Linear Algebra - MAT223 Solutions [2021/08/06 ed.]

Table of contents :
Lessons
The Magic Carpet Ride
1
The Magic Carpet Ride, Hide and Seek
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
The Magic Carpet Ride, Getting Back Home
18
19
20
21
22
Linear Independence and Dependence, Creating Examples
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
Italicizing N
44
Beyond the N
45
46
47
48
Pat and Jamie
49
50
51
52
53
54
55
56
57
58
59
60
61
62
Getting back N
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
The Green and the Black
82
83
84
85
86
87
88
89
90
91

Citation preview

Solutions

u ⃗

w ⃗ α = αu ⃗ + (1 − α)v⃗

⃗v

Jason Siefken

The Magic Carpet Ride 1

You are a young adventurer. Having spent most of your time in the mythical city of Oronto, you decide to leave home for the first time. Your parents want to help you on your journey, so just before your departure, they give you two gifts. Specifically, they give you two forms of transportation: a hover board and a magic carpet. Your parents inform you that both the hover board and the magic carpet have restrictions in how they operate: We denote • ˜ the restriction on the hover board’s movement by the 3 vector . By this we mean that if the hover board traveled “forward” 1 for one hour, it would move along a “diagonal” path that would result in a displacement of 3 km East and 1 km North of its starting location. We •denote the restriction on the magic carpet’s movement by the vec˜ 1 tor . By this we mean that if the magic carpet traveled “forward” 2 for one hour, it would move along a “diagonal” path that would result in a displacement of 1 km East and 2 km North of its starting location.

Scenario One: The Maiden Voyage Your Uncle Cramer suggests that your first adventure should be to go visit the wise man, Old Man Gauss. Uncle Cramer tells you that Old Man Gauss lives in a cabin that is 107 km East and 64 km North of your home. Task: Investigate whether or not you can use the hover board and the magic carpet to get to Gauss’s cabin. If so, how? If it is not possible to get to the cabin with these modes of transportation, why is that the case?

Drawings by @DavidsonJohnR (twitter)

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© IOLA Team iola. math. vt. edu

Hands-on experience with vectors as displacements. ■ Internalize vectors as geometric objects representing displacements. ■ Use column vector notation to write vectors. ■ Use pre-existing knowledge of algebra to answer vector questions.

The Magic Carpet Ride, Hide and Seek 2

You are a young adventurer. Having spent most of your time in the mythical city of Oronto, you decide to leave home for the first time. Your parents want to help you on your journey, so just before your departure, they give you two gifts. Specifically, they give you two forms of transportation: a hover board and a magic carpet. Your parents inform you that both the hover board and the magic carpet have restrictions in how they operate: We denote • ˜ the restriction on the hover board’s movement by the 3 vector . By this we mean that if the hover board traveled “forward” 1 for one hour, it would move along a “diagonal” path that would result in a displacement of 3 km East and 1 km North of its starting location. We •denote the restriction on the magic carpet’s movement by the vec˜ 1 tor . By this we mean that if the magic carpet traveled “forward” 2 for one hour, it would move along a “diagonal” path that would result in a displacement of 1 km East and 2 km North of its starting location.

Scenario Two: Hide-and-Seek Old Man Gauss wants to move to a cabin in a different location. You are not sure whether Gauss is just trying to test your wits at finding him or if he actually wants to hide somewhere that you can’t visit him. Are there some locations that he can hide and you cannot reach him with these two modes of transportation? Describe the places that you can reach using a combination of the hover board and the magic carpet and those you cannot. Specify these geometrically and algebraically. Include a symbolic representation using vector notation. Also, include a convincing argument supporting your answer.

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© IOLA Team iola. math. vt. edu

Address an existential question involving vectors: “Is it possible to find a linear combination that does. . . ?” The goal of this problem is to ■ Formalize geometric questions using the language of vectors. ■ Find both geometric and algebraic arguments to support the same conclusion. ■ Establish what a “negative multiple” of a vector should be.

Sets and Set Notation Set A set is a (possibly infinite) collection of items and is notated with curly braces (for example, {1, 2, 3} is the set containing the numbers 1, 2, and 3). We call the items in a set elements.

DEFINITION

If X is a set and a is an element of X , we may write a ∈ X , which is read “a is an element of X .” If X is a set, a subset Y of X (written Y ⊆ X ) is a set such that every element of Y is an element of X . Two sets are called equal if they are subsets of each other (i.e., X = Y if X ⊆ Y and Y ⊆ X ). We can define a subset using set-builder notation. That is, if X is a set, we can define the subset Y = {a ∈ X : some rule involving a}, which is read “Y is the set of a in X such that some rule involving a is true.” If X is intuitive, we may omit it and simply write Y = {a : some rule involving a}. You may equivalently use “|” instead of “:”, writing Y = {a | some rule involving a}.

DEFINITION

Some common sets are N = {natural numbers} = {non-negative whole numbers}. Z = {integers} = {whole numbers, including negatives}. R = {real numbers}. Rn = {vectors in n-dimensional Euclidean space}.

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Practice reading sets and set-builder notation.

3.1 Which of the following statements are true?

The goal of this problem is to ■ Become familiar with ∈, ⊆, and = in the context of sets.

(a) 3 ∈ {1, 2, 3}. True

■ Distinguish between ∈ and ⊆.

(b) 1.5 ∈ {1, 2, 3}. False

■ Use quantifiers with sets.

(c) 4 ∈ {1, 2, 3}. False (d) “b”∈ {x : x is an English letter}. True (e) “ò”∈ {x : x is an English letter}. False (f) {1, 2} ⊆ {1, 2, 3}. True (g) For some a ∈ {1, 2, 3}, a ≥ 3. True (h) For any a ∈ {1, 2, 3}, a ≥ 3. False (i) 1 ⊆ {1, 2, 3}. False (j) {1, 2, 3} = {x ∈ R : 1 ≤ x ≤ 3}. False Practice writing sets using set-builder notation.

(k) {1, 2, 3} = {x ∈ Z : 1 ≤ x ≤ 3}. True

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The goal of this problem is to ■ Express English descriptions using math notation.

Write the following in set-builder notation 4.1 The subset A ⊆ R of real numbers larger than



x ∈R : x >

■ Recognize there is more than one correct way to write formal math.

⎷ 2.

⎷ 2 .

■ Preview vector form of a line.

4.2 The subset B ⊆ R2 of vectors whose first coordinate is twice the second.

§

• ˜ ª § • ˜ ª a 2t ⃗v ∈ R2 : ⃗v = with a = 2b or ⃗v ∈ R2 : ⃗v = for some t ∈ R b t §• ˜ ª a 2 or ∈ R : a = 2b . b 3

© Jason Siefken, 2015–2021

DEFINITION

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Unions & Intersections Let X and Y be sets. The union of X and Y and the intersection of X and Y are defined as follows. (union) X ∪ Y = {a : a ∈ X or a ∈ Y }. (intersection) X ∩ Y = {a : a ∈ X and a ∈ Y }.

Let X = {1, 2, 3} and Y = {2, 3, 4, 5} and Z = {4, 5, 6}. Compute

Apply the definition of ∪ and ∩.

5.1 X ∪ Y {1, 2, 3, 4, 5} 5.2 X ∩ Y {2, 3} 5.3 X ∪ Y ∪ Z {1, 2, 3, 4, 5, 6} 5.4 X ∩ Y ∩ Z ∅ = {}

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Visualize sets of vectors.

Draw the following subsets of R2 . § • ˜ ª 0 2 6.1 V = ⃗ x ∈ R : ⃗x = for some t ∈ R . t § • ˜ ª t 6.2 H = ⃗ x ∈ R2 : ⃗x = for some t ∈ R . 0 § • ˜ ª 1 6.3 D = ⃗ x ∈ R2 : ⃗x = t for some t ∈ R . 1

The goal of this problem is to ■ Apply set-builder notation in the context of vectors. ■ Distinguish between “for all” and “for some” in set builder notation. ■ Practice unions and intersections. ■ Practice thinking about set equality.

V

D

⃗ 0

§

H

• ˜ ª 1 6.4 N = ⃗ x ∈ R : ⃗x = t for all t ∈ R . N = {}. 1 2

6.5 V ∪ H. V ∪ H looks like a “+” going through the origin.

⃗} is just the origin. 6.6 V ∩ H. V ∩ H = {0 6.7 Does V ∪ H = R2 ?

No. V ∪ H does not contain

• ˜ • ˜ 1 1 while R2 does contain . 1 1

DEFINITION

Vector Combinations Linear Combination A linear combination of the vectors ⃗v 1 , ⃗v 2 , . . . , ⃗v n is a vector w ⃗ = α1 ⃗v 1 + α2 ⃗v 2 + · · · + αn ⃗v n . The scalars α1 , α2 , . . . , αn are called the coefficients of the linear combination. 4

© Jason Siefken, 2015–2021

Practice linear combinations. The goal of this problem is to ■ Practice using the formal term linear combination. ■ Foreshadow span.

• ˜ • ˜ 1 1 Let ⃗v 1 = , ⃗v 2 = , and w ⃗ = 2v⃗1 + ⃗v 2 . 1 −1

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7.1 Write w ⃗ as a column vector. When w ⃗ is written as a linear combination of ⃗v 1 and ⃗v 2 ,

what are the coefficients of ⃗v 1 and ⃗v 2 ? • ˜ 3 w ⃗= ; the coefficients are (2, 1). 1 • ˜ • ˜ 3 3 7.2 Is a linear combination of ⃗v 1 and ⃗v 2 ? Yes. = 3v⃗1 + 0v⃗2 . 3 3 • ˜ 0 ⃗ = 0v⃗1 + 0v⃗2 . 7.3 Is a linear combination of ⃗v 1 and ⃗v 2 ? Yes. 0 0 • ˜ • ˜ 4 4 7.4 Is a linear combination of ⃗v 1 and ⃗v 2 ? Yes. = 2v⃗1 + 2v⃗2 . 0 0

7.5 Can you find a vector in R2 that isn’t a linear combination of ⃗ v 1 and ⃗v 2 ?

No.

