + 36 = 5 k 49 = 5 7 k =5 k=

5 7

5 Thus, k = ± . 7 3. u = (− 1, 3, 2) and v = (4 , 2,1) u u v = (− 1, 3, 2) u (4 , 2,1) = − 4 + 6 + 2 = 4 ≠ 0 Hence, the vectors are not orthogonal. 4. u = (− 4 , 6, − 10,1) and v = (2,1, − 2, 9) u u v = (− 4 , 6, − 10,1) u (2,1, − 2, 9) = − 8 + 6 + 20 + 9 = 27 ≠ 0 Hence, the vectors are not orthogonal. 5. u = (a, b, c ) and v = (0, 0, 0) uu v =0 The vectors are orthogonal. Tip: The zero vector is orthogonal to all vectors. 6. u = (2,1, 3) and v = (1, 7 , k ) If u and v are orthogonal, then u u v = 0. In this case, 2 + 7 + 3k = 0, which yields k = −3. 7. u = (k , k ,1) and v = (k , 5, 6) If u and v are orthogonal, then u u v = 0. In this case, k 2 + 5k + 6 = 0, which has two solutions: k = −2 and k = −3. 8. Find a vector of norm 2 that is orthogonal to the three vectors, u = (2,1, 0), v = (− 1, − 1, 0), and w = (4 , 0, 0). Let x = (x1, x2, x3). If x = 2, then x12 + x 22 + x32 = 4 ; and if x is orthogonal to all three of the given vectors, then 2 x1 + x 2 = 0 − x1 − x 2 = 0 4 x1 = 0 Thus, x1 = x 2 = x3 = 0 . Hence, there is no solution because the norm of the zero vector is 0. Answer key

183

9. Find scalars a, b, and c such that a(2, 0,1) + b(1, 3, 2) + c(1, 0,1) = (7 , 3, 4 ). Solve the system of equations: 2a + b + c = 7 =3

3b

a + 2b + c = 4 This system reduces to solving 2a + c = 6 a+c = 2

, which has solution b = 1, a = 4 , and c = −2.

10. Show that u = (4 , 0,1) can be expressed as a linear combination of u1 = (2, 0, 0), u 2 = (0, 3, 0), and u3 = (0, 0,1). If (4 , 0,1) = a(2, 0, 0) + b(0, 3, 0) + c(0, 0,1) , then you solve the system: 2a = 4 3b = 0 c =1 The solution is (4 , 0,1) = 2(2, 0, 0) + 0(0, 3, 0) + 1(0, 0,1). 11. If a(2, 2, 0) + b(2, − 2, 0) + c(0, 0, 2) = 0, then solve for a, b, and c. 2a + 2b

=0

2a − 2b

=0 2c = 0

The solution to this system is a = b = c = 0. 12. Show that (8, 0, 4 ) can be expressed as a linear combination of (2, 2, 0),(2, − 2, 0), and (0, 0, 2). If (8, 0, 4 ) = a(2, 2, 0) + b(2, − 2, 0) + c(0, 0, 2) , then 2a + 2b

=8

2a − 2b

=0 2c = 4

The solution to this system is a = b = c = 2 . Thus, (8, 0, 4 ) = 2(2, 2, 0) + 2(2, − 2, 0) + 2(0, 0, 2) . 13. Show that (3,1, 5) can be expressed as a linear combination of (2, 2, 0),(2, − 2, 0), and (0, 0, 2). If (3,1, 5) = a(2, 2, 0) + b(2, − 2, 0) + c(0, 0, 2) , then 2a + 2b

=3

2a − 2b

=1 2c = 5

1 5 The solution to this system is a = 1, b = , and c = . 2 2 1 5 Thus, (3,1, 5) = 1(2, 2, 0) + (2, − 2, 0) + (0, 0, 2). 2 2 14. Show that (4 , 0, 3, 7 ) can be expressed as a linear combination of (1, 0, 0, 0), (0,1, 0, 0), (0, 0,1, 0), and (0, 0, 0,1). (4 , 0, 3, 7 ) = 4(1, 0, 0, 0) + 0(0,1, 0, 0) + 3(0, 0,1, 0) + 7(0, 0, 0,1)

184

Answer key

15. Find the vector that is the normalization of v = (4 , 0, 3, 7 ) . v v (4,0,3,7) = = v 16 + 0 + 9 + 49 74 16. If u = (2, k ,1, 4 ) and v = (3, − 1, 6, − 3) , find k such that u − v = 6. If u − v = 6, then (2 − 3)2 + (k + 1)2 + (1 − 6)2 + (4 + 3)2 = 6. Simplifying yields 1 + (k + 1)2 + 25 + 49 = 36 for which there is no solution. 17. If S = {(2, 2, 0),(2, − 2, 0),(0, 0, 2)}, find the set T of vectors that are the normalizations of the vectors in S. Let u1 =

1 1 ( 2, 2, 0) (0 , 0 , 2 ) 1 ( 2, − 2, 0) = = (0, 0, 2), then (2, 2, 0), u 2 = = (2, − 2, 0), and u3 = ( 2, 2, 0) (0 , 0 , 2 ) 2 ( 2, − 2, 0) 8 8

T = {u1 , u 2 , u3 } is the normalized set. 18. If u = (1, 2, 3) and v = (0, 2, 2), show that u u v ≤ u v (Note: This inequality is true for all vectors and is called the Cauchy-Schwarz Inequality.) u u v = 0 + 4 + 6 = 10 u

v = ( 1 + 4 + 9 )( 0 + 4 + 4 ) = ( 14 )( 8 ) = 112 ≈ 10 .6 .

Thus, u u v ≤ u

v

19. If u = (1, 2, 0, 3) and v = (0, 3, 0, 2), calculate (u u v )(4 , 4 , 3,1). (u u v )(4 , 4 , 3,1) = 12(4 , 4 , 3,1) = (48, 48, 36,12) 20. If u = (1, 2, 0, 3) and v = (0, 3, 0, 2), calculate u( v u (4 , 4 , 3,1)). u( v u (4 , 4 , 3,1)) = u(12 + 2) = 14(1, 2, 0, 3) = (14 , 28, 0, 42)

7

Vector spaces

7·1

1. Let V be the set of all real-valued functions with domain [0,1]. Let addition, f + g , scalar multiplication, kf , and − f , be defined by ( f + g )( x ) = f ( x ) + g ( x ), (kf )( x ) = kf ( x ), and − f = (− 1) f , respectively. Prove that V is a vector space over the real numbers. Proof: If f , g, and h are in V , then i. ( f + g )( x ) = f ( x ) + g ( x ) is real. Thus, f + g is in V. ii. ( f + g )( x ) = f ( x ) + g ( x ) = g ( x ) + f ( x ) = ( g + f )( x ). Thus, f + g = g + f . iii. ( f + ( g + h))( x ) = f ( x ) + ( g + h)( x ) = f ( x ) + ( g ( x ) + h( x )) = ( f ( x ) + g ( x )) + h( x ) = ( f + g )( x ) + h( x ) = (( f + g ) + h)( x ) . Hence, f + ( g + h) = ( f + g ) + h . iv. As the function 0( x ) = 0 is in V, ( f + 0 )( x ) = f ( x ) + 0( x ) = f ( x ) + 0 = f ( x ). Thus, f + 0 = f . By ii., 0 + f = 0. v. ( f + (− f ))( x ) = f ( x ) + (− f )( x ) = f ( x ) + (−1) f ( x ) = 1 f ( x ) + (− 1) f ( x ) = (1 + (− 1)) f ( x ) = 0 f ( x ) = 0. Thus, f + (− f ) = 0. By ii., (−f ) + f = 0. vi. (af )( x ) = af ( x ) is real and therefore af is in V. vii. (a( f + g ))( x ) = a( f + g )( x ) = a( f ( x ) + g ( x )) = af ( x ) + ag ( x ) = (af + ag )( x ) . Hence, a( f + g ) = af + ag . viii. ((a + b) f )( x ) = (a + b) f ( x ) = af ( x ) + bf ( x ) = (af + bf )( x ). Thus, (a + b) f = af + bf . ix. (a(bf ))( x ) = a((bf )( x )) = a(b( f ( x ))) = (ab) f ( x ). Thus, a(bf ) = (ab) f . x. (1 f )( x ) = 1 f ( x ) = f ( x ) . Hence, 1 f = f . All the axioms are satisfied, so V is a vector space. Answer key

185

⎫ ⎧ ⎡x ⎤ 2. Let H = ⎨ ⎢ ⎥ : x 2 + y 2 ≤ 1, where x , y ∈R⎬ . Show that H is not a vector space under the standard y ⎭ ⎩⎣ ⎦ operations of addition and scalar multiplication. ⎡1⎤ ⎡9⎤ ⎡1⎤ 2 2 2 2 ⎢⎣0⎥⎦ is in H because 1 + 0 ≤ 1, but 9 ⎢⎣0⎥⎦ = ⎢⎣0⎥⎦ is not in H because 9 + 0 ≤ 1. Thus, H is not a vector space because axiom vi is not satisfied. ⎧ ⎡ 3t ⎤ ⎫ ⎪⎢ ⎥ ⎪ 3. Let W = ⎨ 0 , where t ∈R⎬ . Show that W is a subspace of R3. ⎢ ⎥ ⎪ ⎣− 7t⎦ ⎪ ⎩ ⎭ ⎡ 3(0) ⎤ ⎡0⎤ (1) ⎢ 0 ⎥ = ⎢0⎥ is in W. ⎢ ⎥ ⎢⎥ ⎣− 7(0)⎦ ⎣0⎦ ⎡d ⎤ ⎡ 3t2 ⎤ ⎡a⎤ ⎡d ⎤ ⎡ 3(t1 + t2 ) ⎤ ⎡a⎤ ⎡ 3t1 ⎤ ⎥ is in W. 0 (2) If ⎢b⎥ = ⎢ 0 ⎥ is in W and ⎢ e ⎥ = ⎢ 0 ⎥ is in W, then ⎢b⎥ + ⎢ e ⎥ = ⎢ ⎢ ⎥ ⎢ ⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢⎥ ⎢ ⎥ f − t c f 7 7 − ( t + t ) 7 c − t 1 2 ⎦ ⎣ ⎦ ⎣ 2⎦ ⎣⎦ ⎣ ⎦ ⎣ ⎣ ⎦ ⎣ 1⎦ ⎡x ⎤ ⎡ 3(kt ) ⎤ ⎡x ⎤ ⎡x ⎤ ⎡ 3t ⎤ (3) If ⎢ y⎥ is in W, then ⎢ y⎥ = ⎢ 0 ⎥ for some real t. If k is real, then k ⎢ y⎥ = ⎢ 0 ⎥ is in W. ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣z ⎦ ⎣− 7(kt )⎦ ⎣z ⎦ ⎣z ⎦ ⎣− 7t⎦ Thus, W is a subspace of R3. ⎫ ⎧⎡ x ⎤ , where x ∈R⎬. Show that W is not a subspace of R 2 . 4. Let W = ⎨ ⎢ ⎥ 1 x + ⎦ ⎭ ⎩⎣ ⎡0⎤ ⎡ x ⎤ ⎡0⎤ 2 ⎢⎣0⎥⎦ ≠ ⎢⎣x + 1⎥⎦ for any real x, and so 0 = ⎢⎣0⎥⎦ is not in W. Hence, W is not a subspace of R . 5. Let v 1 = (0, 2, 3), v 2 = (1,1,1), and S = { v 1 , v 2 }. Construct the Span of S. The span of S is all possible linear combinations of the vectors in S. In this case, a v 1 + b v 2 = a(0, 2, 3) + b(1,1,1), so Span(S ) = {(b, 2a + b, 3a + b)| a, b real}. 6. Show that the span of S , where S = {(1, 0),(0,1)}, is R 2 . If (a, b) is in R 2 , then (a, b) = a(1, 0) + b(0,1) , which is a linear combination of the vectors in S. Hence, span of S is R 2 . 7. If a(1, 0, 0, 0) + b(0,1, 0, 0) + c(0, 0,1, 0) + d(0, 0, 0,1) = 0 = (0, 0, 0, 0), then solve for a, b, c, and d. By inspection, a = b = c = d = 0. 8. Express the vector v = (1, − 2, 5) as a linear combination of u1 = (1,1,1), u 2 = (1, 2, 3), and u3 = (2, − 1,1). Let v = (1, − 2, 5) = a(1,1,1) + b(1, 2, 3) + c(2, − 1,1) , then a + b + 2c = 1 a + 2b − c = − 2 a + 3b + c = 5 This system has augmented matrix that reduces to ⎡1 0 0 − 6⎤ ⎢0 1 0 3 ⎥ ⎢ ⎥ ⎣0 0 1 2 ⎦ The solution to this system is a = −6, b = 3, and c = 2. Thus, v = (1, − 2, 5) = − 6(1,1,1) + 3(1, 2, 3) + 2(2, − 1,1).

186

Answer key

⎡3 1 ⎤ ⎡1 1⎤ ⎡0 0⎤ 9. Express the matrix A = ⎢ , U2 = ⎢ , and as a linear combination of the matrices U1 = ⎢ ⎥ ⎥ − 1 0 1 1 ⎣ ⎦ ⎣ ⎦ ⎣1 1⎥⎦ ⎡0 2 ⎤ U3 = ⎢ . ⎣0 − 1⎥⎦ ⎡3 1 ⎤ ⎡1 1⎤ ⎡0 0⎤ ⎡0 2 ⎤ Let A = ⎢ +b +c , then = a⎢ ⎥ − 1 1 ⎣ ⎦ ⎣1 0⎥⎦ ⎢⎣1 1⎥⎦ ⎢⎣0 − 1⎥⎦ =3

a a+

2c = 1

a+b

=1

b − c = −1 The solution to this system is a = 3, b = −2, and c = −1. ⎡3 1 ⎤ ⎡1 1⎤ ⎡0 0⎤ ⎡0 2 ⎤ Thus, A = ⎢ = 3⎢ −2 −1 . ⎣1 − 1⎥⎦ ⎣1 0⎥⎦ ⎢⎣1 1⎥⎦ ⎢⎣0 − 1⎥⎦ 10. Show that W = {(a, b, 0), where a, b ∈R} is generated by the vectors u = (2,1, 0) and v = (0,1, 0). Let (a, b, 0) = x (2,1, 0) + y(0,1, 0) , then 2x

=a

x + y =b a a and y = b − . 2 2 a ⎛ a⎞ Thus, (a, b, 0) = (2,1, 0) + ⎜b − ⎟ (0,1, 0). Hence, W is in the span of u and v. 2 ⎝ 2⎠ This system has solution x =

Now, let x , y ∈R , then xu + yv = x (2,1, 0) + y(0,1, 0) = (2 x , x + y , 0), which is in W. Therefore, W = {(a, b, 0), where a, b ∈R} is generated by the vectors u = (2,1, 0) and v = (0,1, 0).

7·2

1. u = (− 1, 2, 4 ), v = (5, − 10, − 20) in R3 Let a(− 1, 2, 4 ) + b(5, − 10, − 20) = 0, from which you get the system of equations −a + 5b = 0 2a − 10b = 0 4a − 20b = 0 Equations 2 and 3 are multiples of equation 1, so there are infinitely many solutions. Consequently, the vectors are linearly dependent. ⎡− 3 4⎤ ⎡ 3 − 4⎤ 2. A = ⎢ and B = ⎢ in M 22 ⎣ 2 0⎥⎦ ⎣− 2 0 ⎥⎦ ⎡− 3 4⎤ ⎡ 3 − 4⎤ ⎡0 0⎤ +b = Let a ⎢ from which you get the system of equations ⎣ 2 0⎥⎦ ⎢⎣− 2 0 ⎥⎦ ⎢⎣0 0⎥⎦ − 3a + 3b = 0 4a − 4b = 0 2a − 2b = 0 Equations 2 and 3 are multiples of equation 1, so there are infinitely many solutions. Consequently, the vectors are linearly dependent. Answer key

187

3. u = (− 3, 0, 4 ), v = (5, − 1, 2), w = (1,1, 3) in R3 Let a(− 3, 0, 4 ) + b(5, − 1, 2) + c(1,1, 3) = (0, 0, 0) , then − 3a + 5b + c = 0 −b+ c = 0 4a + 2b + 3c = 0 This system has augmented matrix that reduces to ⎡1 0 0 0⎤ ⎢0 1 0 0⎥ , ⎢ ⎥ ⎣0 0 1 0⎦ which has only the trivial solution. Thus, the vectors are linearly independent. 4. u = 6 − x 2 , v = 1 + x + 4 x 2 in P2 Let a(6 − x 2 ) + b(1 + x + 4 x 2 ) = 0 , then (6a + b) + bx + (−a + 4b)x 2 = 0, then 6a + b = 0 b=0 −a + 4b = 0 This system has only the trivial solution a = b = 0, so the vectors are linearly independent. 5. u = 2 − x + 4 x 2 , v = 3 + 6 x + 2 x 2 , w = 2 + 10 x − 4 x 2 in P2 Let a(2 − x + 4 x 2 ) + b(3 + 6 x + 2 x 2 ) + c(2 + 10 x − 4 x 2 ) = 0, then (2a + 3b + 2c ) + (−a + 6b + 10c )x + (4a + 2b − 4c )x 2 = 0, then 2a + 3b + 2c = 0 −a + 6b + 10c = 0 4a + 2b − 4c = 0 This system has augmented matrix that reduces to ⎡1 0 0 0⎤ ⎢0 1 0 0⎥ , ⎢ ⎥ ⎣0 0 1 0⎦ which has only the trivial solution. Thus, the vectors are linearly independent. 6. u = (0, 3,1, − 1), v = (6, 0, 5,1), w = (4 , − 7 ,1, 3) in R3 Let a(0, 3,1, − 1) + b(6, 0, 5,1) + c(4 , − 7 ,1, 3) = (0, 0, 0, 0), then a + 5b + c = 0 −a + b + 3c = 0 3a −

7c = 0 6b + 4c = 0

188

Answer key

Reduce the coefficient matrix of this system: 1 R 2 2 1 1 5 1 5 1⎤ ⎡ ⎤ ⎡ 4 ⎥ 1 ⎢0 1 3 ⎥ 1R1 + R2 ⎢0 6 − R 0 − 7⎥⎥ − 3R + R ⎢⎢0 − 15 − 10⎥⎥ 5 3 ⎢⎢0 1 3 4 ⎥⎦ 1 6 4 ⎥⎦ ⎢⎣0 6 ⎢⎣0 R4 2