• ˜ • ˜ 1 0 = 12 ⃗v 1 + 12 ⃗v 2 and = 12 ⃗v 1 − 12 ⃗v 2 . Therefore 0 1

• ˜ • ˜ • ˜ a 1 0 =a +b = a( 12 ⃗v 1 + 12 ⃗v 2 ) + b( 12 ⃗v 1 − 12 ⃗v 2 ) = ( a+b ⃗1 + ( a−b ⃗2 . 2 )v 2 )v b 0 1 Therefore any vector in R2 can be written as linear combinations of ⃗v 1 and ⃗v 2 . 7.6 Can you find a vector in R2 that isn’t a linear combination of ⃗ v1?

Yes. •All˜ linear combinations of ⃗v 1 have equal x and y coordinates, therefore 2 w ⃗= is not a linear combination of ⃗v 1 . 1 Practice formal writing.

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Recall in the direction • ˜the Magic Carpet Ride task where the hover board could• travel ˜ 3 1 ⃗= h and the magic carpet could move in the direction m ⃗= . 1 2 8.1 Rephrase the sentence “Gauss can be reached using just the magic carpet and the hover

board” using formal mathematical language. ⃗. Gauss’s location can be written as a linear combination of m ⃗ and h 8.2 Rephrase the sentence “There is nowhere Gauss can hide where he is inaccessible by magic

carpet and hover board” using formal mathematical language.

DEFINITION

⃗. Every vector in R2 can be written as a linear combination of m ⃗ and h ⃗ and m 8.3 Rephrase the sentence “R2 is the set of all linear combinations of h ⃗ ” using formal mathematical language. ⃗ for some t, s ∈ R}. R2 = {v⃗ : ⃗v = t m ⃗ + sh

Non-negative & Convex Linear Combinations Let w ⃗ = α1 ⃗v 1 + α2 ⃗v 2 + · · · + αn ⃗v n . The vector w ⃗ is called a non-negative linear combination of ⃗v 1 , ⃗v 2 , . . . , ⃗v n if α1 , α2 , . . . , αn ≥ 0. The vector w ⃗ is called a convex linear combination of ⃗v 1 , ⃗v 2 , . . . , ⃗v n if α1 , α2 , . . . , αn ≥ 0

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α1 + α2 + · · · + αn = 1.

and

Let • ˜ 1 a ⃗= 1

• ˜ ⃗b = −1 1

• ˜ 0 ⃗c = 1

• ˜ 0 d⃗ = 2

• ˜ −1 ⃗e = . −1

Geometric meaning of non-negative and convex linear combinations. The goal of this problem is to ■ Read and apply the definition of nonnegative and convex linear combinations. ■ Gain geometric intuition for nonnegative and convex linear combinations. ■ Learn how to describe line segments using convex linear combinations.

9.1 Out of a ⃗ , ⃗b, ⃗c , d⃗ , and ⃗e, which vectors are

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© Jason Siefken, 2015–2021

(a) linear combinations of a ⃗ and ⃗b? All of them, since any vector in R2 can be written as a linear combination of a ⃗ and ⃗b. (b) non-negative linear combinations of a ⃗ and ⃗b? a ⃗ , ⃗b, ⃗c , d⃗ . (c) convex linear combinations of a ⃗ and ⃗b? a ⃗ , ⃗b, ⃗c . 9.2 If possible, find two vectors u ⃗ and ⃗v so that

(a) a ⃗ and ⃗c are non-negative linear combinations of u ⃗ and ⃗v but ⃗b is not. Let u ⃗=a ⃗ and ⃗v = ⃗c . (b) a ⃗ and ⃗e are non-negative linear combinations of u ⃗ and ⃗v . Let u ⃗=a ⃗ and ⃗v = ⃗e. (c) a ⃗ and ⃗b are non-negative linear combinations of u ⃗ and ⃗v but d⃗ is not. ⃗ Impossible. If a ⃗ and b are non-negative linear combinations of u ⃗ and ⃗v , then every non-negative linear combination of a ⃗ and ⃗b is also a non-negative linear combination of u ⃗ and ⃗v . And, we already concluded that d⃗ is a non-negative linear combination of a ⃗ and ⃗b. (d) a ⃗ , ⃗c , and d⃗ are convex linear combinations of u ⃗ and ⃗v . Impossible. Convex linear combinations all lie on the same line segment, but a ⃗ , ⃗c , and d⃗ are not collinear. Otherwise, explain why it’s not possible.

Lines and Planes 10

Let L be the set of points (x, y) ∈ R2 such that y = 2x + 1. 10.1 Describe L using set-builder notation.

§

• ˜ ª t ⃗v ∈ R2 : ⃗v = for some t ∈ R 2t + 1 §• ˜ ª §• ˜ ª x t or ∈ R2 : y = 2x + 1 or ∈ R2 : t ∈ R y 2t + 1 10.2 Draw L as a subset of R2 . 10.3 Add the vectors a ⃗=

•

˜ • ˜ −1 ⃗ 1 , b= and d⃗ = ⃗b − a ⃗ to your drawing. −1 3

d⃗ ⃗b

a ⃗

L

⃗ ∈ L? Explain. 10.4 Is d • ˜ 2 ⃗ No. d = and so its entries don’t satisfy y = 2x + 1. 4 10.5 For which t ∈ R is it true that a ⃗ + t d⃗ ∈ L? Explain using your picture.

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© Jason Siefken, 2015–2021

Link prior knowledge to new notation/concepts. The goal of this problem is to ■ Convert between y = mx + b form of a line and the set-builder definition of the same line. ■ Think about lines in terms of vectors rather than equations.

DEFINITION

a ⃗ + t d⃗ ∈ L for any t ∈ R. We can see this because if we start at the vector a ⃗ and the displace by t d⃗ , we will always be on the line L.

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Vector Form of a Line Let ℓ be a line and let d⃗ and ⃗p be vectors. If ℓ = {x⃗ : ⃗x = t d⃗ + ⃗p for some t ∈ R}, we say the vector equation ⃗x = t d⃗ + ⃗p is ℓ expressed in vector form. The vector d⃗ is called a direction vector for ℓ.

Let ℓ ⊆ R2 be the line with equation 2x + y = 3, and let L ⊆ R3 be the line with equations 2x + y = 3 and z = y. 11.1 Write ℓ in vector form. Is vector form of ℓ unique?

Practice with vector form. The goal of this problem is to ■ Express lines in R2 and R3 in vector form. ■ Produce direction vectors by subtracting two points on a line.

•

˜ • ˜ 1 0 ⃗x = t + −2 3

■ Recognize vector form is not unique.

•

˜ 1 The vector form is not unique, as any non-zero scalar multiple of can serve −2 as • a direction vector. Additionally, ˜ • ˜ any • ˜other point on the line can be used in place 0 −4 1 of . For example, ⃗x = t + is another vector form of ℓ. 3 8 1 ⎡ ⎤ ⎡ ⎤ 1 0 11.2 Write L in vector form. ⃗ x = t ⎣−2⎦ + ⎣3⎦. This is obtained by finding two points: one −2 3 when x = 0 and one when x = 1 and subtracting them to find a direction vector for L. ⃗ ” and “p 11.3 Find another vector form for L where both “d ⃗ ” are different from before. ⎡ ⎤ ⎡ ⎤ −3 1 ⃗x = t ⎣ 6⎦ + ⎣1⎦. 6 1 Again, any non-zero scalar multiple of the direction vector will work for d⃗ , as will any other point on the line work for ⃗p.

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Intersect lines in vector form.

Let A, B, and C be given in vector form by A ⏟ ⎡ ⏞⏞ ⎤ ⎡ ⎤ 1 0 ⃗x = t ⎣2⎦ + ⎣0⎦ 3 1

⏟

B ⏟ ⎡ ⏞⏞ ⎤ ⎡ ⎤ −1 −1 ⃗x = t ⎣ 1⎦ + ⎣ 1⎦ 1 2

⏟

C ⏟ ⎡ ⏞⏞⎤ ⎡ ⎤ 2 1 ⃗x = t ⎣−1⎦ + ⎣1⎦ . 1 1

⏟

12.1 Do the lines A and B intersect? Justify your conclusion.

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 0 −1 −1 Yes. (0) ⎣2⎦ + ⎣0⎦ = ⎣0⎦ = (−1) ⎣ 1⎦ + ⎣ 1⎦. 3 1 1 1 2 To find the intersection, if there is one, we must solve the vector equation: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 −1 −1 t ⎣2⎦ + ⎣0⎦ = s ⎣ 1⎦ + ⎣ 1⎦ . 3 1 1 2 One solution is when t = 0 and s = −1. 12.2 Do the lines A and C intersect? Justify your conclusion.

No. The vector equation ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 2 1 t ⎣2⎦ + ⎣0⎦ = s ⎣−1⎦ + ⎣1⎦ 3 1 1 1 7 © Jason Siefken, 2015–2021

The goal of this problem is to ■ Practice computing the intersection between lines in vector form. ■ Recognize “ t ” as a dummy variable as used in vector form and that, when comparing lines in vector form, “ t ” needs to be replaced with nondummy variables.

has no solutions. This is equivalent to saying that the following system of equations has no solutions: t = 2s + 1 2t = −s + 1 3t + 1 = s + 1 The third equation tells us that s = 3t, which when substituted into the first equation forces t = − 15 and therefore s = − 35 . However, these two numbers don’t satisfy the second equation. 12.3 Let ⃗ p ̸= q⃗ and suppose X has vector form ⃗x = t d⃗ + ⃗p and Y has vector form ⃗x = t d⃗ + q⃗. Is it possible that X and Y intersect? Yes. If q⃗ = ⃗p + ad⃗ for a ̸= 0, then X and Y will actually be the same line, since in this case ⃗x = t d⃗ + q⃗ = t d⃗ + (p ⃗ + ad⃗ ) = (t + a)d⃗ + ⃗p.

DEFINITION

For example, the following two vector equations represent the same line. ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 1 7 ⃗x = t ⎣1⎦ + ⎣0⎦ and ⃗x = t ⎣1⎦ + ⎣7⎦ . 1 0 1 7

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Vector Form of a Plane A plane P is written in vector form if it is expressed as ⃗x = t d⃗ 1 + sd⃗ 2 + ⃗p for some vectors d⃗ 1 and d⃗ 2 and point ⃗p. That is, P = {x⃗ : ⃗x = t d⃗ 1 + sd⃗ 2 + ⃗p for some t, s ∈ R}. The vectors d⃗ 1 and d⃗ 2 are called direction vectors for P .

Recall the intersecting lines A and B given in vector form by A ⏟ ⏟ ⎡ ⏞⏞ ⎤ ⎡ ⎤ 1 0 ⃗x = t ⎣2⎦ + ⎣0⎦ 3 1

B ⏟ ⏟ ⎡ ⏞⏞ ⎤ ⎡ ⎤ −1 −1 ⃗x = t ⎣ 1⎦ + ⎣ 1⎦ . 1 2

Let P the plane that contains the lines A and B. 13.1 Find two direction vectors for P .

Two possible answers are: ⎡ ⎤ 1 d⃗ 1 = ⎣2⎦ 3

⎡

and

⎤ −1 d⃗ 2 = ⎣ 1⎦ . 1

These are the two direction vectors we already know are in the plane—the ones from the two lines: Note that neither of these is a multiple of the other, so they really are two unique direction vectors in P . 13.2 Write P in vector form.

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 −1 0 ⃗x = t d⃗ 1 + sd⃗ 2 + ⃗p = t ⎣2⎦ + s ⎣ 1⎦ + ⎣0⎦ . 3 1 1 We already have two direction vectors, so we just needed a point on the plane. ⎡ ⎤ 0 We used the point ⃗p = ⎣0⎦ that we already know is on line A. 1 8 © Jason Siefken, 2015–2021

Apply vector form of a plane. The goal of this problem is to ■ Use direction vectors for lines given in vector form. ■ Think about planes in terms of vectors rather than equations. ■ Combine direction vectors in a plane to produce new direction vectors.

13.3 Describe how vector form of a plane relates to linear combinations.

The vector form of a plane says that a vector ⃗x is on the plane exactly when it is equal to some linear combination of d⃗ 1 and d⃗ 2 , plus ⃗p. Another way of saying the same thing is that the vector ⃗x is on the plane exactly when ⃗x − ⃗p is equal to some linear combination of d⃗ 1 and d⃗ 2 . 13.4 Write P in vector form using different direction vectors and a different point.

One possible answer: ⎡

⎤ ⎡ ⎤ ⎡ ⎤ −1 −7 −1 ⃗x = t ⎣−2⎦ + s ⎣ 7⎦ + ⎣ 1⎦ . −3 7 2 As with the equations of lines from before, we can use any non-zero scalar multiple ⎡ ⎤ −1 of either direction vector and get the same plane. We also used the point q⃗ = ⎣ 1⎦ 2 that we already knew is on line B.

14

Connect vector form and scalar form of a plane.

Let Q ⊆ R be a plane with equation x + y + z = 1. 3

The goal of this problem is to ■ Produce direction vectors for a plane defined by an equation.

14.1 Find three points in Q.

There are many choices here, of course. Three natural ones are: ⎡ ⎤ 1 ⃗p1 = ⎣0⎦ 0

⎡ ⎤ 0 ⃗p2 = ⎣1⎦ 0

⎡ ⎤ 0 ⃗p3 = ⎣0⎦ 1

■ Generalize the procedure for finding direction vectors that was used for lines.

14.2 Find two direction vectors for Q.

Now that we have three points on the plane, we can use the direction vectors joining any two pairs of them. For example: ⎡

⎤ 1 d⃗ 1 = ⃗p1 − ⃗p2 = ⎣−1⎦ 0

⎡

⎤ 1 d⃗ 2 = ⃗p1 − ⃗p3 = ⎣ 0⎦ . −1

14.3 Write Q in vector form.

Using the point ⃗p1 from above, one possible answer is: ⎡

⎤ ⎡ ⎤ ⎡ ⎤ 1 1 1 ⃗x = t d⃗ 1 + sd⃗ 2 + ⃗p1 = t ⎣−1⎦ + s ⎣ 0⎦ + ⎣0⎦ . 0 −1 0

DEFINITION

Span Span The span of a set of vectors V is the set of all linear combinations of vectors in V . That is, Apply the definition of span. The goal of this problem is to span V = {v⃗ : ⃗v = α1 ⃗v 1 +α2 ⃗v 2 +· · ·+αn ⃗v n for some ⃗v 1 , ⃗v 2 , . . . , ⃗v n ∈ V and scalars α1 , α2 , . . . , α n }.

■ Practice applying a new definition in a familiar context (R2 ). ■ Recognize spans as lines and planes through the origin.

⃗}. Additionally, we define span{} = {0 9

© Jason Siefken, 2015–2021

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• ˜ • ˜ • ˜ 1 1 2 Let ⃗v 1 = , ⃗v 2 = , and ⃗v 3 = . 1 −1 2 15.1 Draw span{v ⃗1 }. 15.2 Draw span{v ⃗2 }.

span{v⃗2 }

span{v⃗1 }

15.3 Describe span{v ⃗1 , ⃗v 2 }.

span{v⃗1 , ⃗v 2 } = R2 • ˜ x We can see this since for any ∈ R2 , y • ˜ • ˜ • ˜‹ • ˜ • ˜‹ y 1 x+y x−y x x 1 1 1 = + + − = ⃗v 1 + ⃗v 2 y −1 −1 2 1 2 1 2 2 15.4 Describe span{v ⃗1 , ⃗v 3 }. span{v⃗1 , ⃗v 3 } = span{v⃗1 }, a line through the origin with direction

vector ⃗v 1 .

15.5 Describe span{v ⃗1 , ⃗v 2 , ⃗v 3 }. span{v⃗1 , ⃗v 2 , ⃗v 3 } = span{v⃗1 , ⃗v 2 } = R2

16

Let ℓ1 ⊆ R2 be the line with equation x − y = 0 and ℓ2 ⊆ R2 the line with equation x − y = 4. 16.1 If possible, describe ℓ1 as a span. Otherwise explain why it’s not possible.

§• ˜ª • ˜ 1 x ℓ1 = span , since ∈ ℓ1 if and only if x = y, which in turn is true if and 1 y • ˜ 1 only if it is a scalar multiple of . 1 16.2 If possible, describe ℓ2 as a span. Otherwise explain why it’s not possible.

• ˜ 0 is an element of the span of any set of vectors, since we 0 • ˜ 0 can use all zeroes as the scalars in a linear combination, but ∈ / ℓ2 . 0 This is not possible.

16.3 Does the expression span(ℓ1 ) make sense? If so, what is it? How about span(ℓ2 )?

Both of these expressions do make sense. One can compute the span of any set of vectors, and these lines are just special set of points in R2 which we are already used to thinking of as vectors. • ˜ 1 span(ℓ1 ) = ℓ1 , since all of the vectors on the line ℓ1 are already multiples of , 1 as we discovered earlier. • ˜ • ˜ 4 0 span(ℓ2 ) equals all of R2 . It’s easy to see that the vectors v = and w = 0 −4 are both on ℓ2 , and the span of these two vectors alone is all of R2 . 10

© Jason Siefken, 2015–2021

Connect geometric figures to spans. The goal of this problem is to ■ Identify a relationship between lines and spans. ■ Describe a line through the origin as a span. ■ Identify when a line cannot be described as a span. ■ Apply the definition of span X even when X is infinite.

DEF

17

Set Addition If A and B are sets of vectors, then the set sum of A and B, denoted A + B, is A + B = {x⃗ : ⃗x = a ⃗ + ⃗b for some a ⃗ ∈ A and ⃗b ∈ B}.

Let A =

§• ˜ª §• ˜ • ˜ª §• ˜ª 1 1 1 1 ,B= , , and ℓ = span . 2 1 −1 −1

17.1 Draw A, B, and A + B in the same picture.

Describing geometry using sets. The goal of this problem is to ■ Practice applying a new definition in a familiar context (R2 ). ■ Gain an intuitive understanding of set addition. ■ Describe lines that don’t pass ⃗ using a combination of set through 0 addition and spans.

A A+ B

B

17.2 Is A + B the same as B + A?

Yes. Since A and B are such small sets we could just compute all the vectors in A + B and B + A and see that they’re equal. However, we know that real numbers can be added up in any order, and the coordinates of an element of A + B or B + A are simply sums of the corresponding coordinates of elements of A and B. 17.3 Draw ℓ + A.

A

ℓ+A ℓ

˜ • ˜ 1 1 + . Can ℓ2 be described using −1 2 only a span? What about using a span and set addition?

17.4 Consider the line ℓ2 given in vector form by ⃗ x=t

•

ℓ2 cannot be described using only a span, for the same • ˜ reason as the line ℓ2 in 0 Problem 16.2 couldn’t be. We know that the origin must be an element of 0 any span, but it is not a point on ℓ2 . ℓ2 can be described as a span plus a set addition though. Specifically, ℓ2 = ℓ + A.

11

© Jason Siefken, 2015–2021

The Magic Carpet, Getting Back Home 18

Suppose you are now in a three-dimensional world for the carpet ride problem, and you have three modes of transportation: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 6 4 ⃗v 1 = ⎣1⎦ ⃗v 2 = ⎣3⎦ ⃗v 3 = ⎣1⎦ 1 8 6 You are only allowed to use each mode of transportation once (in the forward or backward direction) for a fixed amount of time (c1 on ⃗v 1 , c2 on ⃗v 2 , c3 on ⃗v 3 ).