⎡1 ⎢− 1 ⎢3 ⎢ ⎢⎣ 0

5 3 3 3

1⎤ ⎡1 2⎥ − 1R2 + R3 ⎢0 2⎥⎥ − 1R + R ⎢⎢0 2 4 2⎥⎦ ⎢⎣0

5 3 0 0

1⎤ 2⎥ 0⎥⎥ 0⎥⎦

This system has infinitely many solutions, so the vectors are linearly dependent. 7. For which value(s) of k are the following vectors linearly dependent? 1 1⎞ 1⎞ ⎛ ⎛ 1 ⎛ 1 1 ⎞ u = ⎜ k, − , − ⎟ , v = ⎜ − , k , − ⎟ , w = ⎜ − , − ,k⎟ 2 2⎠ 2⎠ ⎝ ⎝ 2 ⎝ 2 2 ⎠ 1 k cannot be − because if so, there is just one vector and it is linearly independent. 2 If the vectors are to be linearly dependent, then one must be a linear combination of the others. Without 1 1⎞ 1⎞ ⎛ 1 1 ⎞ ⎛ 1 ⎛ loss of generality, assume ⎜ k , − , − ⎟ = a ⎜ − , k , − ⎟ + b ⎜ − , − , k⎟ . This yields the system 2 2⎠ 2⎠ ⎝ 2 2 ⎠ ⎝ 2 ⎝ a b − − =k 2 2 ka −

1 b =− 2 2

a 1 − + kb = − 2 2 Multiplying by −2 or 2 gives the equivalent system a+ 2ka −

b = − 2k b = −1

− a + 2kb = − 1 Form the augmented matrix and do the indicated row transformations. 1 ⎡ 1 1 − 2k⎤ 1R + R ⎡ 1 1 − 2k ⎤ 2k + 1 R2 1 2 ⎢2k − 1 − 1 ⎥ ⎢2k + 1 0 − 2k − 1⎥ ⎢ ⎥ R +R ⎢ ⎥ 1 3 − − 1 2 k 1 0 2 k + 1 − 2k − 1⎦ 1 R ⎢⎣ ⎦ ⎢⎣ k + 1 3 2

⎡1 1 − 2k⎤ − 1R + R ⎡0 0 − 2k + 2⎤ 2 1 ⎢1 0 − 1 ⎥ ⎢1 0 − 1 ⎥ ⎢ ⎥ − 1R + R ⎢ ⎥ 3 1 − 0 1 1 ⎢⎣ ⎦ ⎢⎣0 1 − 1 ⎦

1 For this to have a nontrivial solution, − 2k + 2 = 0 or k = 1. (Notice you use the fact that k ≠ − when 2 dividing.) Thus, k = 1 is the only value. 8. Prove that for any vectors u, v, and w, the vectors u − v , v − w , and w − u form a linearly dependent set. u − v = − 1( v − w ) − 1( w − u ) is a linear combination of the other two, so the vectors are linearly dependent. 9. Prove that vectors e1 = (1, 0, 0), e 2 = (0,1, 0), and e 3 = (0, 0,1) are orthogonal to each other and each has norm 1. First, show the vectors are orthogonal. e1 u e 2 = 0 + 0 + 0 = 0 , e1 u e 3 = 0 + 0 + 0 = 0, and e 3 u e 2 = 0 + 0 + 0 = 0. Hence, the vectors are orthogonal. Next, show each has norm 1. e1 = e 2 = e 3 = 1 = 1 10. Prove that vectors e1 = (1, 0, 0), e 2 = (0,1, 0), and e 3 = (0, 0,1) span (generate) R3. If (a, b, c ) is in R3, then (a, b, c ) = a(1, 0, 0) + b(0,1, 0) + c(0, 0,1) . Thus, the vectors span R3. Answer key

189

7·3

1. u = (3,1), v = (1, 0), w = (0, 2) for R 2 Any three vectors in 2-space are linearly dependent. 2. u = (0,1,1), v = (4 , 0, 0) for R3 Two vectors cannot form a basis for R3 because every basis for R3 has three vectors (given that the standard basis has three vectors). ⎡1 1⎤ ⎡ 6 0⎤ ⎡0 3⎤ ⎡5 1⎤ ⎡7 1⎤ 3. A = ⎢ , B=⎢ ,C = ⎢ ,D=⎢ ,E=⎢ for M 22 ⎥ ⎥ ⎥ ⎥ − 2 3 1 4 7 1 3 2 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣2 5⎥⎦ Any five vectors are linearly dependent in M 22 . 4. u = (0, 0), v = (1, 2) for R 2 Any set containing the zero vector is a linearly dependent set. 5. {(2,1),(3, 0)} for R 2 (1) If u = (a, b) is any vector in R 2 , then find coefficients, if any, such that (a, b) = x (2,1) + y(3, 0). That is, solve (a , b ) = ( 2 x , x ) + ( 3 y , 0 ) This equality yields the system of equations: 2x + 3 y = a x

=b

This system has a unique solution x = b and y = (a, b) = b(2,1) +

a − 2b (3, 0) 3

a − 2b , so u can be expressed as 3

Therefore, {(2,1),(3, 0)} spans R 2 . (2) Suppose x (2,1) + y(3, 0) = (0, 0), then 2 x + 3 y = 0 and x = 0. Thus, y = 0 also. Because this result is the trivial solution, the vectors are linearly independent. Hence, (1) and (2) are satisfied, so {(2,1),(3, 0)} is a basis for R 2 . 6. {(4 ,1),(− 7 , − 8)} for R 2 (1) If u = (a, b) is any vector in R 2 , then find coefficients, if any, such that (a, b) = x (4 ,1) + y(− 7 , − 8). That is, solve (a , b ) = ( 4 x , x ) + ( − 7 y , − 8 y ) This equality yields the system of equations: 4x − 7 y = a x − 8y = b This system has a unique solution x =

8(a − 4b) a − 4b + b and y = , so u can be expressed as 25 25

a − 4b ⎡8(a − 4b) ⎤ (a , b ) = ⎢ + b⎥ (4 ,1) + ( − 7 , − 8) 25 25 ⎣ ⎦ (2) Suppose x (4 ,1) + y(− 7 , − 8) = (0, 0) , then 4x − 7 y = 0 x − 8y = 0 This system only has the trivial solution because the determinant of the coefficient matrix is not 0. Thus, the vectors are linearly independent. Hence, (1) and (2) are satisfied, so {(4 ,1),(− 7 , − 8)} is a basis for R 2 .

190

Answer key

7. (1, 0, 0),(2, 0, 0),(3, 3, 3) for R3 (2, 0, 0) is a multiple of (1, 0, 0) , and thus the vectors are linearly dependent and cannot be a basis for R3. 8. {1 + x + x 2 , x + x 2 , x 2 } for P2 (1) If u = a0 + a1x + a2 x 2 is any vector in P2 , then find coefficients, if any, such that a0 + a1x + a2 x 2 = r (1 + x + x 2 ) + s( x + x 2 ) + tx 2 . That is, solve a0 + a1x + a2 x 2 = (r + rx + rx 2 ) + (sx + sx 2 ) + tx 2 This equality yields the system of equations: r

= a0

r+s

= a1

r + s + t = a2 This system has the unique solution r = a0 , s = a1 − a0 , and t = a2 − a1, so u can be expressed as a0 + a1x + a2 x 2 = a0 (1 + x + x 2 ) + (a1 − a0 )( x + x 2 ) + (a2 − a1 )x 2. Thus, {1 + x + x 2 , x + x 2 , x 2 } spans P2 . (2) Suppose r (1 + x + x 2 ) + s( x + x 2 ) + tx 2 = 0, then r

=0

r+s

=0

r + s +t = 0 Clearly, the solution to this system is r = s = t = 0, so the vectors are linearly independent. Hence, (1) and (2) are satisfied, so {1 + x + x 2 , x + x 2 , x 2 } is a basis for P2 . 9. {1, x , x 3 } for P3 Three vectors cannot form a basis for P3 because every basis for P3 has four elements (given that the standard basis has four elements). Thus, {1, x , x 3 } cannot form a basis for P3. ⎧ ⎡1 3 ⎤ ⎡0 1⎤ ⎡ 0 − 2⎤ ⎡ 1 0⎤ ⎫ 10. ⎨ ⎢ ⎥,⎢ ⎥,⎢ ⎥,⎢ ⎥ ⎬ for M 22 ⎩ ⎣1 − 3⎦ ⎣1 0⎦ ⎣− 3 − 1⎦ ⎣− 1 2⎦ ⎭ ⎡a b⎤ is any vector in M 22, then find coefficients, if any, such that (1) If ⎢ ⎣c d⎥⎦ ⎡ 0 − 2⎤ ⎡ 1 0⎤ . That is, solve ⎡a b⎤ = ⎡x z⎢ +t ⎢⎣c d⎥⎦ ⎢⎣x ⎣− 3 − 1⎥⎦ ⎢⎣− 1 2⎥⎦

3x ⎤ ⎡0 + − 3x⎥⎦ ⎢⎣ y

⎡a b⎤ ⎡1 3 ⎤ ⎡0 1⎤ ⎢⎣c d⎥⎦ = x ⎢⎣1 − 3⎥⎦ + y ⎢⎣1 0⎥⎦ +

y⎤ ⎡ 0 + 0⎥⎦ ⎢⎣− 3z

− 2z ⎤ ⎡ t + − z ⎥⎦ ⎢⎣−t

0⎤ . 2t⎥⎦

This equality yields the system of equations: + t =a

x 3x + y − 2 z

=b

x + y − 3z − t = c − 3x −

z + 2t = d

with augmented matrix that reduces as follows: ⎡1 ⎢3 ⎢ ⎢1 ⎣⎢− 3

0 0 1 a⎤ − 3R1 + R2 ⎡1 ⎢0 1 − 2 0 b⎥ − 1R1 + R3 ⎢ ⎥ 1 −3 −1 c ⎥ ⎢0 0 − 1 2 d⎥⎦ 3R1 + R4 ⎢⎣0

0 0 1 a ⎤ ⎡1 ⎢0 1 − 2 − 3 − 3a + b⎥ 1R2 − + R3 ⎢ ⎥ 1 − 3 − 2 −a + c ⎥ ⎢0 0 − 1 5 3a + d ⎥⎦ ⎢⎣0

0 0 1 a ⎤ 1 − 2 − 3 − 3a + b ⎥ 0 − 1 1 2a − b + c⎥⎥ 0 − 1 5 3a + d ⎥⎦ Answer key

191

a 0 0 1 ⎡1 ⎤ 1 − 2 − 3 − 3a + b ⎥ 2 R3 + R2 ⎢0 ⎢ 0 1 − 1 − 2a + b − c⎥⎥ 1R + R ⎢0 3 4 0 − 1 5 3a + d ⎥⎦ ⎢⎣0

⎡1 ⎢0 − 1R3 ⎢ ⎢0 ⎣⎢0 ⎡1 ⎢ 1 ⎢0 R4 0 4 ⎢ ⎢0 ⎢⎣

0 1 0 0

0 1 0 −5 1 −1 0 1

⎡ ⎢ a ⎤ − 1R + R ⎢1 4 1 − 7a + 3b − 2c⎥ ⎢0 − 2a + b − c ⎥⎥ 5 R4 + R2 ⎢0 ⎢ a+b−c +d ⎥ ⎢0 1 + R R 4 3 ⎥⎦ 4 ⎢ ⎢ ⎣

0 1 0 0

0 1 a ⎤ 0 − 5 − 7a + 3b − 2c⎥ 1 − 1 − 2a + b − c ⎥⎥ 0 4 a + b − c + d ⎥⎦ a+b−c +d ⎤ ⎥ 4 0 ⎛ a + b − c + d⎞⎥ ⎟⎠ ⎥ 0 − 7a + 3b − 2c + 5 ⎜⎝ 4 ⎥ 0 a+b−c +d ⎥ − 2 + b c a − + ⎥ 1 4 ⎥ a+b−c +d ⎥ 4 ⎦ a−

0 1 0 0

0 0 1 0

⎡a b⎤ This system has a unique solution, from which you can express ⎢ as ⎣c d⎥⎦ ⎡a b⎤ ⎛ a + b − c + d ⎞ ⎡1 3 ⎤ ⎛ ⎛ a + b − c + d ⎞ ⎞ ⎡0 1⎤ ⎟⎠ ⎟ ⎢1 0⎥ ⎟⎠ ⎢1 − 3⎥ + ⎜ − 7a + 3b − 2c + 5 ⎜⎝ ⎢⎣c d⎥⎦ = ⎜⎝ a − 4 4 ⎠⎣ ⎝ ⎦ ⎣ ⎦ a + b − c + d ⎞ ⎡ 0 − 2⎤ ⎛ a + b − c + d ⎞ ⎡ 1 0⎤ ⎛ + ⎜ − 2a + b − c + ⎟⎠ ⎢− 3 − 1⎥ + ⎜⎝ ⎟⎠ ⎢− 1 2⎥ 4 4 ⎝ ⎣ ⎦ ⎣ ⎦ ⎧ ⎡1 3 ⎤ ⎡0 1⎤ ⎡ 0 − 2⎤ ⎡ 1 0⎤ ⎫ So ⎨ ⎢ ⎥,⎢ ⎥,⎢ ⎥,⎢ ⎥ ⎬ spans M 22. ⎩ ⎣1 − 3⎦ ⎣1 0⎦ ⎣− 3 − 1⎦ ⎣− 1 2⎦ ⎭ ⎡1 3 ⎤ ⎡0 1⎤ ⎡ 0 − 2⎤ ⎡ 1 0⎤ ⎡0 0⎤ (2) Suppose x ⎢ + y⎢ +z +t = , then ⎥ 1 − 3 ⎣ ⎦ ⎣1 0⎥⎦ ⎢⎣− 3 − 1⎥⎦ ⎢⎣− 1 2⎥⎦ ⎢⎣0 0⎥⎦ + t =0

x 3x + y − 2 z

=0

x + y − 3z − t = 0 − 3x

− z + 2t = 0

This system has only the trivial solution because the determinant of the coefficient matrix = ⎛⎡1 ⎜⎢3 det ⎜ ⎢ ⎜⎢1 ⎝ ⎣⎢− 3

0 0 1 ⎤⎞ 1 −2 0 ⎥ ⎟ = − 4 ≠ 0. Thus, the vectors are linearly independent. 1 − 3 − 1⎥⎥ ⎟⎟ 0 − 1 2 ⎥⎦ ⎠

⎧ ⎡1 3 ⎤ ⎡0 1⎤ ⎡ 0 − 2⎤ ⎡ 1 0⎤ ⎫ Hence, (1) and (2) are satisfied, so ⎨ ⎢ ⎥,⎢ ⎥,⎢ ⎥,⎢ ⎥ ⎬ is a basis for M 22 . ⎩ ⎣1 − 3⎦ ⎣1 0⎦ ⎣− 3 − 1⎦ ⎣− 1 2⎦ ⎭

7·4

1. The set of all vectors of the form (a, b, c , 0) For any vector in the subspace, (a, b, c , 0) = a(1, 0, 0, 0) + b(0,1, 0, 0) + c(0, 0,1, 0). Hence, the three linearly independent vectors (1, 0, 0, 0), (0,1, 0, 0), and (0, 0,1, 0) span the subspace. Thus, the dimension of the subspace is 3. 2. The set of all vectors of the form (a, a, a, a) For any vector in the subspace, (a, a, a, a) = a(1,1,1,1) . Hence, the subspace is spanned by (1,1,1,1); and, therefore, has dimension 1. 3. Determine the dimension of the subspace of P3 consisting of all polynomials of the form a1x + a2 x 2 + a3 x 3 . For any polynomial in the subspace, a1x + a2 x 2 + a3 x 3 = a1 ( x ) + a2 ( x 2 ) + a3 ( x 3 ). Hence, the linearly independent vectors x, x 2 , and x 3 span the subspace. Thus, the dimension of the subspace is 3.

192

Answer key

⎡a − 2b⎤ 4. Determine the dimension of the subspace of M31 consisting of vectors of the form ⎢ a + b ⎥ . ⎢ ⎥ ⎣ 3b ⎦ ⎧ ⎫ ⎡a − 2b⎤ ⎡1⎤ ⎡− 2⎤ ⎡1⎤ ⎡− 2⎤ ⎪ ⎪ For any vector in the subspace, ⎢ a + b ⎥ = a ⎢1⎥ + b ⎢ 1 ⎥ and thus ⎨ ⎢1⎥ , ⎢ 1 ⎥ ⎬ spans the subspace. Also, ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎣0⎦ ⎣ 3 ⎦ ⎪ ⎣ 3b ⎦ ⎣0⎦ ⎣ 3 ⎦ ⎩ ⎭ ⎡1⎤ ⎡− 2⎤ ⎡0⎤ these vectors are linearly independent because a ⎢1⎥ + b ⎢ 1 ⎥ = ⎢0⎥ , yields ⎢⎥ ⎢ ⎥ ⎢⎥ ⎣0⎦ ⎣ 3 ⎦ ⎣0⎦ a − 2b = 0 a+ b=0 3b = 0 ⎧ ⎡1⎤ ⎡− 2⎤ ⎫ ⎪ ⎪ which has only has the trivial solution. Hence, ⎨ ⎢1⎥ , ⎢ 1 ⎥ ⎬ is a basis for the subspace. Thus, the dimension ⎢ ⎥ ⎢ ⎥ ⎪ ⎣0⎦ ⎣ 3 ⎦ ⎪ of the subspace is 2. ⎩ ⎭ ⎡1⎤ ⎡− 2⎤ ⎡− 3⎤ 5. Find the dimension of the subspace of R3 spanned by ⎢5⎥ , ⎢10 ⎥ , and ⎢15⎥ . ⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎣0⎦ ⎣ 0 ⎦ ⎣0⎦ ⎡1⎤ ⎡− 2⎤ The vectors ⎢5⎥ , ⎢10 ⎥ , and ⎢0⎥ ⎢ 0 ⎥ ⎣⎦ ⎣ ⎦

⎡− 3⎤ ⎢15⎥ are linearly dependent because ⎢ ⎥ ⎣0⎦

⎡1⎤ ⎡− 2⎤ ⎡− 3⎤ ⎡0⎤ if a ⎢5⎥ + b ⎢10 ⎥ + c ⎢15⎥ = ⎢0⎥ , then ⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎣0⎦ ⎣ 0 ⎦ ⎣ 0 ⎦ ⎣0⎦ a − 2b − 3c = 0 5a + 10b + 15c = 0 which has more unknowns than equations and, thus, nontrivial solutions. Examination of the three ⎡− 2⎤ 2 vectors shows that ⎢10 ⎥ = ⎢ ⎥ 3 ⎣0⎦

⎡− 2⎤ ⎡− 3⎤ ⎡1⎤ ⎢15⎥ , so consider the two vectors ⎢5⎥ and ⎢10 ⎥ . ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎣0⎦ ⎣0⎦ ⎣0⎦

⎡1⎤ ⎡− 2⎤ ⎡0⎤ If a ⎢5⎥ + b ⎢10 ⎥ = ⎢0⎥ , then ⎢⎥ ⎢ ⎥ ⎢⎥ ⎣0⎦ ⎣ 0 ⎦ ⎣0⎦ a − 2b = 0 5a + 10b = 0 ⎡1⎤ This system has only the trivial solution. Thus, the vectors ⎢5⎥ , and ⎢⎥ ⎣0⎦ ⎧ ⎡1⎤ ⎡− 2⎤ ⎫ ⎪⎢ ⎥ ⎢ ⎥⎪ 3 ⎨ 5 , 10 ⎬ is a basis for a subspace of R of dimension 2. ⎢ ⎥ ⎢ ⎥ ⎪ ⎣0⎦ ⎣ 0 ⎦ ⎪ ⎩ ⎭

⎡− 2⎤ ⎢10 ⎥ are linearly independent. Hence, ⎢ ⎥ ⎣0⎦

Answer key

193

7·5

⎡1 3 ⎤ 1. A = ⎢ ; ⎣2 − 3⎥⎦

⎡− 1⎤ b=⎢ ⎥ ⎣5⎦

det( A) = − 3 − 6 = − 9 ≠ 0, so the system is consistent and, thus, b is in the column space of A. ⎡1 0 4 3 ⎤ The augmented matrix can be reduced to ⎢ , from which you have ⎣0 1 − 7 9⎥⎦ 1 3 − 1 4 7 ⎡ ⎤ ⎡⎤ ⎡ ⎤ b= ⎢ ⎥ = ⎢ ⎥− ⎢ ⎥ ⎣ 5 ⎦ 3 ⎣2⎦ 9 ⎣− 3⎦ ⎡1 1 2⎤ 2. A = ⎢1 0 1⎥ ; ⎢ ⎥ ⎣2 1 3⎦

⎡− 1⎤ b=⎢0 ⎥ ⎢ ⎥ ⎣4⎦

det( A) = 0 and the augmented matrix can be reduced to ⎡1 1 2 − 1⎤ ⎢0 1 1 − 1⎥ The system has no solution, so b is not in the column space of A. ⎢ ⎥ ⎣0 0 0 5 ⎦ ⎡1 − 1 1⎤ 3. A = ⎢9 3 1⎥ ; ⎢ ⎥ ⎣1 1 1⎦

⎡5⎤ b=⎢1 ⎥ ⎢ ⎥ ⎣− 1⎦

det( A) = 16 ≠ 0, so the system is consistent, and, thus, b is in the column space of A. The system has ⎡ 5 ⎤ ⎡1⎤ ⎡− 1⎤ ⎡1⎤ ⎡1⎤ solution ⎢− 3⎥ from which you have b = ⎢ 1 ⎥ = ⎢9⎥ − 3 ⎢ 3 ⎥ + ⎢1⎥ . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎣− 1⎦ ⎣1⎦ ⎣ 1 ⎦ ⎣1⎦ ⎣1⎦ ⎡− 1 − 1 1 ⎤ 4. A = ⎢ 1 1 − 1⎥ ; ⎢ ⎥ ⎣ 1 −1 1 ⎦

⎡1⎤ b = ⎢0⎥ ⎢⎥ ⎣0⎦

The augmented matrix quickly reduces to ⎡− 1 − 1 1 1⎤ ⎢ 0 0 0 1⎥ ⎢ ⎥ ⎣ 1 − 1 1 0⎦ This system has no solution, so b is not in the column space of A. 5.

x1 − 3x 2 = 1 2 x1 − 6 x 2 = 2 The coefficients are proportional, so the solution is x1 = 3x 2 + 1 where, x 2 , is a free variable. The general solution in vector form is ⎡x1 ⎤ ⎡3x 2 + 1⎤ ⎡1⎤ ⎡3⎤ ⎢⎣x 2 ⎥⎦ = ⎢⎣ x 2 ⎥⎦ = ⎢⎣0⎥⎦ + x 2 ⎢⎣1⎥⎦ The general solution of the homogenous system is ⎡x1 ⎤ ⎡3⎤ ⎢⎣x 2 ⎥⎦ = x 2 ⎢⎣1⎥⎦ x + y + 2z = 5

6.

x

+ z = −2

2 x + y + 3z = 3

194

Answer key

Reduce the augmented matrix: 2 5⎤ 2 5 ⎤ 1R2 + R1 ⎡1 1 2 5 ⎤ − 1R1 + R2 ⎡1 1 ⎡1 1 ⎢1 0 1 − 2⎥ ⎢0 − 1 − 1 − 7⎥ − 1R + R ⎢0 − 1 − 1 − 7⎥ 2 3 ⎢ ⎢ ⎥ −2R + R ⎢ ⎥ ⎥ − 1R2 1 3 ⎣0 − 1 − 1 − 7⎦ ⎣2 1 3 3 ⎦ ⎣0 0 0 0 ⎦

⎡1 0 1 − 2⎤ ⎢0 1 1 7 ⎥ ⎢ ⎥ ⎣0 0 0 0 ⎦

The solution is x = − z − 2 and y = − z + 7, where z is a free variable. The general solution in vector form is ⎡x ⎤ ⎡− z − 2⎤ ⎡− 2⎤ ⎡− 1⎤ ⎢ y⎥ = ⎢− z + 7⎥ = ⎢ 7 ⎥ + z ⎢− 1⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣z ⎦ ⎣ z ⎦ ⎣ 0 ⎦ ⎣ 1 ⎦ The general solution of the homogenous system is ⎡x ⎤ ⎡− 1⎤ ⎢ y⎥ = z ⎢− 1⎥ ⎢ ⎥ ⎢ ⎥ ⎣z ⎦ ⎣1⎦ ⎡2 0 − 1⎤ 7. A = ⎢4 0 − 2⎥ ⎢ ⎥ ⎣0 1 0 ⎦ The basis for the null space is found by solving the homogeneous system Ax = 0 . Because the reduced row-echelon form of A is ⎡1 0 − 1 ⎤ 2⎥ ⎢ 0 ⎥ ⎢0 1 ⎢0 0 0 ⎥ ⎣ ⎦

⎡x ⎤ ⎡1/ 2⎤ 1 The solution is x = z , y = 0, where z is a free variable. The vector form of the solution is ⎢ y⎥ = z ⎢ 0 ⎥ . 2 ⎢ ⎥ ⎢ ⎥ ⎣z ⎦ ⎣1⎦ ⎧ ⎡1 2⎤ ⎫ ⎪⎢ ⎥⎪ Thus, the null space has basis S = ⎨ 0 ⎬ , which can also be written as S = {[1/ 2 0 1]}. ⎪ ⎢⎣ 1 ⎥⎦ ⎪ ⎩ ⎭ ⎡ 1 4 5 2⎤ 8. A= ⎢ 2 1 3 0⎥ ⎢ ⎥ ⎣− 1 3 2 2⎦ The basis for the null space is found by solving the homogeneous system Ax = 0 . Because the reduced row-echelon form of A is ⎡1 0 1 − 2 ⎤ 7⎥ ⎢ A = ⎢0 1 1 4 7 ⎥ ⎢0 0 0 0 ⎥⎥ ⎢⎣ ⎦ 2 4 The solution is x = − z + w and y = − z − w , where z and w are free variable. The vector form of the 7 7 2 ⎡2 ⎤ ⎡ x ⎤ ⎡⎢− z + 7 w⎤⎥ ⎡− 1⎤ ⎢ 7⎥ ⎢ y ⎥ ⎢− z − 4 w⎥ ⎢− 1⎥ 4 solution is ⎢ ⎥ = 7 ⎥ = z ⎢ ⎥ + w ⎢⎢− 7⎥⎥ . Thus, the null space has basis ⎢ z 1 ⎢ ⎥ ⎢ ⎢ ⎥ z ⎥ ⎢ 0 ⎥ 0 ⎥⎦ ⎢⎣w⎥⎦ ⎢ ⎢ ⎣ w ⎢⎣ 1 ⎥⎦ ⎣ ⎦⎥

⎧ −1 ⎡ 2 ⎤ ⎫ ⎪⎡ ⎤ ⎢ 7 ⎥⎪ ⎪ ⎢− 1⎥ ⎢− 4 ⎥ ⎪ ⎨ ⎢ 1 ⎥ , ⎢ 7⎥ ⎬ . ⎪⎢ ⎥ ⎢ 0 ⎥⎪ ⎪ ⎣⎢ 0 ⎦⎥ ⎣⎢ 1 ⎥⎦ ⎪ ⎩ ⎭

Answer key

195

⎡1 0 2⎤ 9. A = ⎢0 0 1⎥ ⎢ ⎥ ⎣0 0 0⎦ ⎧ ⎡1⎤ ⎡2⎤ ⎫ ⎪ ⎪ A basis for the row space is S = {[1 0 2],[0 0 1]} and for the column space is T = ⎨ ⎢0⎥ , ⎢1⎥ ⎬ ⎢ ⎥ ⎢ ⎥ ⎪ ⎣0⎦ ⎣0⎦ ⎪ ⎩ ⎭ ⎡1 2 − 1 5 ⎤ ⎢0 1 4 3⎥ 10. A = ⎢ ⎥ ⎢0 0 1 − 7⎥ 1 ⎦⎥ ⎣⎢0 0 0 A basis for the row space is S = {[1 2 − 1 5],[0 1 4 3],[0 0 1 − 7],[0 0 0 1]} and for the ⎧ ⎡1⎤ ⎡2⎤ ⎡− 1⎤ ⎡ 5 ⎤ ⎫ ⎪⎪ ⎢0⎥ ⎢1⎥ ⎢ 4 ⎥ ⎢ 3 ⎥ ⎪⎪ column space T = ⎨ ⎢ ⎥ , ⎢ ⎥ , ⎢ ⎥ , ⎢ ⎥ ⎬ 0 0 1 −7 ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪⎩ ⎣⎢0⎦⎥ ⎣⎢0⎦⎥ ⎣⎢ 0 ⎦⎥ ⎣⎢ 1 ⎦⎥ ⎪⎭

7·6

1. Determine the condition(s) under which the following system is consistent. x1 − 3x 2 = b1 x1 − 2 x 2 = b2 x1 + x 2 = b3 This is an overdetermined system and thus cannot have a solution for every possible vector b. Using Gauss-Jordan elimination, the augmented matrix is ⎡1 0 − 2b1 + 3b2 ⎤ ⎢0 1 b2 − b1 ⎥ ⎢ ⎥ b 0 0 3 1 − 4b2 + b3 ⎦ ⎣ Hence, the original system is consistent when 3b1 − 4b2 + b3 = 0 . 2. Verify that rank ( A) = rank ( AT ) for the following matrix. 4 ⎤ ⎡1 0 − 6 7 − 7⎥ ⎡ 1 2 4 0⎤ ⎢ A = ⎢− 3 1 5 2⎥ is row equivalent to ⎢0 1 17 7 2 7 ⎥ . ⎢ ⎥ ⎥ ⎢ 0 0 ⎥ ⎣− 2 3 9 2⎦ ⎢⎣0 0 ⎦ Thus, rank( A) = 2. Reduce the matrix AT . ⎡1 − 3 − 2⎤ ⎡1 ⎢2 1 ⎥ ⎢0 3 AT = ⎢ is row equivalent to ⎢ 9 ⎥⎥ ⎢4 5 ⎢0 2 ⎥⎦ ⎢⎣0 2 ⎢⎣0

0 1 0 0

1⎤ 1⎥ 0⎥⎥ 0⎥⎦

Thus, rank( AT ) = 2. Because rank A = rank ( AT ) = 2, verification is established. ⎡1 − 1 3 ⎤ 3. A = ⎢5 − 4 − 4⎥ ⎢ ⎥ ⎣7 − 6 2 ⎦ ⎡1 − 1 3 ⎤ − 5 R1 + R2 ⎢5 − 4 − 4⎥ ⎢ ⎥ −7R + R 1 3 ⎣7 − 6 2 ⎦

⎡1 − 1 3 ⎤ ⎢0 1 − 19⎥ − 1R + R 2 3 ⎢ ⎥ ⎣0 1 − 19⎦

rank( A) = 2 and nullity( A) = 1

196

Answer key

⎡1 − 1 3 ⎤ ⎢0 1 − 19⎥ 1R + R 2 1 ⎢ ⎥ 0 ⎦ ⎣0 0

⎡1 0 − 16⎤ ⎢0 1 − 19⎥ ⎢ ⎥ ⎣0 0 0 ⎦

⎡2 0 1⎤ 4. A = ⎢4 0 2⎥ ⎢ ⎥ ⎣0 1 0⎦ ⎡1 0 ⎡2 0 1⎤ ⎢4 0 2⎥ is row equivalent to ⎢0 1 ⎢ ⎢ ⎥ ⎢0 0 ⎣0 1 0⎦ ⎣

1 ⎤ 2⎥ 0⎥ 0⎥ ⎦ rank( A) = 2 and nullity( A) = 1 (because the solution of the homogeneous system has exactly 1 free variable). ⎡ 1 4 5 2⎤ 5. A = ⎢ 2 1 3 0⎥ ⎢ ⎥ ⎣− 1 3 2 2⎦

⎡1 0 1 − 2 ⎤ 7⎥ ⎡ 1 4 5 2⎤ ⎢ ⎢ 2 1 3 0⎥ is row equivalent to ⎢0 1 1 4 ⎥ 7 ⎢ ⎥ ⎢0 0 0 0 ⎥⎥ ⎣− 1 3 2 2⎦ ⎢⎣ ⎦ rank ( A) = 2 and nullity ( A) = 2 (because the solution of the homogeneous system has exactly 2 free variables.)

⎡1 1 t ⎤ 6. Discuss how the rank of the matrix A = ⎢1 t 1⎥ varies with t. ⎢ ⎥ ⎣1 1 t ⎦ t ⎤ t ⎤ ⎡1 1 ⎡1 1 t ⎤ 1R1 + R2 ⎡1 1 ⎢2 t + 1 t + 1⎥ − 1R + R ⎢2 t + 1 t + 1⎥ − 2 R + R ⎢1 t 1⎥ 2 1 3 ⎢ 2 ⎥ ⎥ ⎢ ⎥ 1R + R ⎢ 2 0 ⎦ 3 ⎣2 t + 1 t + 1⎦ ⎣0 0 ⎣1 1 t ⎦

t ⎤ ⎡1 1 ⎢0 t − 1 −t + 1⎥ ⎢ ⎥ 0 ⎦ ⎣0 0

If t = 1 the rank is 1 and if t ≠ 1 the rank is 2. ⎡2 0 2⎤ 7. Solve the homogeneous system Ax = 0 if matrix A = ⎢1 1 1⎥ . ⎢ ⎥ ⎣1 0 1⎦ ⎡2 0 2⎤ − 1R3 + R2 ⎡0 0 0⎤ ⎢0 1 0⎥ ⎢1 1 1⎥ ⎥ ⎢ ⎥ −2R + R ⎢ 3 1 ⎣1 0 1⎦ ⎣1 0 1⎦ The solution is x = − z, y = 0. In vector form ⎡x ⎤ ⎡− 1⎤ ⎢ y⎥ = z ⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣z ⎦ ⎣1⎦ 8. If A is an m × n matrix, what is the largest possible value for its rank and the smallest possible value for its nullity? The largest possible value for rank is minimum (m,n). The smallest possible value for nullity is 0. 9. Express (4a, a − b, a + 2b) as a linear combination of (4 ,1,1) and (0, − 1, 2) . (4a, a − b, a + 2b) = (4a, a, a) + (0, −b, 2b) = a(4 ,1,1) + b(0, − 1, 2) 10. Find a basis for the vector space of all 2 × 2 symmetric matrices. ⎡1 0 ⎤ ⎡ 0 1 ⎤ ⎡ 0 0 ⎤ ⎡a b ⎤ For any 2 × 2 symmetric matrix ⎢ = a⎢ ⎥+c⎢ ⎥ . Thus, the vectors in ⎥ ⎥+b⎢ ⎣b c ⎦ ⎣ 0 0 ⎦ ⎣1 0 ⎦ ⎣ 0 1 ⎦ ⎧ ⎡1 0⎤ ⎡0 1⎤ ⎡0 0⎤ ⎫ ⎨ ⎢0 0⎥ , ⎢1 0⎥ , ⎢0 1⎥ ⎬ span the space. In addition, these vectors are linearly independent because if ⎦ ⎣ ⎦ ⎣ ⎦⎭ ⎩⎣ ⎧ ⎡1 0⎤ ⎡0 1⎤ ⎡0 0⎤ ⎫ ⎡1 0⎤ ⎡0 1⎤ ⎡0 0⎤ ⎡0 0⎤ +c⎢ =⎢ , then a = b = c = 0. Consequently, ⎨ ⎢ a⎢ +b⎢ ⎥,⎢ ⎥,⎢ ⎥ ⎬ is a basis for ⎥ ⎥ ⎥ ⎥ 0 0 1 0 0 1 0 0 ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎩ ⎣0 0⎦ ⎣1 0⎦ ⎣0 1⎦ ⎭ the vector space of all 2 × 2 symmetric matrices. Answer key

197

8

Inner product spaces

8·1

1. Prove that 〈u , kw 〉 = k 〈u , w 〉. 〈u ,kw 〉 = 〈kw , u 〉

By the symmetry property of inner products.

= k〈w , u〉

By the proof in an example problem.

= k 〈u , w 〉

By the symmetry property of inner products.

2. Prove that 〈ku , kw 〉 = k 2 〈u , w 〉 . 〈ku , kw 〉 = k 〈u , kw 〉

By property (b).

= k( k 〈 u , w 〉 )

By property (b).

= k 2 〈u , w 〉

By real number property.

3. Prove that 〈u ,av + bw 〉 = a〈u , v 〉 + b 〈u , w 〉. 〈u ,av + bw 〉 = 〈av + bw , u 〉

By symmetry.

= a〈 v , u 〉 + b〈 w , u 〉

By linearity.

= a〈u , v 〉 + b 〈u , w 〉

By symmetry.

4. Prove that 〈u − v , w 〉 = 〈u , w 〉 − 〈 v , w 〉 . 〈u − v , w 〉 = 〈1u + (−1)v , w 〉

By scalar multiplication property.

= 1〈u , w 〉 + (−1)〈 v , w 〉

By Prob. 3.

= 〈u , w 〉 − 〈 v , w 〉

By scalar multiplication property.