1. Find the amounts of time on each mode of transportation (c1 , c2 , and c3 , respectively) needed to go on a journey that starts and ends at home or explain why it is not possible to do so. 2. Is there more than one way to make a journey that meets the requirements described above? (In other words, are there different combinations of times you can spend on the modes of transportation so that you can get back home?) If so, how? 3. Is there anywhere in this 3D world that Gauss could hide from you? If so, where? If not, why not? ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ 6 4 ⎬ ⎨ 1 4. What is span ⎣1⎦ , ⎣3⎦ , ⎣1⎦ ? ⎩ 1 8 6 ⎭

12

© IOLA Team iola. math. vt. edu

Span in higher dimensions. The goal of this problem is to ■ Examine subtleties that exist in three dimensions that are missing in two dimensions. ■ Apply linear algebra tools to answer open-ended questions.

DEFINITION

Linearly Dependent & Independent (Geometric) We say the vectors ⃗v 1 , ⃗v 2 , . . . , ⃗v n are linearly dependent if for at least one i, ⃗v i ∈ span{v⃗1 , ⃗v 2 , . . . , ⃗v i−1 , ⃗v i+1 , . . . , ⃗v n }. Otherwise, they are called linearly independent.

Apply the (geometric) definition of linear independence/dependence.

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 1 Let u ⃗ = ⎣0⎦, ⃗v = ⎣1⎦, and w ⃗ = ⎣1⎦. 0 0 0

19

Geometric definition of linear independence/dependence.

The goal of this problem is to ■ Develop a mental picture linking linear dependence and “redundant” vectors.

19.1 Describe span{u ⃗, ⃗v , w ⃗ }.

The x y-plane in R3 . That is, the set of all vectors in R3 with z-coordinate equal to zero.

■ Practice applying a new definition. ■ Find multiple linearly independent subsets of a linearly dependent set.

19.2 Is {u ⃗, ⃗v , w ⃗ } linearly independent? Why or why not?

No. w ⃗ =u ⃗ + ⃗v , and so w ⃗ ∈ span{u ⃗, ⃗v }. Let X = {u ⃗, ⃗v , w ⃗ }. 19.3 Give a subset Y ⊆ X so that span Y = span X and Y is linearly independent.

Y = {u ⃗, ⃗v } is one example that works. 19.4 Give a subset Z ⊆ X so that span Z = span X and Z is linearly independent and Z ̸= Y .

DEF

Z = {u ⃗, w ⃗ } and Z = {v⃗, w ⃗ } both have the same span as Y above.

Trivial Linear Combination The linear combination α1 ⃗v 1 + · · · + αn ⃗v n is called trivial if α1 = · · · = αn = 0. If at least one αi ̸= 0, the linear combination is called non-trivial. Link trivial/non-trivial linear combinations to linear independence/dependence.

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 1 Recall u ⃗ = ⎣0⎦, ⃗v = ⎣1⎦, and w ⃗ = ⎣ 1⎦. 0 0 0

20

20.1 Consider the linearly dependent set {u ⃗, ⃗v , w ⃗ } (where u ⃗, ⃗v , w ⃗ are defined as above). Can

⃗ as a non-trivial linear combination of vectors in this set? 0 ⃗=u you write 0 ⃗ + ⃗v − w ⃗.

⃗ as a non-trivial linear 20.2 Consider the linearly independent set {u ⃗, ⃗v }. Can you write 0 combination of vectors in this set? No. Suppose ⃗ a1 u ⃗ + a2 ⃗v = 0 was a non-trivial linear combination. Then at least one of a1 or a2 is non-zero. If a1 is non-zero, then a u ⃗ = − 2 ⃗v a1 and so u ⃗ ∈ span{v⃗}. If a2 is non-zero, then ⃗v = −

a1 u ⃗. a2

and so ⃗v ∈ span{u ⃗}. In either case, {u ⃗, ⃗v } would be linearly dependent.

DEF

We now have an equivalent definition of linear dependence.

Linearly Dependent & Independent (Algebraic) The vectors ⃗v 1 , ⃗v 2 , . . . , ⃗v n are linearly dependent if there is a non-trivial linear combination of ⃗v 1 , . . . , ⃗v n that equals the zero vector. Otherwise they are linearly independent. 13

© Jason Siefken, 2015–2021

Link algebraic and geometric definitions of linear independence/dependence. The goal of this problem is to ■ Understand how the algebraic and geometric definitions of linear independence/dependence relate. ■ Practice writing mathematical arguments.

21

21.1 Explain how the geometric definition of linear dependence (original) implies this alge-

braic one (new). Suppose that ⃗v 1 , ⃗v 2 , . . . , ⃗v n are linearly dependent according to the geometric definition. Fix i so that ⃗v i ∈ span{v⃗1 , . . . , ⃗v i−1 , ⃗v i+1 , . . . , ⃗v n }. By the definition of span, we know that ⃗v i = β1 ⃗v 1 + · · · + βi−1 ⃗v i−1 + βi+1 ⃗v i+1 + · · · + βn ⃗v n . Thus

⃗ = −v⃗ i + β1 ⃗v 1 + · · · + βi−1 ⃗v i−1 + βi+1 ⃗v i+1 + · · · + βn ⃗v n , 0

and this is a non-trivial linear combination since the coefficient of ⃗v i is −1 ̸= 0. 21.2 Explain how this algebraic definition of linear dependence (new) implies the geometric

one (original). Suppose the vectors ⃗v 1 , ⃗v 2 , . . . , ⃗v n is linearly dependent in this new sense. That means there are scalars a1 , a2 , . . . , an , at least one of which is non-zero, such that a1 ⃗v 1 + · · · + an ⃗v n = 0. Suppose ai ̸= 0. Then ⃗v i =

−a1 v1 ai ⃗

+ ··· +

−ai−1 v i−1 ai ⃗

+

−ai+1 v i+1 ai ⃗

+ ··· +

−an vn. ai ⃗

This means ⃗v i ∈ span{v⃗1 , . . . , ⃗v i−1 , ⃗v i+1 , . . . , ⃗v n }, which is precisely the geometric definition of linear dependence. Since we have geometric def =⇒ algebraic def, and algebraic def =⇒ geometric def ( =⇒ should be read aloud as ‘implies’), the two definitions are equivalent (which we write as algebraic def ⇐⇒ geometric def).

22

Linear dependence and infinite solutions.

Suppose for some unknown u ⃗, ⃗v , w ⃗ , and a ⃗, a ⃗ = 3u ⃗ + 2v⃗ + w ⃗

and

a ⃗ = 2u ⃗ + ⃗v − w ⃗.

■ Connect linear independence with unique solutions.

22.1 Could the set {u ⃗, ⃗v , w ⃗ } be linearly independent?

No. If both equations are true, they would combine to show 3u ⃗ + 2v⃗ + w ⃗ = 2u ⃗ + ⃗v − w ⃗. Collecting all the terms on the left side, we get: ⃗, u ⃗ + ⃗v + 2w ⃗ =0 which is a non-trivial linear combination of vectors in the given set equalling the zero vector. Suppose that a ⃗=u ⃗ + 6r⃗ − ⃗s is the only way to write a ⃗ using u ⃗, ⃗r , ⃗s. 22.2 Is {u ⃗, ⃗r , ⃗s} linearly independent?

Yes. If it were not, there would exist scalars a1 , a2 , a3 , not all of which are zero, such that: ⃗. a1 u ⃗ + a2 ⃗r + a3⃗s = 0 But then u ⃗ + 6r⃗ − ⃗s + (a1 u ⃗ + a2 ⃗r + a3⃗s) would be another way to write a ⃗ using only the same three vectors. 22.3 Is {u ⃗, ⃗r } linearly independent?

14

The goal of this problem is to ■ Connect linear dependence with infinite solutions.

© Jason Siefken, 2015–2021

Yes. If it were not, we would necessarily have u ⃗ = β ⃗r for some scalar β. But then (β + 6)r⃗ − ⃗s would be another way to write a ⃗ using only the same three vectors. 22.4 Is {u ⃗, ⃗v , w ⃗ , ⃗r } linearly independent?

⃗, and so u ⃗ is a No. We know from earlier that u ⃗ + ⃗v + 2w ⃗ =0 ⃗ + ⃗v + 2w ⃗ + 0r⃗ = 0 non-trivial linear combination of the vectors in this set that equals the zero vector.

15

© Jason Siefken, 2015–2021

Linear Independence and Dependence, Creating Examples 23 1. Fill in the following chart keeping track of the strategies you used to generate examples. Linearly independent

Linearly dependent

2

A set of 2 vectors in R A set of 3 vectors in R2 A set of 2 vectors in R3 A set of 3 vectors in R3 A set of 4 vectors in R3 2. Write at least two generalizations that can be made from these examples and the strategies you used to create them.

16

© IOLA Team iola. math. vt. edu

Dot Product Norm

DEFINITION

DEFINITION

⎡ ⎤ v1 . ⎣ The norm of a vector ⃗v = .. ⎦ is the length/magnitude of ⃗v . It is written ∥v⃗∥ and vn q can be computed from the Pythagorean formula ∥v⃗∥ = v12 + · · · + vn2 .

Dot Product ⎡ ⎤ ⎡ ⎤ a1 b1 a ⎢ 2⎥ ⎢b ⎥ ⎥ and ⃗b = ⎢ .2 ⎥ are two vectors in n-dimensional space, then the dot If a ⃗=⎢ . ⎣.⎦ ⎣.⎦ . . an bn product of a ⃗ an ⃗b is a ⃗ · ⃗b = a1 b1 + a2 b2 + · · · + an bn . Equivalently, the dot product is defined by the geometric formula ⃗ ∥ cos θ a ⃗ · ⃗b = ∥a ⃗ ∥∥b where θ is the angle between a ⃗ and ⃗b. Practicing dot products.

⎡ ⎤ • ˜ • ˜ 1 1 ⃗ 3 Let a ⃗= , b= , and u ⃗ = ⎣ 2⎦. 1 1 1

24

24.1

The goal of this problem is to ■ Use both the algebraic and geometric definitions of the dot product as appropriate to compute dot products.

(a) Draw a picture of a ⃗ and ⃗b.

■ Gain an intuition that positive dot product means “pointing in similar directions”, negative dot product means “pointing in opposite directions”, and zero dot product means “pointing in orthogonal directions”.

⃗b

a ⃗

(b) Compute a ⃗ · ⃗b. a ⃗ · ⃗b = (1)(3) + (1)(1) = 4. ⃗ ∥ and use your knowledge of the multiple ways to compute the dot (c) Find ∥a ⃗ ∥ and ∥b product to find θ , the angle between a ⃗ and ⃗b. Label θ on your picture. p p ⎷ ⎷ 2 2 ⃗ ∥ = (3)2 + (1)2 = 10. ∥a ⃗ ∥ = (1) + (1) = 2 and ∥b Using the two definitions of the dot product we have: ⃗ ∥ cos θ a ⃗ · ⃗b = ∥a ⃗ ∥∥b ⎷ ⎷ =⇒ 4 = ( 2)( 10) cos θ  ‹ 2 =⇒ θ = arccos ⎷ 5 24.2 Draw the graph of cos and identify which angles make cos negative, zero, or positive.

π 2

π

3π 2

2π


Cosine is positive for angles in the interval − π2 , π2 , as well as all shifts of this interval by a multiple of 2π in either direction.
cos is positive for angles in the interval π2 , 3π 2 , as well as all shifts of this interval by a multiple of 2π in either direction. 17

© Jason Siefken, 2015–2021

24.3 Draw a new picture of a ⃗ and ⃗b and on that picture draw

(a) a vector ⃗c where ⃗c · a ⃗ is negative. (b) a vector d⃗ where d⃗ · a ⃗ = 0 and d⃗ · ⃗b < 0. (c) a vector ⃗e where ⃗e · a ⃗ = 0 and ⃗e · ⃗b > 0. (d) Could you find a vector ⃗f where ⃗f · a ⃗ = 0 and ⃗f · ⃗b = 0? Explain why or why not.

a ⃗

d⃗

⃗b

⃗c ⃗e

• ˜ ⃗ is the only possibility. For any vector ⃗f = x , we can compute: (d) ⃗f = 0 y ⃗f · a ⃗=x+y

and

⃗f · ⃗b = 3x + y.

If these both equal zero, the first equation says that y = −x, and in turn the second one says x = 0 (and so y = 0 as well). 24.4 Recall the vector u ⃗ whose coordinates are given at the beginning of this problem.

(a) Write down a vector ⃗v so that the angle between u ⃗ and ⃗v is π/2. (Hint, how does this relate to the dot product?) ⎡ ⎤ 1 ⃗v = ⎣ 1⎦ is one such vector. −3 Since cos(π/2) = 0, from the second definition of the dot product above we know we are looking for a ⃗v such that u ⃗ · ⃗v = 0. Using the first definition of the dot product, we can see that the ⃗v given above is one possibility. (b) Write down another vector w ⃗ (in a different direction from ⃗v ) so that the angle between w ⃗ and u ⃗ is π/2. ⎡ ⎤ −1 w ⃗ = ⎣ 1⎦ is a possible answer. −1 u ⃗·w ⃗ = 0, and w ⃗ is clearly not parallel to ⃗v from above. (c) Can you write down other vectors different than both ⃗v and w ⃗ that still form an angle of π/2 with u ⃗? How many such vectors are there? ⎡ ⎤ 0 Yes. ⎣ 2⎦ is one possibility. −4 There are actually infinitely many such vectors; any linear combination of w ⃗ and ⃗v will work. To see this, note that any such vector ⃗x is of the form ⎡

⎤ ⎡ ⎤ ⎡ ⎤ 1 −1 t −s ⃗x = t ⎣ 1⎦ + s ⎣ 1⎦ = ⎣ t + s ⎦ , −3 −1 −3t − s 18 © Jason Siefken, 2015–2021

for scalars t and s. We can then compute u ⃗ · ⃗x = (1)(t − s) + (2)(t + s) + (1)(−3t − s) = 0,

For a vector ⃗v ∈ Rn , the formula ∥v⃗∥ =

p

⃗v · ⃗v

DEF

always holds.

Distance The distance between two vectors u ⃗ and ⃗v is ∥u ⃗ − ⃗v ∥.

DEF

THEOREM

and so any such vector ⃗x forms an angle of π/2 with u ⃗.

Unit Vector A vector ⃗v is called a unit vector if ∥v⃗∥ = 1. ⎡ ⎤ ⎡ ⎤ 1 1 Let u ⃗ = ⎣2⎦ and ⃗v = ⎣1⎦. 1 3

25

Practice using norms. The goal of this problem is to ■ Practice finding the distance between two vectors.

25.1 Find the distance between u ⃗ and ⃗v .

⎡

■ Produce a unit vector pointing in the same direction as another vector.

⎤

0 ⎷ ⎣ 1⎦, and so ∥u u ⃗ − ⃗v = ⃗ − ⃗v ∥ = 5. −2

■ Intuitively apply the triangle inequal⃗ ∥. ity: ∥a⃗ + ⃗b∥ ≤ ∥a⃗ ∥ + ∥b

25.2 Find a unit vector in the direction of u ⃗.

⎤ ⎷1 6 ⎣ ⎷2 ⎦. 6 ⎷1 6 ⎡

⎷1 u ⃗ 6

=

∥u ⃗∥ = be 1.

⎷ 6, and so if we multiply u ⃗ by

⎷1 , 6

the length of the resulting vector will

25.3 Does there exist a unit vector ⃗ x that is distance 1 from u ⃗?

⎷ No. ∥u ⃗∥ = ⎷6, and so the shortest length that a vector whose distance from u ⃗ is 1 can have is 6 − 1, which is greater than 1.

25.4 Suppose ⃗ y is a unit vector and the distance between ⃗y and u ⃗ is 2. What is the angle

between ⃗y and u ⃗?

The angle between u ⃗ and ⃗y is arccos

€

3 ⎷ 2 6

Š

.

By assumption, 2 = ∥u ⃗ − ⃗y ∥, and so 4 = ∥u ⃗ − ⃗y ∥2 = (u ⃗ − ⃗y ) · (u ⃗ − ⃗y ) =u ⃗·u ⃗ − 2(u ⃗ · ⃗y ) + ⃗y · ⃗y = ∥u ⃗∥2 − 2u ⃗ · ⃗y + ∥ ⃗y ∥2 = 6 − 2u ⃗ · ⃗y + 1. Then we rearrange to find that u ⃗ · ⃗y = 32 . Using this in the second definition of the dot product, we see: 3 €p Š = 6 (1) cos θ , 2

DEF

where θ is the angle between u ⃗ and ⃗y .

Orthogonal Two vectors u ⃗ and ⃗v are orthogonal to each other if u ⃗ · ⃗v = 0. The word orthogonal is synonymous with the word perpendicular. 19

© Jason Siefken, 2015–2021

Apply the definition of orthogonal. The goal of this problem is to ■ Gain an intuitive understanding of orthogonal vectors. ■ Produce orthogonal vectors via guessand-check. ■ Apply the Pythagorean theorem to orthogonal vectors to find lengths.

26

26.1 Find two vectors orthogonal to a ⃗=

•

˜ 1 . Can you find two such vectors that are not −3

parallel? Two such vectors are

• ˜ • ˜ 3 −6 and . 1 −2

• ˜ x ⃗ It is impossible for two non-parallel vectors to both be orthogonal to a ⃗ . If b = y is orthogonal to a ⃗ , then we must have that x• −˜3 y = 0, or in other words that 3 x = 3 y. Any ⃗b satisfying this is a multiple of . 1 ⎡ ⎤ 1 26.2 Find two vectors orthogonal to ⃗ b = ⎣−3⎦. Can you find two such vectors that are not 4 parallel? ⎡ ⎤ ⎡ ⎤ 7 2 Two such vectors are ⎣ 1⎦ and ⎣2⎦. −1 1 These two vectors are not parallel. 26.3 Suppose ⃗ x and ⃗y are orthogonal to each other and ∥x⃗ ∥ = 5 and ∥ ⃗y ∥ = 3. What is the

distance between ⃗x and ⃗y ?

The distance between them must be

⎷ 34.

One way to see this is with Pythagoras’ theorem. Two perpendicular line segments of lengths 3 and 5 form the ⎷two shorter ⎷sides of a right angle triangle, and so the length of the third side is 52 + 32 = 34. An equivalent way to see this is to use what we know about dot products to calculate ∥x⃗ − ⃗y ∥ as follows: ∥x⃗ − ⃗y ∥ =

Æ

(x⃗ − ⃗y ) · (x⃗ − ⃗y ) =

Æ

∥x⃗ ∥2 − 2(x⃗ · ⃗y ) + ∥ ⃗y ∥2 =

Æ

52 + 2(0) + 32 ,

where in the last step we’ve used the fact that ⃗x and ⃗y are orthogonal, so ⃗x · ⃗y = 0. Generate lines using orthogonality.

27

• ˜ 2 27.1 Draw u ⃗= and all vectors orthogonal to it. Call this set A. 3

The goal of this problem is to ■ Visually see how the set of all vectors orthogonal to a given vector forms a line. ■ Given a line defined as the set of all vectors orthogonal to a given vector, express the line using an equation or span.

u ⃗

A

• ˜ x and ⃗x is orthogonal to u ⃗, what is ⃗x · u ⃗? y orthogonality.

27.2 If ⃗ x =

⃗x · u ⃗ = 0, by the definition of

27.3 Expand the dot product u ⃗ · ⃗x to get an equation for A.

20

© Jason Siefken, 2015–2021

˜ 3 A is the line with vector equation ⃗x = t . −2 • ˜ x If ⃗x = ∈ A, then ⃗x · u ⃗ = 2x + 3 y = 0. y §• ˜ª 3 27.4 If possible, express A as a span. A = span . −2 DEF

•

28

Normal Vector A normal vector to a line (or plane or hyperplane) is a non-zero vector that is orthogonal to all direction vectors for the line (or plane or hyperplane).