5. Verify that the following is an inner product in R 2 : 〈u , v 〉 = x1 y1 − x1 y 2 − x 2 y1 + 2 x 2 y 2. i. 〈u , u 〉 = x1 x1 − x1 x 2 − x 2 x1 + 2 x 2 x 2 = x12 − 2 x1 x 2 + x 22 + x 22 = ( x1 − x 2 )2 + x 22 ≥ 0; and the equality holds (that is, ( x1 − x 2 )2 + x 22 = 0 ) if and only if x1 = x 2 = 0 , i.e., u = 0. ii. 〈u , v 〉 = x1 y1 − x1 y 2 − x 2 y1 + 2 x 2 y 2 = y1 x1 − y1 x 2 − y 2 x1 + 2 y 2 x 2 = 〈 v , u 〉 iii. 〈au + bv , w 〉 = 〈(ax1 + by1 , ax 2 + by 2 ),(w1 , w 2 )〉 = (ax1 + by1 )w1 − (ax1 + by1 )w 2 − (ax 2 + by 2 )w1 + 2(ax 2 + by 2 )w 2 = a( x1w1 − x1w 2 − x 2 w1 + 2 x 2 w 2 ) + b( y1w1 − y1w 2 − y 2 w1 + 2yy 2 w 2 ) = a〈u , w 〉 + b 〈 v , w 〉 6. Show that 〈u , v 〉 = x1 y1 x 2 y 2 is not an inner product in R 2. A counterexample will suffice. Let k = 2, u = (1,1), and v = (1, 4 ). k〈u , v 〉 = 2(1 ⋅1 ⋅1 ⋅ 4 ) = 8 〈ku, v 〉 = 〈(2, 2),(1, 4 )〉 = 2 ⋅1 ⋅ 2 ⋅ 4 = 16 Because k〈u , v 〉 ≠ 〈ku, v 〉, property (b) does not hold.

198

Answer key

7. Find 〈u , v 〉 with respect to the usual inner product (dot product) of R 2. 〈u , v 〉 = 1 ⋅ 3 + 4 ⋅ 5 = 23 8. Find 〈 v , w 〉 with respect to the inner product of Prob. 5. 〈 v , w 〉 = 3 ⋅ 2 − 3 ⋅ 3 − 5 ⋅ 2 + 2 ⋅ 5 ⋅ 3 = 17 9. Find 〈u , w 〉 with respect to the usual inner product of R 2. 〈u , w 〉 = 2 + 12 = 14 10. Find 〈u , w 〉 with respect to the inner product of Prob. 5. 〈u , w 〉 = 1 ⋅ 2 − 1 ⋅ 3 − 4 ⋅ 2 + 2 ⋅ 4 ⋅ 3 = 15 11. Find 〈3u + 2 v , w 〉 with respect to the inner product of Prob. 5. 〈3u + 2 v , w 〉 = 3〈u , w 〉 + 2〈 v , w 〉 = 3(15) + 2(17 ) = 79 , using the results of Probs. 8 and 10. 12. Find 〈 A, B〉. 〈 A, B〉 = 1(−1) + 3(0) + 2(0) + 7(2) = 13 13. Find 〈 A, C 〉 . 〈 A, C 〉 = 1(1) + 3(1) + 2(−1) + 7(1) = 9 14. Find 〈 A, A〉. ⎛ ⎡ 1 3 ⎤ ⎡1 2 ⎤⎞ 2 2 2 2 〈 A, A〉 = tr ( AT A) = tr ⎜ ⎢ ⎥⎟ = 1 + 2 + 3 + 7 = 63 ⎥⎢ ⎝ ⎣ 2 7 ⎦ ⎣ 3 7 ⎦⎠ 15. In the realm of complex numbers an inner product on C n , the space of ordered n-tuples of complex numbers, can be defined as follows: n

If z = (z1 , z 2 ,…, z n ) and w = (w1 , w 2 ,…, wn ) then 〈z ,w 〉 = ∑ z k w k . Use this definition to find the inner k =1

product of z = (3, i , 2 − i ) and w = (5, −i ,1) in C 3. 3

〈z, w 〉 = ∑ z k w k = 3( 5 ) + i ( −i ) + (2 − i )( 1 ) = 15 + i 2 + 2 − i = 16 − i k =1

8·2

1. Find the norm of u = (5, 6,1) with respect to the usual inner product (dot product) of R 3. (5, 6,1) = (5, 6,1)@(5, 6,1) = 25 + 36 + 1 = 62 ⎡1 0 ⎤ 2. If A = ⎢ ⎥ , find the norm of A with respect to the trace inner product. ⎣3 4 ⎦ ⎡1 0 ⎤ ⎥ = ⎢ ⎣3 4 ⎦

⎡1 0 ⎤ ⎡1 0 ⎤ T 2 2 2 2 ⎥ = tr( A A) = 1 + 3 + 0 + 4 = 26 ⎥,⎢ ⎢ 3 4 3 4 ⎦ ⎦ ⎣ ⎣

3. Let u = ( x1 , x 2 ) = (4 , 3), find the norm of u with respect to the inner product 〈u , v 〉 = x1 y1 − x1 y 2 − x 2 y1 + 2 x 2 y 2, where u = ( x1 , x 2 ) and v = ( y1 , y 2 ), as defined in Prob. 5, Exercise 8.1. u = ( x1 , x 2 ) = (4 , 3) = 〈(4 , 3),(4 , 3)〉 = x1 x1 − x1 x 2 − x 2 x1 + 2 x 2 x 2 = (4 )(4 ) − (4 )(3) − (3)(4 ) + 2(3)(3) = 10 4. Find the norm of u = (4 , 3) with respect to the usual inner product (dot product) of R 2. (4 , 3) = 16 + 9 = 5 5. Normalize u = (4 , 3) with respect to the inner product 〈u , v 〉 = x1 y1 − x1 y 2 − x 2 y1 + 2 x 2 y 2 , where u = (x1, x2) and v = (y1,y 2), as defined in Prob. 5, Exercise 8.1. ⎛ 4 1 1 1 3 ⎞ u= u= (4 , 3) = ⎜ , ⎟ , using the result of Prob. 3. u 10 ⎝ 10 10 10 ⎠

Answer key

199

⎡1 0 ⎤ 6. Normalize A = ⎢ ⎥ with respect to the trace inner product. ⎣3 4 ⎦ ⎡1 0 ⎤ 1 1 ⎡1 0 ⎤ ⎢ 26 ⎥ A= = ⎢ ⎥ ⎥, using the result of Prob. 2. 4 A 26 ⎣3 4 ⎦ ⎢ 3 ⎢⎣ 26 26 ⎥⎦ 7. Find the norm of z = (i ,1 − i , 5i ) with respect to the inner product of Prob. 15, Exercise 8.1. z = 〈z , z 〉 =

3

∑z z

k k

= ii + (1 − i )(1 − i ) + (5i )(5i ) = 1 + (12 + (−1)2 ) + 25 = 28

k=1

8. Normalize z = (i ,1 − i , 5i ) with respect to the inner product of Prob. 15, Exercise 8.1. 1 1 z= (i ,1 − i , 5i ), using the result of Prob. 7. z 28 9. Find the norm of u = (3, 5, 0) with respect to the usual inner product (dot product) of R 3. (3, 5, 0) = 32 + 52 = 34 10. Normalize u = (3, 5, 0) with respect to the usual inner product (dot product) of R 3. 1 1 5 ⎛ 3 ⎞ u= (3, 5, 0) = ⎜ , , 0⎟ , using the result of Prob. 9. ⎝ u 34 34 34 ⎠

8·3

1. Verify that −4 u = 4 u . −4 u = (−20, −20) = (−20)2 + (−20)2 = 202 (2) = 20 2 4 u = 4 ( 52 + 52 ) = 4 (5 2 ) = 20 2 Thus, −4 u = 4 u is verified. 2. Verify that u + v ≤ u + v . u + v = (5, 5) + (3, −2) = (8, 3) = 82 + 32 = 73 ≈ 8.54 u + v = (5, 5) + (3, −2) = 52 + 52 + 32 + (−2)2 = 5 2 + 13 ≈ 10.68 Thus u + v ≤ u + v is verified. 3. Calculate 〈 A, B〉 . 〈 A, B〉 = tr( AT B) = 1(3) + 5(3) + (−2)3 + 3(3) = 21 4. Calculate 〈 A, A〉 . 〈 A, A〉 = 12 + 52 + (−2)2 + 32 = 39 5. Calculate 〈 B, B〉 〈 B, B〉 = 32 + 32 + 32 + 32 = 36 = 6

200

Answer key

8·4

1. Prove that if u ⊥ v , then ku ⊥ v for every real number k. If u ⊥ v , then 〈u , v 〉 = 0. 〈ku , v 〉 = k 〈u , v 〉

By inner product property.

k 〈 u , v 〉 = k( 0 ) = 0

Because 〈u , v 〉 = 0.

Thus, ku ⊥ v for every real number k. 2. Prove or disprove: If u ⊥ v and v ⊥ w , then u ⊥ w . Disprove by showing a counterexample. Let u = (1, 0), v = (0,1), and w = ( −1, 0 ). Then with respect to the Euclidean inner product of R 2, u ⊥ v and v ⊥ w but 〈u ,w 〉 = 1(−1) + 0(0) = −1 ≠ 0 . 3. S = {(1, 0),(0, 2),(1,1)} with respect to the Euclidean inner product of R 2. Because 〈(0, 2),(1,1)〉 = 2, the set is not orthogonal. 4. S = {(1,1, 0),(1, −1, 0),(−1,1, 0)} with respect to the Euclidean inner product of R 3. 〈(1,1, 0),(1, −1, 0)〉 = 1 − 1 = 0 〈(1,1, 0),(−1,1, 0)〉 = −1 + 1 = 0 〈(1, −1, 0),(−1,1, 0)〉 = −1 − 1 = −2 ≠ 0 The set is not orthogonal. 5. S = {(1,1, 0),(−1,1, 0),(0, 0,1)} with respect to the Euclidean inner product of R 3. 〈(1,1, 0),(−1,1, 0)〉 = −1 + 1 = 0 〈(1,1, 0),(0, 0,1)〉 = 0 + 0 + 0 = 0 〈(−1,1, 0),(0, 0,1)〉 = 0 + 0 + 0 = 0 The set is orthogonal. ⎪⎧ ⎡1 0 ⎤ ⎡0 1 ⎤ ⎡0 0 ⎤ ⎡0 0 ⎤ ⎫⎪ 6. S = ⎨ ⎢ ⎥ ⎬ , with respect to the trace inner product. ⎥,⎢ ⎥,⎢ ⎥,⎢ ⎩⎪ ⎣0 0 ⎦ ⎣0 0 ⎦ ⎣1 1 ⎦ ⎣1 −1⎦ ⎭⎪ ⎡1 0 ⎤ ⎡0 1 ⎤ ⎡0 0 ⎤ ⎡0 0 ⎤ Let u1 = ⎢ ⎥, u 2 = ⎢ ⎥ , u3 = ⎢ ⎥ , and u 4 = ⎢ ⎥. ⎣0 0 ⎦ ⎣0 0 ⎦ ⎣1 1 ⎦ ⎣1 −1⎦ 〈u1 , u 2 〉 = 0 + 0 + 0 + 0 = 0 ; 〈u1 , u 3 〉 = 0 + 0 + 0 + 0 = 0; 〈u1 , u 4 〉 = 0 + 0 + 0 + 0 = 0 ; 〈u 2 , u 3 〉 = 0 + 0 + 0 + 0 = 0; 〈u 2 , u 4 〉 = 0 + 0 + 0 + 0 = 0; 〈u 3 , u 4 〉 = 0 + 0 + 1 − 1 = 0 Thus, the set is orthogonal. 7. S = {(1, 0),(0,1)} with respect to the Euclidean inner product of R 2. The vectors are orthogonal, and because (1, 0) = 1 and (0,1) = 1, the set is orthonormal. ⎧⎪ ⎡ 2 0 ⎤ ⎡ 0 1 ⎤ ⎡ 0 0 ⎤ ⎡0 0 ⎤ ⎫⎪ 8. S = ⎨ ⎢ ⎥ ⎬ , with respect to the trace inner product. ⎥,⎢ ⎥,⎢ ⎥,⎢ ⎪⎩ ⎣ 0 0 ⎦ ⎣ 0 0 ⎦ ⎣ 2 2 ⎦ ⎣1 −1⎦ ⎪⎭ By inspection, the set is orthogonal, but because

2 0 = 4 + 0 + 0 + 0 = 2 ≠ 1, the set is not orthonormal. 0 0

Answer key

201

9. S = {(2, 0, 0),(0, 2, 0),(0, 0, 3)} with respect to the Euclidean inner product of R 3. 1 1 (2, 0, 0) = (2, 0, 0) = (1, 0, 0) ( 2, 0, 0) 2 1 (0, 2, 0) = (0,1, 0) (0 , 2 , 0 ) 1 1 (0, 0, 3) = (0, 0, 3) = (0, 0,1) (0, 0, 3) 3 ⎧⎪ ⎡ 2 0 ⎤ ⎡ 0 1 ⎤ ⎡ 0 0 ⎤ ⎡0 0 ⎤ ⎫⎪ 10. S = ⎨ ⎢ ⎥ ⎬ , with respect to the trace inner product. ⎥,⎢ ⎥,⎢ ⎥,⎢ ⎩⎪ ⎣ 0 0 ⎦ ⎣ 0 0 ⎦ ⎣ 2 2 ⎦ ⎣1 −1⎦ ⎭⎪ ⎡ 2 0 ⎤ 1 ⎡ 2 0 ⎤ ⎡1 0 ⎤ ⎥ ⎥=⎢ ⎥= ⎢ ⎢ ⎡ 2 0 ⎤ ⎣ 0 0 ⎦ 2 ⎣ 0 0 ⎦ ⎣0 0 ⎦ ⎥ ⎢ ⎣0 0 ⎦ 1

⎡0 1 ⎤ ⎡0 1 ⎤ ⎡0 1 ⎤ ⎥ ⎥=⎢ ⎥ = 1⎢ ⎢ ⎡0 1 ⎤ ⎣0 0 ⎦ ⎣0 0 ⎦ ⎣0 0 ⎦ ⎥ ⎢ ⎣0 0 ⎦ 1

⎡ 0 ⎡0 0 ⎤ 1 ⎡0 0 ⎤ ⎢ ⎥= 1 ⎢ ⎥= ⎢ ⎡0 0 ⎤ ⎣2 2 ⎦ 2 2 ⎣2 2 ⎦ ⎢ ⎢⎣ 2 ⎥ ⎢ ⎣2 2⎦

0 ⎤ ⎥ 1 ⎥ 2 ⎥⎦

1

⎡ 0 ⎡0 0 ⎤ 1 ⎡0 0 ⎤ ⎢ = = 1 ⎥ ⎢ ⎥ ⎢ ⎡0 0 ⎤ ⎣1 −1⎦ 2 ⎣1 −1⎦ ⎢⎢ ⎣ 2 ⎥ ⎢ ⎣1 −1⎦

0 ⎤ ⎥ 1 ⎥ − 2 ⎦⎥

1

⎧ ⎡ 0 ⎪ ⎡1 0 ⎤ ⎡ 0 1 ⎤ ⎢ S0 = ⎨ ⎢ ⎥,⎢ 1 ⎥,⎢ ⎪ ⎣0 0 ⎦ ⎣0 0 ⎦ ⎢ ⎣ 2 ⎩

8·5

0 ⎤ ⎡ 0 ⎥ ⎢ 1 ⎥,⎢ 1 2 ⎥⎦ ⎢⎣ 2

⎤⎫ ⎥ ⎪ is the normalized set. ⎥⎬ − ⎪ 2 ⎥⎦ ⎭ 0 1

1. orthogonalization, orthonormal 2. v 1, v 2 −

〈 v 2 , w1 〉 w1

2

2

w 1 Note: Recall that w 1 = 〈 w 1 , w 1 〉.

3. normalize 4. Use the Gram-Schmidt process to compute an orthonormal basis for R 2 from the basis S = {(2, 2),(3, 0)} . Let w 1 = v 1 = (2, 2). Then w 2 = (3, 0) −

〈(3, 0),(2, 2)〉 ⎛ 3 3⎞ ⎛ 3 3⎞ (2, 2) = (3, 0) − ⎜ , ⎟ = ⎜ ,− ⎟ . 2 ⎝ 2 2⎠ ⎝ 2 2⎠ ( 2, 2)

3 3 Clear fractions to get a modified w 2 = (3, −3), which is a positive multiple of ⎜⎛ ,− ⎞⎟ . Normalize the ⎝ 2 2⎠ w vectors to get u1 =

w 1 ( 2, 2) 1 ⎛ 1 1 ⎞ = = ( 2, 2) = ⎜ , ⎝ 2 2 ⎟⎠ w1 8 2 2

u2 =

w 2 (3, −3) (3, −3) ⎛ 1 1 ⎞ = = =⎜ ,− ⎟ w2 18 3 2 ⎝ 2 2⎠

The u vectors are the orthonormal basis.

202

Answer key

5. Use the Gram-Schmidt process to compute an orthonormal basis for R 3 from the basis S = {(0,1,1),(0, 0,1),(1,1,1)}. Let v 1 = w 1 = (0,1,1). Then w 2 = v 2 −

〈 v 2 ,w 1 〉 w1

2

w 1 = (0, 0,1) −

〈(0, 0,1),(0,1,1)〉 1 (0,1,1) = (0, 0,1) − (0,1,1) = 2 2 (0,1,1)

⎛ 1 1⎞ ⎜⎝ 0, − 2 , 2 ⎟⎠ . Clear fractions to get a modified w 2 = (0, −1,1). w3 = v3 −

〈 v 3 , w1 〉 w1

2

w1 −

〈v 3 , w 2 〉 w2

2

w 2 = (1,1,1) −

〈(1,1,1),(0,1,1)〉 〈(1,1,1),(0, −1,1)〉 (0,1,1) − (0, −1,1) = 2 2 (0,1,1) (0, −1,1)

(1,1,1) − (0,1,1) = (1, 0, 0) w 3 = (1, 0, 0). Finally, normalize each one to get u1 =

⎛ 1 1 ⎞ w1 (0,1,1) 1 = = (0,1,1) = ⎜ 0, , ⎟ w1 (0,1,1) 2 ⎝ 2 2⎠

u2 =

⎛ w2 1 1 ⎞ (0, −1,1) 1 , ⎟ = (0, −1,1) = ⎜ 0, − = w2 (0, −1,1) 2 ⎝ 2 2⎠

u 3 = (1, 0, 0) The u vectors are the orthonormal basis.