Express lines in normal form.

• ˜ • ˜ 1 1 Let d⃗ = and ⃗p = , and define the lines 2 −1 ℓ1 = span{d⃗ }

ℓ2 = span{d⃗ } + {p ⃗ }.

and

28.1 Find a vector n ⃗ that is a normal vector for both ℓ1 and ℓ2 .

n ⃗=

•

The goal of this problem is to ■ Express lines, including lines that ⃗, in normal form. don’t pass through 0 ■ See the q⃗ in n ⃗ · (x⃗ − q⃗) = 0 is similar to the ⃗p in the vector form ⃗x = t d⃗ + ⃗p in that it accommodates lines that ⃗. don’t pass through 0 ■ Use the dot product to represent a line compactly.

˜ 2 is one possibility. −1

This vector is orthogonal to d⃗ , which is a direction vector for both lines. 28.2 Let ⃗ v ∈ ℓ1 and u ⃗ ∈ ℓ2 . What is n ⃗ · ⃗v ? What about n ⃗ · (u ⃗ − ⃗p)? Explain using a picture.

n ⃗ · ⃗v = n ⃗ · (u ⃗ − ⃗p) = 0. This is because any ⃗v ∈ ℓ1 is a multiple of d⃗ , which is orthogonal to n ⃗ . Similarly, for any u ⃗ ∈ ℓ2 , the vector u ⃗ − ⃗p is a direction vector for ℓ2 , and so it is orthogonal to n ⃗. n ⃗·u ⃗ = 3, since any such u ⃗ is of the form u ⃗ = ⃗p + t d⃗ for some scalar t, and so n ⃗·u ⃗=n ⃗ · (p ⃗ + t d⃗ ) = n ⃗ · ⃗p + t(n ⃗ · d⃗ ) = 3 + t(0) = 3. 28.3 A line is expressed in normal form if it is represented by an equation of the form n ⃗ ·(x⃗ −q⃗) =

0 for some n ⃗ and q⃗. Express ℓ1 and ℓ2 in normal form. • ˜ 2 A normal form of ℓ1 is · ⃗x = 0. −1 • ˜  • ˜‹ 2 1 A normal form of ℓ2 is · ⃗x − = 0. In the previous part we saw that −1 −1 n ⃗ · ⃗x = n ⃗ · ⃗p for all ⃗x ∈ ℓ2 , or in other words n ⃗ · (x⃗ − ⃗p) = 0. • ˜ 2 28.4 Some textbooks would claim that ℓ2 could be expressed in normal form as · ⃗x = 3. −1 How does this relate to the n ⃗ · (x⃗ − ⃗p) = 0 normal form? Where does the 3 come from? • ˜ 2 Let n ⃗= and let ⃗x ∈ ℓ2 . From the previous part, we know −1 0=n ⃗ · (x⃗ − ⃗p) = n ⃗ · ⃗x − n ⃗ · ⃗p = n ⃗ · ⃗x − 3. Therefore n ⃗ · ⃗x = 3.

29

⎡ ⎤ 1 Let n ⃗ = ⎣1⎦. 1

Planes in normal form.

29.1 Use set-builder notation to write down the set, X , of all vectors orthogonal to n ⃗ . Describe

this set geometrically.  X = ⃗x ∈ R3 : ⃗x · n ⃗=0 . Geometrically, this is a plane through the origin and orthogonal to n ⃗. 21

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The goal of this problem is to ■ Observe that the set of all vectors orthogonal to another in R3 is a plane. ■ Translate descriptions of sets into precise mathematical statements using set-builder notation. ■ Express a plane in multiple ways.

29.2 Describe X using an equation. x + y + z = 0.

⎧⎡ ⎤ ⎡ ⎤⎫ 1 1 ⎬ ⎨ ⎣ ⎦ ⎣ −1 0⎦ is one way to do this. 29.3 Describe X as a span. X = span , ⎩ 0 −1 ⎭

DEF

Projections Projection Let X be a set. The projection of the vector ⃗v onto X , written projX ⃗v , is the closest point in X to ⃗v . Apply the definition of projection.

30

The goal of this problem is to ■ Use the definition of projection to compute projections onto finite sets and lines.

• ˜ • ˜ • ˜ 1 ⃗ 4 2 Let a ⃗= , b= , ⃗v = and ℓ = span{a ⃗ }. 0 0 2 30.1 Draw a ⃗ , ⃗b, and ⃗v in the same picture.

■ Pick an appropriate representation of a line to solve a projection problem.

⃗v a ⃗

⃗b

30.2 Find proj{b⃗ } ⃗ v , proj{a⃗,b⃗ } ⃗v .

⃗ }, it must be the closest point to proj{b⃗ } ⃗v = ⃗b. Since there is only one point in {b ⃗v . ⎷ ⎷ proj{a⃗,b⃗ } ⃗v = a ⃗ . We can simply compute ∥v⃗ − a ⃗ ∥ = 5 and ∥v⃗ − ⃗b∥ = 8, so a ⃗ is closer to ⃗v . b 30.3 Find projℓ ⃗ v . (Recall that a quadratic at 2 + bt + c has a minimum at t = − 2a ).

• ˜ 2 projℓ ⃗v = 2a ⃗= . 0 Any point in ℓ is of the form t a ⃗ for some scalar t. The distance between such a point and ⃗v is ∥v⃗ − t a ⃗∥ =

Æ

∥v∥2 − 2t(v⃗ · a ⃗ ) + t 2 ∥a∥2 =

p

8 − 4t + t 2

The quadratic inside the square root has a minimum at t = 2, so 2a ⃗ is the closest point in the line to ⃗v . Project onto lines.

30.4 Is ⃗ v − projℓ ⃗v a normal vector for ℓ? Why or why not?

Yes. By the previous part, ⃗v − projℓ ⃗v = ⃗v − 2a ⃗=

• ˜ 0 . This vector is orthogonal to a ⃗, 2

and therefore to ℓ. 22

© Jason Siefken, 2015–2021

The goal of this problem is to ■ Use orthogonality to compute the projection onto a line. ■ Project onto lines that don’t pass ⃗. through 0

• ˜ • ˜ • ˜ 1 1 1 Let K be the line given in vector form by ⃗x = t + and let ⃗c = . 2 0 3

31

31.1 Make a sketch with ⃗c , K, and projK ⃗c (you don’t need to compute projK ⃗c exactly).

⃗c

projK ⃗c

K

31.2 What should (c⃗ − projK ⃗c ) ·

(c⃗ − projK ⃗c ) ·

• ˜ 1 be? Explain. 2

• ˜ 1 = 0. 2

From our picture we can see that c − projK ⃗c is perpendicular to the line K, and so the dot product of this vector with any direction vector for K should be zero. 31.3 Use your formula from the previous part to find projK ⃗c without computing any distances.

˜ 11 12 • ˜ x If projK ⃗c = , the formula from the previous part tells us y projK ⃗c =

1 5

•

• ˜ • ˜‹ • ˜ 1 x 1 − · = 1 − x + 6 − 2y = 0 3 y 2

⇐⇒

x + 2y = 7

So we need a point on K that satisfies this equation. In other words, we need (t + 1) + 2(2t) = 7 The point on K for this value of t is

1 5

=⇒

t=

6 . 5

• ˜ 11 . 12

Vector Components ⃗ be vectors. The vector component of u Let u ⃗ and ⃗v ̸= 0 ⃗ in the ⃗v direction, written v comp⃗v u ⃗, is the vector in the direction of ⃗v so that u ⃗ − v comp⃗v u ⃗ is orthogonal to ⃗v . DEFINITION

u ⃗

u ⃗ − v comp⃗v u ⃗ ⃗v

Component of a vector in the direction of another. The goal of this problem is to ■ Read and apply a new definition. ■ Use orthogonality to obtain a formula for components in terms of dot products.

v comp⃗v u ⃗ 23

© Jason Siefken, 2015–2021

32

Let a ⃗ , ⃗b ∈ R3 be unknown vectors. 32.1 List two conditions that v comp ⃗b a ⃗ must satisfy.

v comp ⃗b a ⃗ must be a scalar multiple of ⃗b. a ⃗ − v comp ⃗b a ⃗ must be orthogonal to ⃗b, or in other words (a ⃗ − v comp ⃗b a ⃗ ) · ⃗b = 0. 32.2 Find a formula for v comp ⃗b a ⃗.

v comp ⃗b a ⃗=

a ⃗ · ⃗b ⃗ b. ⃗b · ⃗b

From the previous part, we should have v comp ⃗b a ⃗ = t ⃗b for some scalar t, and ⃗ (a ⃗ − v comp ⃗b a ⃗ ) · b = 0. Combining these, we get: ⃗ · ⃗b). 0 = (a ⃗ − t ⃗b) · ⃗b = a ⃗ · ⃗b − t ⃗b · ⃗b = a ⃗ · ⃗b − t(b Solving for t, we get t =

33

a ⃗ · ⃗b . ⃗b · ⃗b

Relate components and projections. The goal of this problem is to ■ Find a connection between components and projections onto spans.

• ˜ • ˜ 3 1 Let d⃗ = and u ⃗= . 3 2

■ Recognize v comp−u⃗ ⃗v .