9 9·1

Linear transformations 1. addition, scalar 2. 0 3. aT (u ) + bT ( v ) 4. Prove that T (− v ) = −T ( v ) for all v ∈V. T (− v ) = T ((−1)v ) = (−1)T ( v ) = −T ( v ) 5. Prove that T (u − v ) = T (u ) − T ( v ) for all u , v ∈V . T (u − v ) = T (u + (−1)v ) = T (u ) + (−1)T ( v ) = T (u ) − T ( v ) 6. Prove that T (au − bv ) = aT (u ) − bT ( v ) for all u , v ∈V and all scalars a and b. T (au − bv ) = T (au + (−1)bv ) = aT (u ) + (−1)bT ( v ) = aT (u ) − bT( v ) 7. T : R 2 → R 2 defined by T (( x , y )) = (3x − y , x ) (a) T (( x , y ) + (z , t )) = T (( x + z , y + t )) = (3( x + z ) − ( y + t ), x + z) = ((3x − y ) + (3z − t ), x + z ) = (3x − y , x ) + (3z − t , z ) = T (( x , y )) + T ((z , t )) (b) T (k( x , y )) = T ((kx , ky )) = (3(kx ) − (ky ), kx ) = k(3x − y , x ) = kT (( x , y )) 8. T : R 3 → R 3 defined by T (( x , y , z )) = ( y , 0, z ) (a) T (( x1 , y1 , z1 ) + ( x , y , z )) = T (( x1 + x , y1 + y , z1 + z )) = ( y1 + y , 0, z1 + z ) = ( y1 , 0, z1 ) + ( y , 0, z ) = T (( x1 , y1 , z1 )) + T (( x , y , z )) (b) T (k( x , y , z )) = T ((kx , ky , kz )) = (ky , 0, kz ) = k( y , 0, z ) = kT (( x , y , z ))

Answer key

203

9. T : R → R 2 defined by T ( x ) = (5 x , − x ) (a) T ( x + y ) = (5( x + y ), −( x + y )) = (5 x + 5 y , − x − y ) = (5 x , − x ) + (5 y , − y ) = T ( x ) + T ( y ) (b) T (kx ) = (5(kx ), −(kx )) = k(5 x , − x ) = kT ( x ) 10. T : R 2 → R 3 defined by T (( x , y )) = ( x + y , y , x ) (a) T (( x1 , y1 ) + ( x , y )) = T (( x1 + x , y1 + y )) = (( x1 + x ) + ( y1 + y ), y1 + y , x1 + x ) = (( x1 + y1 ) + ( x + y ), y1 + y , x1 + x ) = ( x1 + y1 , y1 , x1 ) + ( x + y , y , x ) = T (( x1 , y1 )) + T (( x , y )) (b) T (k( x , y )) = T ((kx , ky )) = (kx + ky , ky , kx ) = k( x + y , y , x ) = kT (( x , y )) 11. T : R 3 → R defined by T (( x , y , z )) = x − 2 y + 3z (a) T ((a, b , c ) + ( x , y , z )) = T ((a + x , b + y , c + z )) = (a + x ) − 2(b + y ) + 3(c + z ) = (a − 2b + 3c ) + ( x − 2 y + 3z ) = T ((a, b, c )) + T (( x , y , z )) (b) T (k( x , y , z )) = T ((kx , ky , kz )) = kx − 2(ky ) + 3(kz ) = k( x − 2 y + 3z ) = kT (( x , y , z )) 12. G : R → R defined by G( x ) = 3x + 2 G(0 ) = 2 ≠ 0 . G is not a linear transformation. 13. G : R 2 → R 3 defined by G( v ) = G((v1 , v 2 )) = (v1 , v1v 2 , v 2 ). G((2, 3)) = (2, 6, 3) and G((0,1)) = (0, 0,1) G((2, 3)) + G((0,1)) = (2, 6, 3) + (0, 0,1) = (2, 6, 4 ) G((2, 3) + (0,1)) = G((2, 4 )) = (2, 8, 4 ) G((2, 3) + (0,1)) ≠ G((2, 3)) + G((0,1)). The transformation is not linear.

(

)

14. G : R 2 → R 2 defined by G( v ) = G((v1 , v 2 )) = v12 , v 2 2 . G((2, 3)) = (4 , 9) G((0,1)) = (0,1) G((2, 3) + (0,1)) = G((2, 4 )) = (4 ,16) G((2, 3)) + G((0,1)) = (4 , 9) + (0,1) = (4 ,10) G((2, 3) + (0,1)) ≠ G((2, 3)) + G((0,1)). The transformation is not linear. 15. G : R 3 → R 2 defined by G( v ) = G((v1 , v 2 , v3 )) = (v1 − 5, v 2 + v3 ). G((0, 0, 0)) = (−5, 0) ≠ (0, 0). G is not a linear transformation.

9·2

1. zero 2. T( v ) 3. V, W 4. dim(V ) 5. rank, nullity

204

Answer key

6. Is the vector (12, 6) in the kernel of T? Explain. Yes, because T((12, 6)) = (−36 + 36,12 − 12) = (0, 0) . 7. Is the vector (5, 3) in the kernel of T? Explain. No, because T((5, 3)) = (−15 + 18, 5 − 6) = (3, −1) ≠ (0, 0). 8. Is 0 ∈R 2 in the kernel of T? Explain. Yes, because T((0, 0)) = (0, 0) . 9. Is the vector (14 , 6) in the image of T? Explain. No. If so, the equation T ((v1 , v 2 )) = (−3v1 + 6v 2 , v1 − 2v 2 ) = (14 , 6) must have a solution. This leads to the system of equations −3v1 + 6v 2 = 14 v1 − 2v 2 = 6 Multiply the second equation by −3 to see that the system has no solution. Therefore, (14 , 6) is not in the image of T. 10. Is the vector (−15, 5) in the image of T? Explain. Yes. The set of equations for this ordered pair is −3v1 + 6v 2 = −15 v1 − 2v 2 = 5 The second equation is a multiple of the first, so there are infinitely many solutions of the form v1 = 2v 2 + 5 with the free variable v 2 . One solution is (5, 0). Therefore, (−15, 5) is in the image of T. 11. Is 0 ∈R 2 in the image of T? Explain. Yes, because T((0, 0)) = (0, 0). 12. T : R 4 → R 3 has rank 1. nullity(T ) = dim( R 4 ) − rank (T ). Thus, nullity(T ) = 4 − 1 = 3. 13. T : M 22 → M 22 has rank 4. nullity(T ) = dim( M 22 ) − rank (T ). Thus, nullity(T ) = 4 − 4 = 0. 14. The image of T : R 3 → P2 is P2. nullity(T ) = dim( R 3 ) − rank (T ) = dim( R 3 ) − dim(Im(T )). Thus, nullity(T ) = 3 − 3 = 0. 15. Let T : V → W be a linear transformation. Prove that Im T is a subspace of W. Because T(0 ) = 0 , 0 is in Im T . Suppose w and t are in Im T . Then there exists u and v in V such that T(u ) = w and T( v ) = t . Then w + t = T (u ) + T ( v ) = T (u + v ). Hence, w + t is in Im T . Also, kw = kT(u ) = T (ku ), so kw is in Im T . Thus, Im T is a subspace of W. 16. T : R 4 → R 3 has rank 1. Singular, because by Prob. 12, nullity(T ) = 4 − 1 = 3. Therefore, Ker T ≠ {0} . 17. T : M 22 → M 22 has rank 4. Nonsingular, because by Prob. 13, nullity(T ) = 0 = dim (Ker T ). Therefore, Ker T = {0} .

Answer key

205

18. The image of T : R 3 → P2 is P2. Nonsingular, because by Prob. 13, nullity(T ) = 0. Therefore, Ker T = {0} . ⎛ ⎡ x ⎤⎞ ⎡ x − 2 x 2 ⎤ 19. T : R 2 → R 2 defined by T ⎜ ⎢ 1 ⎥⎟ = ⎢ 1 ⎥. ⎝ ⎣ x 2 ⎦⎠ ⎣ x1 − x 2 ⎦ ⎛ ⎡ x1 ⎤⎞ ⎡ x1 − 2 x 2 ⎤ ⎡0 ⎤ Nonsingular. Proof: Find Ker T by letting T ⎜ x = ⎢ ⎥⎟ = ⎢ ⎥ = ⎢ ⎥ . The equation ⎝ ⎣ x 2 ⎦⎠ ⎣ x1 − x 2 ⎦ ⎣0 ⎦ ⎡ x1 − 2 x 2 ⎤ ⎡0 ⎤ ⎡1 −2 0 ⎤ ⎡1 0 0 ⎤ ⎢ ⎥ = ⎢ ⎥ has augmented matrix ⎢ ⎥, which reduces to ⎢ ⎥ yielding only the trivial − x x 1 − 1 0 0 ⎣ ⎦ ⎣0 1 0 ⎦ ⎣ 1 2 ⎦ ⎣ ⎦ solution x = 0 and y = 0. Thus, Ker T = {0} so T is nonsingular. ⎡ a0 ⎤ ⎢ ⎥ 20. T : P2 → R 3 defined by T a0 + a1 x + a2 x 2 = ⎢ a1 ⎥ . ⎢⎣a2 ⎥⎦

(

)

Nonsingular. Proof: If a0 + a1 x + a2 x 2 ≠ 0 is in P2 then not all of the coefficients can be 0. Assume a1 ≠ 0.

(

Then T a0 + a1 x + a2 x

9·3

2

)

⎡ a0 ⎤ ⎡0 ⎤ ⎡0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 2 = ⎢ a1 ⎥ ≠ ⎢0 ⎥. Consequently, only T 0 + 0 x + 0 x = ⎢0 ⎥ = 0, so Ker T = {0}. ⎢⎣a2 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣0 ⎥⎦

(

)

⎛ ⎡ x ⎤⎞ ⎡ x + y ⎤ ⎧⎪ ⎡ 2 ⎤ ⎡ 3 ⎤ ⎫⎪ ⎧⎪ ⎡1⎤ ⎡ 1 ⎤ ⎫⎪ 1. T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ relative to the bases BV = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ and BW = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎝ ⎣ y ⎦⎠ ⎣ x − 2 y ⎦ ⎪⎩ ⎣ 1 ⎦ ⎣ 2 ⎦ ⎪⎭ ⎪⎩ ⎣1⎦ ⎣ 2 ⎦ ⎪⎭ ⎛ ⎡ 2 ⎤⎞ ⎡ 3 ⎤ ⎛ ⎡ 3 ⎤⎞ ⎡ 5 ⎤ T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ and T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎝ ⎣ 2 ⎦⎠ ⎣ −1⎦ ⎝ ⎣ 1 ⎦⎠ ⎣ 0 ⎦ ⎡1 0 6 11 ⎤ ⎡1 1 3 5 ⎤ ⎥ ⎢1 2 0 −1⎥ ⇒ ⎢ ⎣ ⎦ ⎣0 1 −3 −6 ⎦ ⎡ 6 11 ⎤ [T ]BW ,BV = ⎢ ⎥ ⎣ −3 −6 ⎦ ⎡ ⎤ y ⎛ ⎡ x ⎤⎞ ⎢ ⎧⎪ ⎡ 2 ⎤ ⎡ 3 ⎤ ⎫⎪ ⎥ 2. T ⎜ ⎢ ⎥⎟ = ⎢13 y − 5 x ⎥ relative to the bases BV = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ and BW = ⎝ ⎣ y ⎦⎠ ⎢ ⎩⎪ ⎣ 1 ⎦ ⎣ 2 ⎦ ⎭⎪ ⎥ ⎣16 y − 7 x ⎦ ⎡1 ⎤ ⎡2⎤ ⎛ ⎡ 2 ⎤⎞ ⎢ ⎥ ⎛ ⎡ 3 ⎤⎞ ⎢ ⎥ T ⎜ ⎢ ⎥⎟ = ⎢ 3 ⎥ and T ⎜ ⎢ ⎥⎟ = ⎢11⎥ ⎝ ⎣ 1 ⎦⎠ ⎢ ⎥ ⎝ ⎣ 2 ⎦⎠ ⎢ ⎥ ⎣2⎦ ⎣11⎦ ⎡ 1 −1 0 1 2 ⎤ ⎡1 0 0 1 0 ⎤ ⎢ 0 2 1 3 11⎥ ⇒ ⎢0 1 0 0 −2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ −1 2 1 2 11⎥⎦ ⎢⎣0 0 1 3 15 ⎥⎦ ⎡1 0 ⎤ [T ]BW ,BV = ⎢⎢0 −2 ⎥⎥ ⎢ 3 15 ⎥ ⎣ ⎦

206

Answer key

⎧ ⎡ 1 ⎤ ⎡ −1⎤ ⎡0 ⎤ ⎫ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎨ ⎢ 0 ⎥ , ⎢ 2 ⎥ , ⎢1 ⎥ ⎬ ⎪ ⎢ −1⎥ ⎢ 2 ⎥ ⎢1 ⎥ ⎪ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭

⎛ ⎡ x ⎤⎞ ⎡ x − x ⎤ ⎧⎪ ⎡1 ⎤ ⎡0 ⎤ ⎫⎪ ⎧⎪ ⎡ 2 ⎤ ⎡ 3 ⎤ ⎫⎪ 3. T ⎜ ⎢ 1 ⎥⎟ = ⎢ 1 2 ⎥ relative to the bases BV = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ and BW = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎝ ⎣ x 2 ⎦⎠ ⎣ x1 + x 2 ⎦ ⎪⎩ ⎣0 ⎦ ⎣1 ⎦ ⎪⎭ ⎪⎩ ⎣ 1 ⎦ ⎣ 2 ⎦ ⎪⎭ ⎛ ⎡0 ⎤⎞ ⎡ −1⎤ ⎛ ⎡1 ⎤⎞ ⎡1⎤ T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ and T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎝ ⎣ 1 ⎦⎠ ⎣ 1 ⎦ ⎝ ⎣0 ⎦⎠ ⎣1⎦ ⎡1 0 −1 −5 ⎤ ⎡ 2 3 1 −1⎤ ⎢1 2 1 1 ⎥ ⇒ ⎢0 1 1 3 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ −1 −5 ⎤ [T ]BW ,BV = ⎢ ⎥ ⎣1 3⎦ ⎧ ⎡1⎤ ⎡ 3 ⎤ ⎡ 2 ⎤ ⎫ ⎡ x + 2x 2 ⎤ ⎛ ⎡ x1 ⎤⎞ ⎢ 1 ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪⎧ ⎡1 ⎤ ⎡ −1⎤ ⎪⎫ ⎥ 4. T ⎜ ⎢ ⎥⎟ = ⎢ 0 ⎥ relative to the bases BV = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ and BW = ⎨ ⎢1⎥ , ⎢0 ⎥ , ⎢ 2 ⎥ ⎬ ⎝ ⎣ x 2 ⎦⎠ ⎢ ⎪⎩ ⎣3 ⎦ ⎣ 2 ⎦ ⎭⎪ ⎪ ⎢1⎥ ⎢0 ⎥ ⎢ 0 ⎥ ⎪ ⎥ ⎣ − x1 ⎦ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ ⎡3 ⎤ ⎡7⎤ ⎛ ⎡ −1⎤⎞ ⎢ ⎥ ⎛ ⎡ 1 ⎤⎞ ⎢ ⎥ T ⎜ ⎢ ⎥⎟ = ⎢ 0 ⎥ and T ⎜ ⎢ ⎥⎟ = ⎢0 ⎥ ⎝ ⎣ 2 ⎦⎠ ⎢ ⎥ ⎝ ⎣ 3 ⎦⎠ ⎢ ⎥ ⎣1 ⎦ ⎣ −1⎦ ⎡1 0 0 ⎡1 3 2 7 3 ⎤ ⎢ ⎢ ⎥ ⎢1 0 2 0 0 ⎥ ⇒ ⎢0 1 0 ⎢0 0 1 ⎢1 0 0 −1 1 ⎥ ⎣ ⎦ ⎢⎣

[T ]BW ,BV

⎡ ⎢ −1 = ⎢7 ⎢ 3 ⎢ ⎢⎣ 1 2

−1 1 ⎤ ⎥ 7 3 1 ⎥ ⎥ 1 −1 ⎥ 2⎦ 2

⎤ 1 ⎥ ⎥ 1 ⎥ ⎥ −1 ⎥ 2⎦

⎛ ⎡ x ⎤⎞ ⎡ x + 2 y − z ⎤ ⎧ ⎡1 ⎤ ⎡1 ⎤ ⎡1⎤ ⎫ ⎧ ⎡1 ⎤ ⎡ 2 ⎤ ⎡1 ⎤ ⎫ ⎜ ⎢ y ⎥⎟ = ⎢ y + z ⎥ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ T 5. ⎜ ⎢ ⎥⎟ ⎢ ⎥ relative to the bases BV = ⎨ ⎢1 ⎥ , ⎢0 ⎥ , ⎢1⎥ ⎬ and BW = ⎨ ⎢ 1 ⎥ , ⎢ 2 ⎥ , ⎢ 2 ⎥ ⎬ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ x + y − 2z ⎥ ⎪ ⎢0 ⎥ ⎢1 ⎥ ⎢1⎥ ⎪ ⎪ ⎢ 2 ⎥ ⎢1 ⎥ ⎢ 2 ⎥ ⎪ ⎦ ⎣ ⎦ ⎣ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ ⎛ ⎡1⎤⎞ ⎡ 2 ⎤ ⎛ ⎡ 1 ⎤⎞ ⎡ 3 ⎤ ⎛ ⎡ 1 ⎤⎞ ⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ T ⎜⎜ ⎢1 ⎥⎟⎟ = ⎢ 1 ⎥ , T ⎜⎜ ⎢0 ⎥⎟⎟ = ⎢ 1 ⎥ , and T ⎜⎜ ⎢1⎥⎟⎟ = ⎢ 2 ⎥ ⎜⎝ ⎢1⎥⎟⎠ ⎢ 0 ⎥ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢ 2 ⎥ ⎜⎝ ⎢1 ⎥⎟⎠ ⎢ −1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡1 2 1 ⎢1 2 2 ⎢ ⎢⎣ 2 1 2