⃗, u 33.1 Draw d ⃗, span{d⃗ }, and projspan{d⃗ } u ⃗ in the same picture.

d⃗

span{d⃗ }

u ⃗ projspan{d⃗ } u ⃗

33.2 How do projspan{d⃗ } u ⃗ and v compd⃗ u ⃗ relate? They are equal. 33.3 Compute projspan{d⃗ } u ⃗ and v compd⃗ u ⃗.

Using our formula from the previous problem u ⃗ · d⃗ ⃗ projspan{d⃗ } u ⃗ = v compd⃗ u ⃗= d= ∥d⃗ ∥2

9 ⃗ 18 d

=

1 2

• ˜ 3 . 3

33.4 Compute v comp−d⃗ u ⃗. Is this the same as or different from v compd⃗ u ⃗? Explain.

u ⃗ · (−d⃗ ) ⃗ v comp−d⃗ u ⃗= (−d ) = ∥−d⃗ ∥2

−9 ⃗ 18 (−d )

=

1 2

• ˜ 3 = v compd⃗ u ⃗. 3

We expect them to be equal since d⃗ and −d⃗ are in the same direction as one another. 24

© Jason Siefken, 2015–2021

that

v compu⃗ ⃗v

=

DEFINITION

Subspaces and Bases Subspace A non-empty subset V ⊆ Rn is called a subspace if for all u ⃗, ⃗v ∈ V and all scalars k we have (i) u ⃗ + ⃗v ∈ V ; and (ii) ku ⃗ ∈ V. Subspaces give a mathematically precise definition of a “flat space through the origin.”

34

Visualizing subspaces.

For each set, draw it and explain whether or not it is a subspace of R2 . § • ˜ ª a 2 34.1 A = ⃗ x ∈ R : ⃗x = for some a ∈ Z . 0

The goal of this problem is to ■ Read and apply the definition of subspace. ■ Identify from a picture whether or not a set is a subspace.

• ˜ 1 A is not a subspace, since for example ∈ A but 0

A

34.2 B =

§

⃗x ∈ R2 : ⃗x ̸=

• ˜ 1 1 / A. 2 0 ∈

■ Write formal arguments showing whether or not certain sets are subspaces.

• ˜ª 0 . 0

• ˜ • ˜ 1 −1 B is not a subspace, since for example and are both in B, but their sum 0 0 • ˜ 0 is which is not in B. 0 § • ˜ ª 0 34.3 C = ⃗ x ∈ R2 : ⃗x = for some t ∈ R . t C is a subspace. • ˜ • ˜ 0 0 (i) Let u ⃗, ⃗v ∈ C. Then u ⃗= and ⃗v = for some s, t ∈ R. t s • ˜ 0 But then u ⃗ + ⃗v = ∈ C. s+t • ˜ • ˜ 0 0 (ii) Let u ⃗= ∈ C. For any scalar α we have αu ⃗= ∈ C. t αt § • ˜ • ˜ ª 0 1 34.4 D = ⃗ x ∈ R2 : ⃗x = + for some t ∈ R . t 1

D is not a subspace, since for example

• ˜ • ˜ • ˜ 1 1 0 ∈ D, but 0 = ∈ / 1 1 0

D.

§

• ˜ • ˜ ª 0 t 34.5 E = ⃗ x ∈ R : ⃗x = or ⃗x = for some t ∈ R . t 0 2

• ˜ • ˜ 1 0 E is not a subspace, since for example and are both in E, 0 1 • ˜ 1 but their sum is which is not in E. 1

25

© Jason Siefken, 2015–2021

§

• ˜ ª 3 34.6 F = ⃗ x ∈ R : ⃗x = t for some t ∈ R . 1 2

F is a subspace. • ˜ • ˜ 3 3 (i) Let u ⃗, ⃗v ∈ F . Then u ⃗=t and ⃗v = s for some s, t ∈ R. 1 1 • ˜ 3 But then u ⃗ + ⃗v = (s + t) ∈ F. 1 • ˜ • ˜ 3 3 (ii) Let u ⃗=t ∈ F . For any scalar α we have αu ⃗ = (αt) ∈ 1 1 F. 34.7 G = span

§• ˜ª 1 . 1

G is a subspace. By definition of a ª span, G = § • ˜ 1 ⃗x ∈ R2 : ⃗x = t for some t ∈ R . 1 The proof that G is a subspace now proceeds similarly to the proof for F above. • ˜ • ˜ 1 1 (i) Let u ⃗, ⃗v ∈ G. Then u ⃗=t and ⃗v = s for some s, t ∈ R. 1 1 • ˜ 1 But then u ⃗ + ⃗v = (s + t) ∈ G. 1 • ˜ • ˜ 1 1 (ii) Let u ⃗=t ∈ G. For any scalar α we have αu ⃗ = (αt) ∈ 1 1 G. 34.8 H = span{u ⃗, ⃗v } for some unknown vectors u ⃗, ⃗v ∈ R2 . H is a subspace. (i) Let ⃗x , ⃗y ∈ H. Then ⃗x = α1 u ⃗ + α2 ⃗v and ⃗y = β1 u ⃗ + β2 ⃗v for some scalars α1 , α2 , β1 , β2 . But then ⃗x + ⃗y = α1 u ⃗ + α2 ⃗v + β1 u ⃗ + β2 ⃗v = (α1 + β1 )u ⃗ + (α2 + β2 )v⃗ ∈ H.

35

DEF

Basis A basis for a subspace V is a linearly independent set of vectors, B, so that span B = V .

DEF

(ii) Let ⃗x = α1 u ⃗ + α2 ⃗v ∈ H. For any scalar β we have βu ⃗ = (βα1 )u ⃗ + (βα2 )v⃗ ∈ H.

Dimension The dimension of a subspace V is the number of elements in a basis for V .

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 1 Let u ⃗ = ⎣0⎦, ⃗v = ⎣1⎦, w ⃗ = ⎣1⎦, and V = span{u ⃗, ⃗v , w ⃗ }. 0 0 0

The goal of this problem is to learn ■ To apply the definition of basis and dimension.

35.1 Describe V . V is the x y-plane in R3 . 35.2 Is {u ⃗, ⃗v , w ⃗ } a basis for V ? Why or why not?

No. The set {u ⃗, ⃗v , w ⃗ } is linearly dependent since w ⃗ =u ⃗ + ⃗v . 35.3 Give a basis for V . {u ⃗, ⃗v }.

■ Intuition that a plane is two dimensional. ■ A basis is not unique, but always has the same size (this is not proved).

35.4 Give another basis for V . {u ⃗, w ⃗ } or {v⃗, w ⃗ }.

■ Spans are never bases—you must not confuse a subspace with its basis!

35.5 Is span{u ⃗, ⃗v } a basis for V ? Why or why not?

⃗, so it cannot be linearly No. span{u ⃗, ⃗v } is an infinite set of vectors which includes 0 independent and therefore isn’t a basis. 35.6 What is the dimension of V ?

A basis for V has two vectors so it is two-dimensional. We also know this because V is the x y-plane in R3 and all planes are two-dimensional. 26

Apply the definitions of basis and dimension to a simple example.

© Jason Siefken, 2015–2021

36

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 4 7 ⃗ ⎣ ⎦ ⎣ ⎦ ⎣ 2 5 Let a ⃗= , b= , ⃗c = 8⎦ (notice these vectors are linearly independent) and let 3 6 8 ⃗ , ⃗c }. P = span{a ⃗ , ⃗b} and Q = span{b 36.1 Give a basis for and the dimension of P.

{a ⃗ , ⃗b} is a basis for P, and so its dimension is 2.

The relationship between subspaces, bases, unions, and intersections. The goal of this problem is to learn ■ Recognize intersections of subspaces as subspaces. ■ Recognize the union of subspaces need not be a subspace. ■ Visualize planes in R3 to solve problems without computations.

36.2 Give a basis for and the dimension of Q.

⃗ , ⃗c } is a basis for Q, and so its dimension is 2. {b 36.3 Is P ∩ Q a subspace? If so, give a basis for it and its dimension.

⃗ } is a basis for P ∩ Q, and so its dimension is 1. Yes. {b P and Q are both planes and are not parallel (since a ⃗ , ⃗b, ⃗c are linearly independent). ⃗ The intersection of any two non-parallel planes in R3 is a line. We know that 0 ⃗ ⃗ and b are on this line, and therefore the line is span{b} 36.4 Is P ∪ Q a subspace? If so, give a basis for it and its dimension.

No. For example a ⃗ and ⃗c are both in P ∪ Q, but a ⃗ + ⃗c ∈ / P ∪ Q. Proof: A vector is in P ∪ Q if it is in P or Q, so we must show that a ⃗ + ⃗c ∈ / P and a ⃗ + ⃗c ∈ /Q a ⃗ + ⃗c ∈ / P since if it were, we would also have (a ⃗ + ⃗c ) − a ⃗ = ⃗c ∈ P. We know this is impossible since the vectors a ⃗ , ⃗b, ⃗c are linearly independent, and so ⃗c does not equal a linear combination of a ⃗ and ⃗b. An analogous argument shows that a ⃗ + ⃗c ∈ / Q.

Matrices 37

Let A =

•

1 3

Relate matrix equations and systems of linear equations.

˜ • ˜ • ˜ 2 x −2 , ⃗x = , and ⃗b = . 3 y −1

The goal of this problem is to ■ Use matrix-vector multiplication to represent a system of equations with compact notation.

37.1 Compute the product Ax ⃗.

Ax⃗ =

˜ x + 2y . 3x + 3 y

•

37.2 Write down a system of equations that corresponds to the matrix equation Ax ⃗ = ⃗b.

■ View a matrix equation as a statement about (i) linear combinations of column vectors and (ii) a system of equations coming from the rows.

x + 2 y = −2 3x + 3 y = −1 ˜ x0 be a solution to Ax⃗ = ⃗b. Explain what x 0 and y0 mean in terms of intersecting y0 lines (hint: think about systems of equations).

37.3 Let

•

The lines represented by the equations x + 2 y =•−2 ˜and 3x + 3 y = −1 from the x system of equations above intersect at the point 0 . y0 • ˜ x0 37.4 Let be a solution to Ax⃗ = ⃗b. Explain what x 0 and y0 mean in terms of linear y0 combinations (hint: think about the columns of A). x 0 and y0 , when used as scalars in a linear combination of the columns of A, make the vector ⃗b. In other words: • ˜ • ˜ • ˜ 1 2 −2 x0 + y0 = . 3 3 −1

27

© Jason Siefken, 2015–2021

Rephrase previous questions using matrix equations. The goal of this problem is to ■ Rephrase the question of linear independence as the special matrix equa⃗. tion Ax⃗ = 0

38

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 4 7 ⎣ ⎦ ⎣ ⎦ ⎣ 2 5 Let u ⃗= , ⃗v = ,w ⃗ = 8⎦ . 3 6 9 38.1 How could you determine if {u ⃗, ⃗v , w ⃗ } was a linearly independent set?