[T ]BW ,BV

⎡1 0 0 3 0 2⎤ ⎢ 1 1 2 ⎥⎥ ⇒ ⎢0 1 0 ⎢ 2 −1 0 ⎥⎦ ⎢⎣0 0 1

7 4

3

3 −2

−5

2 ⎤ 3 − 3⎥ 1 4 ⎥ 3 3 ⎥ 1 0 ⎥⎦

⎡ 7 −5 − 2 ⎤ 3 3⎥ ⎢ 3 = ⎢4 1 4 ⎥ 3 3 ⎥ ⎢ 3 ⎢⎣ −2 1 0 ⎥⎦

Answer key

207

⎛ ⎡ x ⎤⎞ ⎡ 2 x − 3 y + 4 z ⎤ ⎢ ⎥ ⎢ ⎥ 6. T ⎜⎜ ⎢ y ⎥⎟⎟ = ⎢ 5 x − y + 2z ⎥ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ 4 x + 7 y ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ 2 −3 4 ⎤ ⎢ ⎥ [T ]B = ⎢ 5 −1 2 ⎥ , which are just the images of the e i vectors. ⎢4 7 0 ⎥ ⎣ ⎦ ⎛ ⎡ x ⎤⎞ ⎡ 2 y + z ⎤ ⎢ ⎥ ⎢ ⎥ 7. T ⎜⎜ ⎢ y ⎥⎟⎟ = ⎢ x − 4 y ⎥ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ 3x ⎥ ⎣ ⎦ ⎣ ⎦ ⎡0 2 1 ⎤ ⎢ ⎥ [T ]B = ⎢1 −4 0 ⎥ , which are just the images of the e i vectors ⎢3 0 0 ⎥ ⎣ ⎦ ⎛ ⎡ x ⎤⎞ ⎡ 3 x − y ⎤ 8. T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ 2 y ⎦ ⎡ 3 −1⎤ [T ]B = ⎢ ⎥ , which are just the images of the e i vectors ⎣0 2 ⎦ ⎛ ⎡ x ⎤⎞ ⎡ x + 4 y ⎤ 9. T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ 3 x − 2 y ⎦ ⎡1 4 ⎤ [T ]B = ⎢ ⎥, which are just the images of the e i vectors ⎣3 −2 ⎦ ⎛ ⎡ x ⎤⎞ ⎡ x ⎤ ⎢ ⎥ ⎢ ⎥ 10. T ⎜⎜ ⎢ y ⎥⎟⎟ = ⎢ y ⎥ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ z ⎥ ⎣ ⎦ ⎣ ⎦ ⎡1 0 0 ⎤ ⎢ ⎥ [T ]B = ⎢0 1 0 ⎥ , which are just the images of the e i vectors. Remark: It should not be surprising to have ⎢0 0 1 ⎥ ⎣ ⎦ the identity matrix for the identity transformation. ⎡2⎤ 11. ⎢ ⎥ ⎣ −3 ⎦ ⎡2 3 2 ⎤ ⎡1 0 ⎢ 1 2 −3 ⎥ ⇒ ⎢0 1 ⎣ ⎣ ⎦

13 ⎤ −8 ⎥⎦

⎡ −3 −4 ⎤ ⎡ −7 ⎤ ⎢ ⎥ ⎡ 13 ⎤ ⎢ ⎥ ⎢ −2 −2 ⎥ ⎢ −8 ⎥ = ⎢ −10 ⎥ ⎢ 6 9 ⎥⎣ ⎦ ⎢ 6 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ −1⎤ ⎡ 1 ⎤ ⎡ 0 ⎤ ⎡ −3 ⎤ ⎛ ⎡ 2 ⎤⎞ ⎢ ⎥ T ( x ) = T ⎜ ⎢ ⎥⎟ = −7 ⎢⎢ 2 ⎥⎥ − 10 ⎢⎢ 0 ⎥⎥ + 6 ⎢⎢ 1 ⎥⎥ = ⎢ −8 ⎥ ⎝ ⎣ −3 ⎦⎠ ⎢⎣ −1⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ 8 ⎥⎦ ⎣⎢ 2 ⎥⎦

208

Answer key

⎡4⎤ 12. ⎢ ⎥ ⎣ −3 ⎦ ⎡2 3 4 ⎤ ⎡1 0 17 ⎤ ⎢ 1 2 −3 ⎥ ⇒ ⎢0 1 −10 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ −3 −4 ⎤ ⎡ −11 ⎤ ⎢ ⎥ ⎡ 17 ⎤ ⎢ ⎥ ⎢ −2 −2 ⎥ ⎢ −10 ⎥ = ⎢ −14 ⎥ ⎦ ⎣ ⎢6 9⎥ ⎢ 12 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ −1⎤ ⎡1⎤ ⎡ 0 ⎤ ⎡ −3 ⎤ ⎛ ⎡ 4 ⎤⎞ ⎢ ⎥ T ( x ) = T ⎜ ⎢ ⎥⎟ = −11 ⎢⎢ 2 ⎥⎥ − 14 ⎢⎢ 0 ⎥⎥ + 12 ⎢⎢ 1 ⎥⎥ = ⎢ −10 ⎥ ⎝ ⎣ −3 ⎦⎠ ⎢⎣ 2 ⎥⎦ ⎢⎣ −1⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ 16 ⎥⎦ ⎡1⎤ 13. ⎢ ⎥ ⎣ −5 ⎦ ⎡2 3 1 ⎤ ⎥⇒ ⎢ ⎣ 1 2 −5 ⎦

⎡1 0 17 ⎤ ⎥ ⎢ ⎣0 1 −11⎦

⎡ −3 −4 ⎤ ⎡ −7 ⎤ ⎢ ⎥ ⎡ 17 ⎤ ⎢ ⎥ 2 2 = − − ⎢ ⎥ ⎢ −11⎥ ⎢ −12 ⎥ ⎦ ⎢ 3 ⎥ ⎢ 6 9 ⎥⎣ ⎣ ⎦ ⎣ ⎦ ⎡ −1⎤ ⎡ 1 ⎤ ⎡ 0 ⎤ ⎡ −5 ⎤ ⎛ ⎡ 1 ⎤⎞ ⎢ ⎥ ⎢ ⎥ T ( x ) = T ⎜ ⎢ ⎥⎟ = −7 ⎢ 2 ⎥ − 12 ⎢⎢ 0 ⎥⎥ + 3 ⎢⎢ 1 ⎥⎥ = ⎢ −11⎥ ⎝ ⎣ −5 ⎦⎠ ⎢⎣ 2 ⎥⎦ ⎢⎣ −1⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ 4 ⎥⎦ 14. Find [T ( v 1 )]B and [T ( v 2 )]B. ⎡3⎤ ⎡1⎤ [T ( v 1 )]B = ⎢ ⎥ and [T ( v 2 )]B = ⎢ ⎥ 5 ⎣ ⎦ ⎣ −2 ⎦ 15. Find T( v 1 ) and T( v 2 ). ⎡ −1⎤ ⎡1 ⎤ ⎡ 2 ⎤ T( v 1 ) = 3 ⎢ ⎥ + 5 ⎢ ⎥ = ⎢ ⎥ ⎣ 4 ⎦ ⎣3 ⎦ ⎣ 27 ⎦ ⎡ −1⎤ ⎡1 ⎤ ⎡ −3 ⎤ T( v 2 ) = 1 ⎢ ⎥ − 2 ⎢ ⎥ = ⎢ ⎥ ⎣ 4 ⎦ ⎣3 ⎦ ⎣ −2 ⎦

Answer key

209

9·4

⎧⎪ ⎡ 3 ⎤ ⎡ 1 ⎤ ⎫⎪ ⎡8⎤ 1. v = ⎢ ⎥ ; B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎣ −4 ⎦ ⎩⎪ ⎣0 ⎦ ⎣ −2 ⎦ ⎭⎪ The augmented matrix corresponding to the system is 8⎤ ⎡3 1 ⎢0 −2 −4 ⎥ ⇒ ⎣ ⎦

⎡1 0 2 ⎤ ⎢0 1 2 ⎥ ⎣ ⎦

⎡2⎤ ⎥ ⎣2⎦

[ v ]B = ⎢

⎡0 ⎤ ⎪⎧ ⎡ −2 ⎤ ⎡ 1 ⎤ ⎪⎫ 2. v = ⎢ ⎥ ; B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ 1 ⎣ ⎦ ⎩⎪ ⎣ 3 ⎦ ⎣ −1⎦ ⎭⎪ The augmented matrix corresponding to the system is ⎡ −2 1 0 ⎤ ⎡1 0 1 ⎤ ⎢ 3 −1 1 ⎥ ⇒ ⎢0 1 2 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡1 ⎤ [ v ]B = ⎢ ⎥ ⎣2⎦ ⎧⎪ ⎡ −1⎤ ⎡ 0 ⎤ ⎫⎪ ⎡ 20 ⎤ 3. v = ⎢ ⎥ ; B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎪⎩ ⎣ 5 ⎦ ⎣ −1⎦ ⎪⎭ ⎣14 ⎦ The augmented matrix corresponding to the system is ⎡ −1 0 20 ⎤ ⎡1 0 −20 ⎤ ⎢ 5 −1 14 ⎥ ⇒ ⎢0 1 −114 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡ −20 ⎤ [ v ]B = ⎢ ⎥ ⎣ −114 ⎦ ⎧ ⎡ 2 ⎤ ⎡0 ⎤ ⎡ −1⎤ ⎫ ⎡5 ⎤ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎢ ⎥ 4. v = ⎢5 ⎥; B = ⎨ ⎢ 0 ⎥ , ⎢0 ⎥ , ⎢ 3 ⎥ ⎬ ⎪ ⎢ −1⎥ ⎢5 ⎥ ⎢ 2 ⎥ ⎪ ⎢5 ⎥ ⎣ ⎦ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ The augmented matrix corresponding to the system is 10 ⎤ ⎡1 0 0 ⎡ 2 0 −1 5 ⎤ 3⎥ ⎢ 0 0 3 5 ⎥ ⇒ ⎢0 1 0 1 ⎥ ⎢ ⎢ ⎥ ⎢⎣0 0 1 5 ⎥⎥ ⎢⎣ −1 5 2 5 ⎥⎦ 3⎦ ⎡10 ⎤ ⎢ 3⎥ [ v ]B = ⎢⎢ 1 ⎥⎥ ⎢5 ⎥ ⎣ 3⎦

210

Answer key

⎧ ⎡ −3 ⎤ ⎡5 ⎤ ⎡ 0 ⎤ ⎫ ⎡0 ⎤ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎢ ⎥ 5. v = ⎢1 ⎥; B = ⎨ ⎢ 1 ⎥ , ⎢0 ⎥ , ⎢ 1 ⎥ ⎬ ⎪ ⎢ −1⎥ ⎢5 ⎥ ⎢ −2 ⎥ ⎪ ⎢⎣0 ⎥⎦ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ The augmented matrix corresponding to the system is ⎡ −3 5 0 ⎢1 0 1 ⎢ ⎢⎣ −1 5 −2

0⎤ ⎡1 0 0 1 ⎥⎥ ⇒ ⎢⎢0 1 0 ⎢⎣0 0 1 0 ⎥⎦

0.5 ⎤ 0.3 ⎥⎥ 0.5 ⎥⎦

⎡0.5 ⎤ ⎢ ⎥ [ v ]B = ⎢ 0.3 ⎥ ⎢0.5 ⎥ ⎣ ⎦ Note: In 6–10, without much discussion, the augmented matrices are set up to solve for the coordinate vectors, C is constructed, and then C ′ is determined. ⎧⎪ ⎡ −1⎤ ⎡ 0 ⎤ ⎪⎫ ⎧⎪ ⎡ −2 ⎤ ⎡ 1 ⎤ ⎫⎪ 6. B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬, B ′ = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎪⎩ ⎣ 3 ⎦ ⎣ −1⎦ ⎪⎭ ⎪⎩ ⎣ 5 ⎦ ⎣ −1⎦ ⎪⎭ ⎡ −2 1 [ v 1 ]B is the solution of the system ⎢ ⎣ 3 −1

−1⎤ ⇒ 5 ⎥⎦

⎡1 0 4 ⎤ ⎢0 1 7 ⎥ ⎣ ⎦

0⎤ ⇒ −1⎥⎦

⎡1 0 ⎢0 1 ⎣

⎡4 ⎤

[ v1 ]B = ⎢7 ⎥ ⎣ ⎦

[ v 2 ]B is the solution of the system

⎡ −2 1 ⎢ 3 −1 ⎣

−1 ⎤ −2 ⎥⎦

⎡ −1 ⎤

[ v 2 ]B = ⎢ −2 ⎥ ⎣

⎦

⎡ 4 −1 ⎤ C=⎢ ⎥ ⎣ 7 −2 ⎦ ⎡ 2 −1 ⎤ C ′ = C −1 = ⎢ ⎥ ⎣ 7 −4 ⎦ ⎧⎪ ⎡0 ⎤ ⎡ 2 ⎤ ⎫⎪ ⎧⎪ ⎡1⎤ ⎡ −1⎤ ⎫⎪ 7. B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬, B ′ = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎪⎩ ⎣ 3 ⎦ ⎣ 0 ⎦ ⎪⎭ ⎪⎩ ⎣1⎦ ⎣ 0 ⎦ ⎪⎭ ⎡1 −1 ⎢1 0 ⎣

0⎤ ⇒ 3 ⎥⎦

⎡1 0 ⎢0 1 ⎣

3⎤ 3 ⎥⎦

Answer key

211

⎡1 −1 2 ⎤ ⎡1 0 0 ⎤ ⎢1 0 0 ⎥ ⇒ ⎢0 1 −2 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡3 0 ⎤ C=⎢ ⎥ ⎣3 −2 ⎦ 0 ⎤ ⎥ ⎥ −1 ⎥ 2⎦

⎡1 C ′ = C −1 = ⎢ 3 ⎢1 ⎣⎢ 2

⎧⎪ ⎡ 2 ⎤ ⎡ 0 ⎤ ⎪⎫ ⎪⎧ ⎡1 ⎤ ⎡ −1⎤ ⎪⎫ 8. B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬, B ′ = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ 3 4 ⎩⎪ ⎣ ⎦ ⎣ ⎦ ⎭⎪ ⎩⎪ ⎣ 2 ⎦ ⎣ −1⎦ ⎭⎪ ⎡ 2 0 1 ⎤ ⎡1 0 0.5 ⎤ ⎢ 2 −1 3 ⎥ ⇒ ⎢0 1 −2 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡1 0 −0.5 ⎤ ⎡ 2 0 −1⎤ ⎥ ⇒ ⎢0 1 −5 ⎥ ⎢ ⎣ ⎦ ⎣ 2 −1 4 ⎦ ⎡0.5 −0.5 ⎤ C=⎢ ⎥ ⎣ −2 −5 ⎦ ⎡ 10 C ′ = C −1 = ⎢ 7 ⎢ 4 − ⎣⎢ 7

−1 ⎤ 7⎥ ⎥ 1 − 7 ⎦⎥

⎧⎪ ⎡ 1 ⎤ ⎡ 3 ⎤ ⎫⎪ ⎧⎪ ⎡1 ⎤ ⎡ 3 ⎤ ⎫⎪ 9. B = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ , B ′ = ⎨ ⎢ ⎥ , ⎢ ⎥ ⎬ ⎩⎪ ⎣ −2 ⎦ ⎣ −4 ⎦ ⎭⎪ ⎩⎪ ⎣ 3 ⎦ ⎣ 8 ⎦ ⎭⎪ ⎡1 3 1 ⎤ ⎢3 8 −2 ⎥ ⇒ ⎣ ⎦

⎡1 0 −14 ⎤ ⎢0 1 5 ⎥ ⎣ ⎦

⎡1 3 3 ⎤ ⎢3 8 −4 ⎥ ⇒ ⎣ ⎦

⎡1 0 −36 ⎤ ⎢0 1 13 ⎥ ⎣ ⎦

⎡ −14 −36 ⎤ C=⎢ ⎥ 13 ⎦ ⎣ 5 ⎡ −6.5 −18 ⎤ C ′ = C −1 = ⎢ ⎥ 7 ⎦ ⎣ 2.5

212

Answer key

⎧ ⎡ 1 ⎤ ⎡ −2 ⎤ ⎡ 2 ⎤ ⎫ ⎧ ⎡ 1 ⎤ ⎡1 ⎤ ⎡0 ⎤ ⎫ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ 10. B = ⎨ ⎢ 0 ⎥ , ⎢ 2 ⎥ , ⎢ 1 ⎥ ⎬, B′ = ⎨ ⎢ 2 ⎥ , ⎢ −3 ⎥ , ⎢ 6 ⎥ ⎬ ⎪ ⎢1 ⎥ ⎢ 1 ⎥ ⎢7 ⎥ ⎪ ⎪ ⎢ −1⎥ ⎢ 2 ⎥ ⎢ 2 ⎥ ⎪ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ ⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎭ ⎡ 1 −2 2 ⎢ 2 −3 6 ⎢ ⎢⎣ 1 1 7

1 ⎤ ⎡1 0 0 0 ⎥⎥ ⇒ ⎢0 1 0 ⎢ −1⎥⎦ ⎢⎣0 0 1

21 ⎤ 6 ⎥⎥ −4 ⎥⎦

⎡ 1 −2 2 ⎢ 2 −3 6 ⎢ ⎢⎣ 1 1 7

1⎤ ⎡1 0 0 ⎥ 2 ⎥ ⇒ ⎢⎢0 1 0 ⎢⎣0 0 1 2 ⎥⎦

7⎤ 2 ⎥⎥ −1⎥⎦

⎡ 1 −2 2 ⎢ 2 −3 6 ⎢ ⎢⎣ 1 1 7

0⎤ ⎡1 0 0 1 ⎥⎥ ⇒ ⎢⎢0 1 0 ⎢⎣0 0 1 2 ⎥⎦

−4 ⎤ −1 ⎥⎥ 1 ⎥⎦

⎡ 21 7 −4 ⎤ C = ⎢⎢ 6 2 −1 ⎥⎥ ⎢⎣ −4 −1 1 ⎥⎦ ⎡ −1 3 −1⎤ ⎥ ⎢ C ′ = C −1 = ⎢ 2 −5 3 ⎥ ⎢ −2 7 0 ⎥ ⎣ ⎦

9·5

1. Prove that the zero transformation is linear; that is, prove that the function O : V → W defined by O( v ) = 0 for every v in V is linear: Proof: By definition, O( v ) = 0 for every v in V. (a) If u and v are in V, then O(u + v ) = 0 = 0 + 0 = O(u ) + O( v ) (b) If k is any scalar, then O(ku ) = 0 = k 0 = kO(u ) Therefore, the zero transformation is linear. 2. Prove that the identity transformation is linear; that is, prove that the function I : V → V defined by I( v ) = v for every v in V is linear. (a) If u and v are in V, then I(u + v ) = u + v = I (u ) + I ( v ) (b) If k is any scalar, then I (ku ) = ku = kI (u ) Therefore, the identity transformation is linear. 3. Prove that −G( v ) = (−1)G( v ) for every G ∈᐀(V ,W ) Because T (V ,W ), is a vector space, this property is a property of vector spaces from Chapter 7. 4. Prove that ( F − G )( v ) = F (v ) − G( v ) for all F , G ∈᐀(V ,W ). If F , G ∈᐀(V ,W ), then ( F − G )( v ) = ( F + (−1)G )(v ) = F ( v ) + (−1)G(v ) = F ( v ) − G( v ) 5. Prove that F + O = F for every F ∈᐀(V ,W ). Because T (V ,W ) is a vector space, this property is a property of vector spaces from Chapter 7. 6. Prove that the composition of two linear transformations is linear; that is, prove that if F : V → U and G : U → W , then F F G defined by ( F F G )(u ) = F (G(u )), where u ∈U, is a linear transformation. Answer key

213

Proof: (a) ( F F G )(u + t ) = F (G(u + t )) = F (G(u ) + G(t )) = F (G(u )) + F (G(t )) = ( F F G )(u ) + ( F F G )(t ) (b) If k is any scalar, then ( F F G )(ku ) = F (G(ku )) = F (kG(u )) = kF (G(u )) = k( F F G )(u ) ⎛ ⎡ x1 ⎤⎞ ⎛ ⎡ x1 ⎤⎞ ⎡ x1 ⎤ ⎡ x1 + x 2 ⎤ ⎡ 2 x1 + x 2 ⎤ ⎛ ⎡ x1 ⎤⎞ ⎥ ⎥ ⎢ ⎜ ⎢ ⎥⎟ + ⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎢ ⎢ ⎥ ⎜ ⎟ 7. ( F + G ) ⎢ x 2 ⎥ = F ⎜ ⎢ x 2 ⎥⎟ G ⎜ ⎢ x 2 ⎥⎟ ⎢ 0 ⎥ + ⎢ x 2 ⎥ = ⎢ x 2 ⎥ ⎜ ⎟ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ x ⎥ ⎢ 0 ⎥ ⎢ x ⎥ ⎣ 3⎦ ⎦ ⎣ 3 ⎣ 3⎦ ⎣ 3⎦ ⎣ 3⎦ ⎣ ⎦ ⎛ ⎡ x1 ⎤⎞ ⎛ ⎡ x1 ⎤⎞ ⎡ x1 ⎤ ⎡ x1 + x 2 ⎤ ⎡ − x1 − 2 x 2 ⎤ ⎛ ⎡ x1 ⎤⎞ ⎢ ⎥⎟ ⎥ ⎥ ⎢ ⎜ ⎜ ⎢ ⎥⎟ ⎢ ⎥ ⎢ ⎢ ⎥ ⎟ ⎜ 8. ( F − 2G ) ⎢ x 2 ⎥ = F ⎜ ⎢ x 2 ⎥⎟ − 2G ⎜ ⎢ x 2 ⎥⎟ = ⎢ 0 ⎥ − 2 ⎢ x 2 ⎥ = ⎢ −2 x 2 ⎥ ⎜ ⎟ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ x ⎥ ⎢ 0 ⎥ ⎢ ⎥ x3 ⎣ 3⎦ ⎦ ⎣ ⎣ 3⎦ ⎣ 3⎦ ⎣ 3⎦ ⎣ ⎦ ⎛ ⎡ x1 ⎤⎞ ⎛ ⎡ x1 ⎤⎞ ⎛ ⎡ x1 ⎤⎞ ⎡ x1 ⎤ ⎡ x1 + x 2 ⎤ ⎡5 x1 + 3x 2 ⎤ ⎥ ⎢ ⎥⎟ ⎢ ⎥⎟ ⎢ ⎥ ⎢ ⎥ ⎢ ⎜ ⎜ ⎢ ⎥ ⎜ ⎟ 9. (2 F + 3G ) ⎢ x 2 ⎥ = 2 F ⎜ ⎢ x 2 ⎥⎟ + 3G ⎜ ⎢ x 2 ⎥⎟ = 2 ⎢ 0 ⎥ + 3 ⎢ x 2 ⎥ = ⎢ 3x 2 ⎥ ⎜ ⎟ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ x ⎥⎟⎠ ⎥ ⎢ x ⎥ ⎢ 0 ⎥ ⎢ 2x ⎣ 3⎦ ⎦ ⎣ 3 ⎦ ⎣ 3⎦ ⎣ 3⎦ ⎣ 3⎦ ⎣ ⎛ ⎛ ⎡ x ⎤⎞ ⎞ ⎛ ⎡ x1 + x 2 ⎤⎞ ⎡ x1 + x 2 ⎤ ⎛ ⎡ x1 ⎤⎞ 1 ⎜ ⎜ ⎢ ⎥⎟ ⎟ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎜ ⎟ 10. ( F F G ) ⎢ x 2 ⎥ = F ⎜ G ⎜ ⎢ x 2 ⎥⎟ ⎟ = F ⎜⎜ ⎢ x 2 ⎥⎟⎟ = ⎢ 0 ⎥ ⎜ ⎟ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ 0 ⎥⎟⎠ ⎢ 0 ⎥ ⎜⎝ ⎜⎝ ⎢ x ⎥⎟⎠ ⎟⎠ ⎣ 3⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 3⎦ ⎛ ⎛ ⎡ x ⎤⎞ ⎞ ⎛ ⎡ x1 ⎤⎞ ⎛ ⎡ x1 ⎤⎞ ⎡ x1 + 0 ⎤ ⎡ x1 ⎤ 1 ⎢ ⎥ ⎜ ⎜ ⎢ ⎥⎟ ⎟ ⎢ ⎥ ⎢ ⎥ ⎜ ⎢ ⎥⎟ 11. (G F F ⎜ ⎢ x 2 ⎥⎟ = G ⎜ F ⎜ ⎢ x 2 ⎥⎟ ⎟ = G ⎜⎜ ⎢ 0 ⎥⎟⎟ = ⎢ 0 ⎥ = ⎢ 0 ⎥ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎜⎝ ⎜⎝ ⎢ x ⎥⎟⎠ ⎟⎠ ⎣ ⎦ ⎣ ⎦ ⎣ 3⎦ ⎣ 3⎦ ⎣ 3⎦

)

⎛ ⎛ ⎡ x ⎤⎞ ⎞ ⎛ ⎡ x1 ⎤⎞ ⎛ ⎡ kx1 ⎤⎞ ⎡ kx1 + 0 ⎤ ⎡ kx1 ⎤ ⎛ ⎡ x1 ⎤⎞ 1 ⎢ ⎥⎟ ⎥⎟ ⎢ ⎜ ⎜ ⎢ ⎥⎟ ⎟ ⎥ ⎢ ⎥ ⎜ ⎜⎢ ⎢ ⎥ ⎜ ⎟ 12. (G F kF ) ⎢ x 2 ⎥ = G ⎜ kF ⎜ ⎢ x 2 ⎥⎟ ⎟ = G ⎜ k ⎢ 0 ⎥⎟ = G ⎜ ⎢ 0 ⎥⎟ = ⎢ 0 ⎥ = ⎢ 0 ⎥ ⎜ ⎟ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ kx ⎥⎟⎠ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎜⎝ ⎜⎝ ⎢ x ⎥⎟⎠ ⎟⎠ ⎣ 3⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 3⎦ ⎣ 3⎦ ⎣ 3⎦ ⎛ ⎛ ⎡ x ⎤⎞ ⎞ ⎛ ⎡ x1 ⎤⎞ ⎛ ⎡ x1 ⎤⎞ ⎡ x1 ⎤ 1 ⎜ ⎜ ⎢ ⎥⎟ ⎟ ⎜ ⎢ ⎥⎟ ⎜ ⎢ ⎥⎟ ⎢ ⎥ 13. ( F F F ⎜ ⎢ x 2 ⎥⎟ = F ⎜ F ⎜ ⎢ x 2 ⎥⎟ ⎟ = F ⎜ ⎢ 0 ⎥⎟ = ⎢ 0 ⎥ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ x ⎥ ⎜⎝ ⎜⎝ ⎢ x ⎥⎟⎠ ⎟⎠ ⎣ 3⎦ ⎣ 3⎦ ⎣ 3⎦ ⎣ 3⎦

)

⎛ ⎡ x1 ⎤⎞ 14. ((G F F ) F F ) ⎜ ⎢⎢ x 2 ⎥⎥⎟ = (G F F ⎜ ⎟ ⎜⎝ ⎢ x ⎥⎟⎠ ⎣ 3⎦

)

⎛ ⎛ ⎡ x ⎤⎞ ⎞ 1 ⎜ ⎜ ⎢ ⎥⎟ ⎟ F x ⎜ ⎜ ⎢ 2 ⎥⎟ ⎟ = ( G F F ⎜⎝ ⎜⎝ ⎢ x ⎥⎟⎠ ⎟⎠ ⎣ 3⎦

)

⎛ ⎛ ⎡ x ⎤⎞ ⎞ ⎛ ⎡ x1 ⎤⎞ ⎛ ⎡ x1 ⎤⎞ ⎡ x1 ⎤ 1 ⎜ ⎜ ⎢ ⎥⎟ ⎟ ⎜ ⎢ ⎥⎟ ⎜ ⎢ ⎥⎟ ⎢ ⎥ ⎜ ⎢ 0 ⎥⎟ = G ⎜ F ⎜ ⎢ 0 ⎥⎟ ⎟ = G ⎜ ⎢ 0 ⎥⎟ = ⎢ 0 ⎥ ⎜⎝ ⎢ x ⎥⎟⎠ ⎜⎝ ⎢ x ⎥⎟⎠ ⎢ 0 ⎥ ⎜⎝ ⎜⎝ ⎢ x ⎥⎟⎠ ⎟⎠ ⎣ ⎦ ⎣ 3⎦ ⎣ 3⎦ ⎣ 3⎦

⎛ ⎡ x ⎤⎞ ⎡ x − 2 x 2 ⎤ 15. Show that T : R 2 → R 2 defined by T ⎜ ⎢ 1 ⎥⎟ = ⎢ 1 ⎥ is an isomorphism. ⎝ ⎣ x 2 ⎦⎠ ⎣ x1 − x 2 ⎦ ⎛ ⎡ x ⎤⎞ ⎡ x − 2 x 2 ⎤ ⎡ 0 ⎤ x1 − 2 x 2 = 0 First show Ker T = {0} : If T ⎜ ⎢ 1 ⎥⎟ = ⎢ 1 . The only solution is x1 = x 2 = 0 . ⎥ = ⎢ ⎥ , then x1 − x 2 = 0 − x x x 0 ⎝ ⎣ 2 ⎦⎠ ⎣ 1 2 ⎦ ⎣ ⎦

214

Answer key

Hence, Ker T = {0} . Then show Im T = R 2: 2 = dim ( R 2 ) = dim (Ker T ) + dim (Im T ) = 0 + dim (Im T ) = dim (Im T ); and, thus, because Im T is a subspace of R 2 and dim (Im T ) = 2 = dim( R 2 ), Im T = R 2. Hence, T is an isomorphism.

9·6

⎛ ⎡ x ⎤⎞ ⎡ − x ⎤ ⎡7⎤ 1. T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ that is a reflection about the y-axis; x = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ y ⎦ ⎣ −3 ⎦ ⎛ ⎡1 ⎤⎞ ⎡ −1⎤ ⎛ ⎡ 0 ⎤⎞ ⎡ 0 ⎤ T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ and T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎝ ⎣ 0 ⎦⎠ ⎣ 0 ⎦ ⎝ ⎣ 1 ⎦⎠ ⎣ 1 ⎦ ⎡ −1 0 ⎤ Thus, [T ] = ⎢ ⎥. ⎣ 0 1⎦ ⎡ −1 0 ⎤ ⎡ 7 ⎤ ⎡ −7 ⎤ T ( x ) = [T ]x = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎣ 0 1 ⎦ ⎣ −3 ⎦ ⎣ −3 ⎦ ⎛ ⎡ x ⎤⎞ ⎡ y ⎤ ⎡4 ⎤ 2. T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ that is a reflection about the line y = x ; x = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ x ⎦ ⎣1 ⎦ ⎛ ⎡ 1 ⎤⎞ ⎡ 0 ⎤ ⎛ ⎡ 0 ⎤⎞ ⎡ 1 ⎤ T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ and T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ . ⎝ ⎣ 0 ⎦⎠ ⎣ 1 ⎦ ⎝ ⎣ 1 ⎦⎠ ⎣ 0 ⎦ ⎡0 1 ⎤ Thus, [T ] = ⎢ ⎥. ⎣1 0 ⎦ ⎡0 1 ⎤ ⎡ 4 ⎤ ⎡ 1 ⎤ T ( x ) = [T ]x = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎣1 0 ⎦ ⎣ 1 ⎦ ⎣ 4 ⎦ ⎛ ⎡ x ⎤⎞ ⎡ 0 ⎤ ⎡3⎤ 3. T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ that is an orthogonal projection on the y-axis; x = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ y ⎦ ⎣ −3 ⎦ ⎛ ⎡ 1 ⎤⎞ ⎡ 0 ⎤ ⎛ ⎡ 0 ⎤⎞ ⎡ 0 ⎤ T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ and T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎝ ⎣ 0 ⎦⎠ ⎣ 0 ⎦ ⎝ ⎣ 1 ⎦⎠ ⎣ 1 ⎦ ⎡0 0 ⎤ Thus, [T ] = ⎢ ⎥. ⎣0 1 ⎦ ⎡0 0 ⎤ ⎡ 3 ⎤ ⎡ 0 ⎤ T ( x ) = [T ]x = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎣0 1 ⎦ ⎣ −3 ⎦ ⎣ −3 ⎦

Answer key

215

⎛ ⎡ x ⎤⎞ ⎡ x ⎤ ⎡4 ⎤ 4. T : R 2 → R 2 defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ that is a reflection about the x-axis; x = ⎢ ⎥ ⎝ ⎣ y ⎦⎠ ⎣ − y ⎦ ⎣1 ⎦ ⎛ ⎡ 1 ⎤⎞ ⎡ 1 ⎤ ⎛ ⎡ 0 ⎤⎞ ⎡ 0 ⎤ T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ and T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ ⎝ ⎣ 0 ⎦⎠ ⎣ 0 ⎦ ⎝ ⎣1 ⎦⎠ ⎣ −1⎦ ⎡1 0 ⎤ Thus, [T ] = ⎢ ⎥. ⎣0 −1⎦ ⎡1 0 ⎤ ⎡ 4 ⎤ ⎡ 4 ⎤ T ( x ) = [T ]x = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎣0 −1⎦ ⎣ 1 ⎦ ⎣ −1⎦ ⎡ 2 ⎤ ⎛ ⎡ x ⎤⎞ ⎡ 5 x ⎤ 5. T : R → R defined by T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ that is a dilation with factor 5; x = ⎢⎢ 5 ⎥⎥ −3 ⎝ ⎣ y ⎦⎠ ⎣5 y ⎦ ⎢⎣ 10 ⎥⎦ 2

2

⎛ ⎡ 1 ⎤⎞ ⎡ 5 ⎤ ⎛ ⎡ 0 ⎤⎞ ⎡ 0 ⎤ T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ and T ⎜ ⎢ ⎥⎟ = ⎢ ⎥ 0 0 ⎝ ⎣ ⎦⎠ ⎣ ⎦ ⎝ ⎣ 1 ⎦⎠ ⎣ 5 ⎦ ⎡5 0 ⎤ Thus, [T ] = ⎢ ⎥. ⎣0 5 ⎦ ⎡ 2 ⎤ ⎡5 0 ⎤ ⎢ 5 ⎥ ⎡ 2 ⎤ ⎥ =⎢ T ( x ) =[T ]x = ⎢ ⎥⎢ ⎥ −3 ⎣0 5 ⎦ ⎢ −3 ⎥ ⎢⎣ 2 ⎥⎦ ⎣ 10 ⎦ ⎛ ⎡ x ⎤⎞ ⎡ x ⎤ ⎡1 ⎤ ⎜ ⎢ ⎥⎟ ⎢ ⎥ ⎢ ⎥ 6. T : R → R defined by T ⎜ ⎢ y ⎥⎟ = ⎢ − y ⎥ that is a reflection about the xz-plane; x = ⎢ 2 ⎥ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ z ⎥ ⎢⎣ 3 ⎥⎦ ⎣ ⎦ ⎣ ⎦ 3

3

⎛ ⎡ 0 ⎤⎞ ⎡ 0 ⎤ ⎛ ⎡ 1 ⎤⎞ ⎡ 1 ⎤ ⎛ ⎡ 0 ⎤⎞ ⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎟ ⎢ ⎥ ⎜ ⎢ ⎥⎟ ⎢ ⎥ ⎜ T ⎜ ⎢0 ⎥⎟ = ⎢0 ⎥ , T ⎜ ⎢1 ⎥⎟ = ⎢ −1⎥ and T ⎜⎜ ⎢0 ⎥⎟⎟ = ⎢0 ⎥ ⎜⎝ ⎢1 ⎥⎟⎠ ⎢1 ⎥ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢0 ⎥ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢ 0 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡1 0 0 ⎤ ⎢ ⎥ Thus, [T ] = ⎢0 −1 0 ⎥ . ⎢⎣0 0 1 ⎥⎦ ⎡1 0 0 ⎤ ⎡ 1 ⎤ ⎡ 1 ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ T ( x ) = [T ]x = ⎢0 −1 0 ⎥ ⎢ 2 ⎥ = ⎢ −2 ⎥ ⎢0 0 1 ⎥ ⎢ 3 ⎥ ⎢ 3 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

216

Answer key

⎛ ⎡ x ⎤⎞ ⎡ 0 ⎤ ⎡ −3 ⎤ ⎢ ⎥ ⎜ ⎢ ⎥⎟ ⎢ ⎥ 7. T : R → R defined by T ⎜ ⎢ y ⎥⎟ = ⎢ y ⎥ that is an orthogonal projection on the yz-plane; x = ⎢ 2 ⎥ ⎢4⎥ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ z ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 3

3

⎛ ⎡ 0 ⎤⎞ ⎡ 0 ⎤ ⎛ ⎡ 1 ⎤⎞ ⎡ 0 ⎤ ⎛ ⎡ 0 ⎤⎞ ⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ T ⎜⎜ ⎢0 ⎥⎟⎟ = ⎢0 ⎥ , T ⎜⎜ ⎢1 ⎥⎟⎟ = ⎢1 ⎥ , and T ⎜⎜ ⎢0 ⎥⎟⎟ = ⎢0 ⎥ ⎜⎝ ⎢1 ⎥⎟⎠ ⎢1 ⎥ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢0 ⎥ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢0 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡0 0 0 ⎤ ⎢ ⎥ Thus, [T ] = ⎢0 1 0 ⎥ ⎢0 0 1 ⎥ ⎣ ⎦ ⎡0 0 0 ⎤ ⎡ −3 ⎤ ⎡ 0 ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ T ( x ) = [T ]x = ⎢0 1 0 ⎥ ⎢ 2 ⎥ = ⎢ 2 ⎥ ⎢0 0 1 ⎥ ⎢ 4 ⎥ ⎢ 4 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎛ ⎡ x ⎤⎞ ⎡ ⎤ x ⎥ ⎜ ⎢ ⎥⎟ ⎢ 8. T : R → R defined by T ⎜ ⎢ y ⎥⎟ = ⎢ y cosθ − z sinθ ⎥ that is a counterclockwise rotation about the positive ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ y sinθ + z cosθ ⎥ ⎣ ⎦ ⎣ ⎦ 3

3

⎡5⎤ ⎢ ⎥ x-axis; x = ⎢ 2 ⎥ , θ = 45° ⎢⎣ −2 ⎥⎦ ⎛ ⎡ 0 ⎤⎞ ⎡ 0 ⎤ ⎛ ⎡ 1 ⎤⎞ ⎡ 1 ⎤ ⎛ ⎡ 0 ⎤⎞ ⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ T ⎜⎜ ⎢0 ⎥⎟⎟ = ⎢0 ⎥ , T ⎜⎜ ⎢1 ⎥⎟⎟ = ⎢cosθ ⎥ , and T ⎜⎜ ⎢0 ⎥⎟⎟ = ⎢ − sinθ ⎥ ⎜⎝ ⎢1 ⎥⎟⎠ ⎢ cosθ ⎥ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢0 ⎥ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢ sinθ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡1 0 ⎢ Thus, [T ] = ⎢0 cosθ ⎢0 sinθ ⎣

0 ⎤ ⎥ − sinθ ⎥ cosθ ⎥⎦

0 0 ⎤⎡ 5 ⎤ ⎡ 5 ⎡1 ⎤ ⎡ 5 ⎤ ⎥ ⎢0 cos 45° − sin 45° ⎥ ⎢ 2 ⎥ ⎢ 2 cos 45° + 2 sin 45° ⎥ ⎢ T ( x ) = [T ]x = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ = ⎢2 2 ⎥ ⎢⎣0 sin 45° cos 45° ⎥⎦ ⎢⎣ −2 ⎥⎦ ⎢⎣ 2 sin 45° − 2 cos 45° ⎥⎦ ⎢ 0 ⎥ ⎣ ⎦ ⎛ ⎡ x ⎤⎞ ⎡ 0.2 x ⎤ ⎡6⎤ ⎢ ⎥ ⎢ ⎥⎟ ⎢ ⎥ ⎜ 9. T : R → R defined by T ⎜ ⎢ y ⎥⎟ = ⎢0.2 y ⎥ that is a contraction with factor 0.2; x = ⎢ −3 ⎥ ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ 0.2z ⎥ ⎢⎣ 8 ⎥⎦ ⎣ ⎦ ⎣ ⎦ 3

3

⎛ ⎡1 ⎤⎞ ⎡0.2 ⎤ ⎛ ⎡0 ⎤⎞ ⎡ 0 ⎤ ⎛ ⎡0 ⎤⎞ ⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ T ⎢0 ⎥ = ⎢ 0 ⎥ , T ⎢1 ⎥ = ⎢0.2 ⎥ , and T ⎜ ⎢⎢0 ⎥⎥⎟ = ⎢⎢ 0 ⎥⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢ 0 ⎥ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢ 0 ⎥ ⎜⎝ ⎢1 ⎥⎟⎠ ⎢0.2 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Answer key

217

0⎤ ⎡0.2 0 ⎢ 0 0.2 0 ⎥ Thus, [T ] = ⎢ ⎥. ⎢⎣ 0 0 0.2 ⎥⎦ 0 ⎤ ⎡ 6 ⎤ ⎡ 1.2 ⎤ ⎡0.2 0 ⎢ 0 0.2 0 ⎥ ⎢ −3 ⎥ ⎢ −0.6 ⎥ T ( x ) = [T ]x = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢⎣ 0 0 0.2 ⎥⎦ ⎢⎣ 8 ⎥⎦ ⎢⎣ 1.6 ⎥⎦ ⎛ ⎡ x ⎤⎞ ⎡ x + 3 ⎤ ⎥ ⎜ ⎢ ⎥⎟ ⎢ 10. Show that T : R → R defined by T ⎜ ⎢ y ⎥⎟ = ⎢ y + 3 ⎥ is not a linear transformation. ⎜⎝ ⎢ z ⎥⎟⎠ ⎢ z + 3 ⎥ ⎦ ⎣ ⎦ ⎣ 3

3

⎛ ⎡ 0 ⎤⎞ ⎡ 3 ⎤ ⎡ 0 ⎤ ⎜ ⎢ ⎥⎟ ⎢ ⎥ ⎢ ⎥ T ⎜ ⎢ 0 ⎥⎟ = ⎢ 3 ⎥ ≠ ⎢ 0 ⎥ ⎜⎝ ⎢0 ⎥⎟⎠ ⎢3 ⎥ ⎢0 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Thus, T is not a linear transformation.

10

Eigenvalues and eigenvectors

10·1

1. λ x 2. eigenvector 3. 0 4. λ n + c1λ n−1 + $ + cn 5. n 6. Find the characteristic polynomial for A.

λI − A =

λ − 10 −4 = λ 2 − 8λ + 16 9 λ+2

7. Find the eigenvalues of A. Because λ 2 − 8λ + 16 = (λ − 4 )2 = 0, the eigenvalues are λ1 = λ2 = 4 . 8. Find eigenvectors corresponding to the eigenvalues of A. ⎡ 4 − 10 −4 ⎤ ⎡ x1 ⎤ ⎡ −6 −4 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ For λ = 4, (λ I − A)x = 0 has the form ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ , which has the augmented 4 + 2 ⎦ ⎣ x 2 ⎦ ⎣ 9 6 ⎦ ⎣ x 2 ⎦ ⎣0 ⎦ ⎣ 9 ⎡1 2 ⎡ −6 −4 0 ⎤ 3 matrix form ⎢ with reduced row-echelon form ⎢ ⎥ ⎢⎣0 0 ⎣ 9 6 0⎦ ⎡ −2 ⎤ eigenvector has the form t ⎢ 3 ⎥ . ⎢ 1 ⎥ ⎣ ⎦

218

Answer key

0⎤ 2 . So x1 = − t and x 2 = t . Thus, any 0 ⎥⎦ 3

9. Find bases for the eigenspaces of A. ⎡2 ⎤ ⎢ 3 ⎥ is a basis for the eigenspace of A ⎢ 1 ⎥ ⎣ ⎦ 10. Find the characteristic polynomial for B.

λI − B =

λ − 3 −2 −2 1 λ−0 1 = λ 3 − 2λ 2 + λ 2 2 λ +1

11. Find the eigenvalues of B. Because λ 3 − 2λ 2 + λ = λ(λ − 1)2, the eigenvalues are λ1 = λ2 = 1, and λ3 = 0. 12. Find eigenvectors corresponding to the eigenvalues of B. ⎡ −2 −2 −2 ⎢ For λ1 = λ2 = 1, the associated augmented matrix has the form ⎢ 1 1 1 ⎢⎣ 2 2 2

0 ⎤ ⎡1 1 1 0 ⎥⎥ ⇒ ⎢⎢0 0 0 0 ⎥⎦ ⎢⎣0 0 0

0⎤ 0 ⎥⎥ . 0 ⎥⎦

Then x1 = −t − s and x 2 = t and x 3 = s are free variables. All the associated eigenvectors have the form ⎡ −t − s ⎤ ⎡ −1⎤ ⎡ −1⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ t ⎥ = t ⎢ 1 ⎥ + s ⎢ 0 ⎥. ⎢ s ⎥ ⎢0⎥ ⎢1⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ −3 −2 −2 ⎢ For λ3 = 0, the associated augmented matrix has the form ⎢ 1 0 1 ⎢⎣ 2 2 1

1 0 ⎤ ⎡1 0 ⎢ ⎥ 0 ⎥ ⇒ ⎢0 1 − 1 2 0 ⎥⎦ ⎢⎣0 0 0

0⎤ 0 ⎥⎥ 0 ⎥⎦

⎡ −t ⎤ ⎡ −1 ⎤ ⎢ ⎥ ⎢ ⎥ 1 Thus, x1 = −t , x 2 = t and x 3 = t . All the associated eigenvectors have the form ⎢ 1 t ⎥ = t ⎢ 1 ⎥ 2 2 ⎢ ⎥ ⎢ 2⎥ ⎢⎣ t ⎥⎦ ⎢⎣ 1 ⎥⎦ 13. Find bases for the eigenspaces of B.

⎧ ⎡ −1⎤ ⎡ −1⎤ ⎫ ⎪⎢ ⎥ ⎢ ⎥ ⎪ A basis for the eigenspace of B associated with λ1 = λ2 = 1 is ⎨ ⎢ 1 ⎥ , ⎢ 0 ⎥ ⎬ ⎪⎢ 0 ⎥ ⎢ 1 ⎥ ⎪ ⎩⎣ ⎦ ⎣ ⎦ ⎭ ⎧ ⎡ −1 ⎤ ⎫ ⎪⎪ ⎢ ⎥ ⎪⎪ A basis for the eigenspace of B associated with λ3 = 0 is ⎨ ⎢ 1 ⎥ ⎬ 2 ⎪⎢ ⎥ ⎪ ⎪⎩ ⎢⎣ 1 ⎥⎦ ⎪⎭

⎡ −4 8 ⎤ 14. ⎢ ⎥ ⎣ 0 3⎦

λI − A =

λ + 4 −8 = (λ + 4 )(λ − 3) = 0 yields the eigenvalues: λ1 = −4 and λ2 = 3. 0 λ −3

⎡ −4 0 ⎤ 15. ⎢ ⎥ ⎣ 8 3⎦

λI − A =

λ+4 0 = (λ + 4 )(λ − 3) = 0 yields the eigenvalues: λ1 = −4 and λ2 = 3. −8 λ − 3 Answer key

219

⎡ −4 0 ⎤ 16. ⎢ ⎥ ⎣ 0 3⎦

λI − A =

λ+4 0 = (λ + 4 )(λ − 3) = 0 yields the eigenvalues: λ1 = −4 and λ2 = 3. 0 λ −3

⎡4 0 ⎤ 17. ⎢ ⎥ ⎣81 −2 ⎦

λI − A = ⎡1 ⎢ 2 18. ⎢ −3 ⎢ ⎢ 6 ⎣

λ−4 0 = (λ − 4 )(λ + 2) = 0 yields the eigenvalues: λ1 = 4 and λ2 = −2. −81 λ + 2

0 5 4

9

0⎤ ⎥ 0⎥ ⎥ −8 ⎥ ⎦

λ− 1 λI − A =

2

3 −6

0

0

λ −5 0 . Expanding by elements in the first row, 4 − λ +8 9

1⎞ 1 ⎛ = ⎜ λ − ⎟ (λ − 5)(λ + 8) = 0 yields the eigenvalues: λ1 = , λ2 = 5, and λ3 = −8. ⎝ ⎠ 2 2 ⎡ −1 5 0 ⎤ ⎢ ⎥ 19. ⎢ 0 11 −10 ⎥ ⎢0 0 6 ⎥⎦ ⎣

λ + 1 −5 0 λI − A = 0 λ − 11 10 . Expanding by elements on the first column, 0 0 λ−6 = (λ + 1)(λ − 11)(λ − 6) = 0 yields the eigenvalues: λ1 = −1, λ2 = 11, and λ3 = 6. ⎡1 0 ⎢ 0 −3 20. ⎢ ⎢0 0 ⎢ ⎢⎣ 0 0

0 0⎤ ⎥ 0 0⎥ 5 0⎥ ⎥ 0 −1⎦⎥

λ −1 0 0 0 0 λ +3 0 0 λI − A = . Expanding by elements in the first row, 0 0 λ −5 0 0 0 0 λ +1 λ+3 0 0 λ −5 0 ; and expanding by the elements in the first row again to get = (λ − 1) 0 0 0 λ +1 (λ − 1)(λ + 3)(λ − 5)(λ + 1) = 0, which yields eigenvalues: λ1 = 1, λ2 = −3, λ3 = 5, and λ4 = −1.

220

Answer key

10·2

1. sum 2. product 1 λ 4. same 3.

5. diagonal 6. Find the eigenvalues of A.

λI − A =

λ−2 1 = λ 2 − 6λ + 9 = (λ − 3)2. Thus, λ1 = λ2 = 3. −1 λ − 4

7. Verify that tr( A) is the sum of the eigenvalues of A. tr( A) = 2 + 4 = 6 and λ1 + λ2 = 3 + 3 = 6 8. Verify that det( A) is the product of the eigenvalues of A. det( A) =

2 −1 1

4

= 9 and λ1 ⋅ λ2 = 3 ⋅ 3 = 9

9. Verify that if λ is an eigenvalue of A, then ⎡ 4 A =⎢ 9 ⎢ 1 ⎢⎣ − 9 −1

1 ⎤ λ − 4 9 − 19 9 ⎥ . The characteristic equation is det(λ I − A−1 ) = = 0. Because 3 is the 1 2 ⎥ λ − 29 9 9 ⎥⎦

repeated eigenvalue for A, substituting

−

1 is an eigenvalue of A−1. λ

1 into this equation gives 3

1 −4 3 9 1 9

−1 9 9 = 1 1 −2 9 3 9 −1

−1 1

9 =

9

1 1 1 + = 0. Thus, is an eigenvalue for A−1. 81 81 3

10. Verify that if λ is an eigenvalue of A, then λ 3 is an eigenvalue of A3. The characteristic equation is λ I − A3 =

λ 3 = 27 into this equation give

27 λ = 0. The eigenvalue for A is λ = 3, Substituting −27 λ − 54

27 27 = −729 + 729 = 0. Thus, λ 3 = 27 is an eigenvalue for A3. −27 −27

11. Verify that A and AT have the same eigenvalues. ⎡ 2 1⎤ λ − 2 −1 = (λ − 2)(λ − 4 ) + 1 = λ 2 − 6λ + 9 which is the AT = ⎢ ⎥, which has characteristic equation 1 4 1 λ − 4 − ⎦ ⎣ same characteristic equation as A. Thus, they have the same eigenvalues. ⎡1 5 ⎤ 12. ⎢ ⎥; λ1 = −1, λ2 = 6 ⎣2 4 ⎦ Trace = 4 + 1 = 5 and λ1 + λ2 = −1 + 6 = 5

Answer key

221

⎡2 1 1⎤ ⎢ ⎥ 13. ⎢ −2 1 3 ⎥ ; λ1 = 1, λ2 = −2, λ3 = 3 ⎢ 3 1 −1⎥ ⎣ ⎦ Trace = 2 + 1 + (−1) = 2 and λ1 + λ2 + λ3 = 1 − 2 + 3 = 2 ⎡1 5 ⎤ 14. ⎢ ⎥ ; λ1 = −1, λ2 = 6 ⎣2 4 ⎦ ⎛ ⎡ 1 5 ⎤⎞ det ⎜ ⎢ ⎥⎟ = 4 − 10 = −6 ⎝ ⎣ 2 4 ⎦⎠

λ1 (λ2 ) = (−1)(6) = −6 ⎡2 1 1⎤ ⎢ ⎥ 15. ⎢ −2 1 3 ⎥ ; λ1 = 1, λ2 = −2, λ3 = 3 ⎢ 3 1 −1⎥ ⎣ ⎦ ⎛ ⎡ 2 1 1 ⎤⎞ ⎢ ⎥ det ⎜⎜ ⎢ −2 1 3 ⎥⎟⎟ = −6 ⎜⎝ ⎢ 3 1 −1⎥⎟⎠ ⎣ ⎦

λ1λ2 λ3 = (1)(−2)(3) = −6 ⎡ − 4.5 16. ⎢ ⎢⎣ 0

7⎤ ⎥ 2 ⎥⎦

λ1 = − 4.5 and λ2 = 2 ⎡ −40 17. ⎢ 1 ⎢ ⎣ 5

0 ⎤ ⎥ 3 ⎥ 5⎦

λ1 = −40 and λ2 = 3

5

⎡0.25 0 ⎤ 18. ⎢ ⎥ ⎣ 0 −8 ⎦

λ1 = 0.25 and λ2 = −8 ⎡ 100 0 0 ⎤ ⎢ ⎥ 19. ⎢ −50 −30 0 ⎥ ⎢ 60 90 −10 ⎥ ⎣ ⎦

λ1 = 100, λ2 = −30, and λ3 = −10

222

Answer key

⎡− 1 ⎢ 3 ⎢ 20. ⎢ 0 ⎢ 0 ⎢ 0 ⎣

5

3 9 0 0

0 ⎤ ⎥ 5 −10 ⎥ ⎥ −2 1 ⎥ 0 5 ⎥⎦

−7

λ1 = − 1 , λ2 = 9 , λ3 = −2, and λ4 = 5 3

10·3

1. similar 2. diagonal 3. distinct ⎡2 4 ⎤ 4. Given A = ⎢ ⎥ that has eigenvalues λ1 = 5 and λ2 = −2 with corresponding eigenvectors ⎣3 1 ⎦ ⎡4 ⎤ ⎡ −1⎤ ⎢ 3 ⎥ and ⎢ ⎥ , respectively, find a nonsingular matrix P such that P −1 AP is diagonal. ⎢ 1 ⎥ ⎣1⎦ ⎣ ⎦ ⎡ 3 P AP = ⎢⎢ 7 3 ⎢⎣ − 7 −1

3 ⎤ ⎡4 7 ⎥ ⎡2 4 ⎤ ⎢ 3 ⎢ ⎥ 4 ⎥ ⎣3 1 ⎦ ⎢ 1 ⎣ 7 ⎥⎦

−1⎤ ⎡5 0 ⎤ ⎥=⎢ ⎥ 1 ⎥⎦ ⎣0 −2 ⎦

⎡2 4 ⎤ 6 6 −1 5. Given A = ⎢ ⎥ , use the results from Prob. 4 to calculate A = PD P . ⎣3 1 ⎦ ⎡4 A6 = PD 6 P −1 = ⎢ 3 ⎢ 1 ⎣

−1⎤ ⎡5 0 ⎤ ⎥⎢ ⎥ 1 ⎥⎦ ⎣0 −2 ⎦

6

⎡ 3 ⎢ 7 ⎢− 3 ⎣ 7

3 ⎤ 7 ⎥ ⎡8956 8892 ⎤ = 4 ⎥ ⎢⎣6669 6733 ⎥⎦ 7⎦

Answer key

223

Notes

Notes

Notes

Notes

Notes

Notes

Notes