The set is linearly independent if and only if no non-trivial linear combination of ⃗. That is, if x, y, z are scalars such that xu ⃗, the vectors u ⃗, ⃗v , w ⃗ equals 0 ⃗ + y ⃗v +zw ⃗ =0 then x = y = z = 0. In other words, the only solution of the following system of equations is x = y = z = 0. x + 4 y + 7z = 0 2x + 5 y + 8z = 0 3x + 6 y + 9z = 0

38.2 Can your method be rephrased in terms of a matrix equation? Explain.

The system of linear⎡equations⎤above can⎡be⎤represented by the matrix equation 1 4 7 x ⃗, where A = ⎣2 5 8⎦ and ⃗x = ⎣ y ⎦. Ax⃗ = 0 3 6 9 z So another way to say the above is that the set is linearly independent if and only ⃗ is ⃗x = 0 ⃗. if the only solution to the equation Ax⃗ = 0

39

Consider the system represented by ⎡ 1 −3 ⎣0 0 0 0

Interpret matrix equations. The goal of this problem is to ■ Use knowledge about systems of linear equations to answer questions about matrix equations.

⎤⎡ ⎤ 0 x 1⎦ ⎣ y ⎦ = ⃗b. 0 z

⎡ ⎤ 1 39.1 If ⃗ b = ⎣2⎦, is the set of solutions to this system a point, line, plane, or other? 3 This system has no solutions, since, if we expand the matrix equation into a system of equations, the third equation would be 0 = 3, which is impossible. ⎡ ⎤ 1 39.2 If ⃗ b = ⎣1⎦, is the set of solutions to this system a point, line, plane, or other? 0 A line. The system would be x − 3y = 1 z=1 0=0 ⎡ ⎤ x A vector ⃗x = ⎣ y ⎦ that satisfies this system must have z = 1, and by the first z equation in the system any value of x determines the value of y, and vice versa. In other words the system has one free variable, and so its set of solutions is a line. Apply matrix equations to planes.

40

⎡ ⎤ ⎡ ⎤ 1 −1 Let d⃗ 1 = ⎣1⎦ and d⃗ 2 = ⎣ 1⎦. Let P be the plane given in vector form by ⃗x = t d⃗ 1 + sd⃗ 2 . 2 0 Further, suppose M is a matrix so that M ⃗r ∈ P for any ⃗r ∈ R2 . 28

© Jason Siefken, 2015–2021

The goal of this problem is to ■ Rephrase properties of a plane in terms of matrix equations. ■ Be able to describe one application of the transpose.

40.1 How many rows does M have?

Three. It must have three rows in order for M ⃗r to be an element of R3 . 40.2 Find such an M .

⎡

1 M = ⎣1 2

⎤ • ˜ −1 a 1⎦ is one possible answer, since if ⃗r = , then M ⃗r = ad⃗ 1 + bd⃗ 2 . b 0

Another less interesting answer is the 3 × 2 zero matrix. 40.3 Find necessary and sufficient conditions (phrased as equations) for n ⃗ to be a normal vector for P .

⃗, n n ⃗ is normal to P if and only if n ⃗ ̸= 0 ⃗ · d⃗ 1 = 0, and n ⃗ · d⃗ 2 = 0 ⃗ are normal vectors for P . How do 40.4 Find a matrix K so that non-zero solutions to K ⃗ x =0 K and M relate? • ˜ 1 1 2 K= . K and M are transposes of one another. −1 1 0 The conditions n ⃗ · d⃗ 1 = 0 and n ⃗ · d⃗ 2 = 0 from the previous part translate to the following system of equations: x + y + 2z = 0 −x + y = 0. This system of equations can be represented by the matrix equation ⎡ ⎤ • ˜ x • ˜ 1 1 2 ⎣ ⎦ 0 y = . −1 1 0 0 z

Change of Basis & Coordinates 41

The mythical town of Oronto is not aligned with the usual compass directions. The streets are laid out as follows:

Motivate change of basis. The goal of this problem is to ■ Describe points in multiple bases when given a visual description of the basis or when given the basis vectors numerically. ■ Recognize ambiguity when faced with the question, “Which basis is better?”

North East

City Hall

Instead, every street is parallel to the vector d⃗ 1 = • ˜ ⃗ = 0 east . center of town is City Hall at 0 0 north

1 5

•

˜ 4 east or d⃗ 2 = 3 north

1 5

• ˜ −3 east . The 4 north

Locations in Oronto are typically specified in street coordinates. That is, as a pair (a, b) where a is how far you walk along streets in the d⃗ 1 direction and b is how far you walk in the d⃗ 2 direction, provided you start at city hall. 29

© Jason Siefken, 2015–2021

41.1 The points A = (2, 1) and B = (3, −1) are given in street coordinates. Find their east-north

coordinates. A = (1, 2) and B = (3, 1) in east-north coordinates. We obtain A for example by finding the vector 2d⃗ 1 + d⃗ 2 . 41.2 The points X = (4, 3) and Y = (1, 7) are given in east-north coordinates. Find their street

coordinates. X = (5, 0) and Y = (5, 5) in street coordinates. • ˜ • ˜ 1 east 0 east 41.3 Define ⃗e1 = and ⃗e2 = . Does span{e⃗1 , ⃗e2 } = span{d⃗ 1 , d⃗ 2 }? 0 north 1 north

Yes. Both of these sets spans all of R2 . ⃗ 1 + 5d⃗ 2 = ⃗e1 + 7e⃗2 . Is the point Y better represented by the pair (5, 5) 41.4 Notice that Y = 5d or by the pair (1, 7)? Explain. It is equally well represented by either pair. For example, the street coordinates might be more useful for a resident of Oronto, while the east-north coordinates might be more useful for someone looking at Oronto on a world map.

Representation in a Basis ⃗ 1 , . . . , ⃗b n } be a basis for a subspace V and let ⃗v ∈ V . The representation Let B = {b of ⃗v in the B basis, notated [v⃗]B , is the column matrix ⎤ α1 . [v⃗]B = ⎣ .. ⎦ αn

DEFINITION

⎡

where α1 , . . . , αn uniquely satisfy ⃗v = α1 ⃗b1 + · · · + αn ⃗b n . Conversely, ⎤ α1 ⎣ .. ⎦ = α1 ⃗b1 + · · · + αn ⃗b n . αn B ⎡

is notation for the linear combination of ⃗b1 , . . . , ⃗b n with coefficients α1 , . . . , αn .

42

• ˜ 2 Let E = {e⃗1 , ⃗e2 } be the standard basis for R and let C = {c⃗1 , ⃗c 2 } where ⃗c 1 = and 1 E • ˜ 5 ⃗c 2 = be another basis for R2 . 3 E 2

42.1 Express ⃗c 1 and ⃗c 2 as a linear combination of ⃗e1 and ⃗e2 . ⃗c 1 = 2e⃗1 + ⃗e2 and ⃗c 2 = 5e⃗1 + 3e⃗2 . 42.2 Express ⃗e1 and ⃗e2 as a linear combination of ⃗c 1 and ⃗c 2 . ⃗e1 = 3c⃗1 − ⃗c 2 and ⃗e2 = −5c⃗1 +2c⃗2 . 42.3 Let ⃗ v = 2e⃗1 + 2e⃗2 . Find [v⃗]E and [v⃗]C .

[v⃗]E =

• ˜ • ˜ 2 −4 and [v⃗]C = . 2 2

The second one is since ⃗v = 2e⃗1 + 2e⃗2 = 2(3c⃗1 − ⃗c 2 ) + 2(−5c⃗1 + 2c⃗2 ) = −4c⃗1 + 2c⃗2 . 42.4 Can you find a matrix X so that X [w ⃗ ]C = [w ⃗ ]E for any w ⃗?

X=

•

2 1

˜ 5 is such a matrix. 3 •

a We know X must be a 2 × 2 matrix, so suppose X = c From the first part above, we know • ˜ • ˜ 1 2 = 0 C 1 E

and 30

˜ b for some a, b, c, d ∈ R. b

• ˜ • ˜ 0 5 = , 1 C 3 E © Jason Siefken, 2015–2021

Change of basis notation. The goal of this problem is to ■ Practice using change-of-basis notation. ■ Compute representations of vectors in different bases. ■ Find a matrix that computes a change of basis.

and so we need X to satisfy X

• ˜ • ˜ 1 2 = 0 1

and

X

• ˜ • ˜ 0 5 = . 1 3

• ˜ • ˜ • ˜ • ˜ 1 a 0 b = and X = , so we can now immediately solve for a, b, c, d 0 c 1 d • ˜ 2 5 to find that X must be the matrix . 1 3 But X

42.5 Can you find a matrix Y so that Y [w ⃗ ]E = [w ⃗ ]C for any w ⃗?

Y=

•

3 −1

˜ −5 is such a matrix. 2

Using similar reasoning • ˜ to the previous part, we know Y must be a 2 × 2 matrix, a b so suppose Y = for some a, b, c, d ∈ R. c b From the second part above, we know • ˜ • ˜ 1 3 = 0 E −1 C

and

• ˜ • ˜ 0 −5 = , 1 E 2 C

and

Y

and so we need Y to satisfy Y

• ˜ • ˜ 1 3 = 0 −1

• ˜ • ˜ 0 −5 = , 1 2

• ˜ • ˜ • ˜ • ˜ 1 a 0 b But Y = and Y = , so we can now immediately solve for a, b, c, d 0 c 1 d • ˜ 3 −5 to find that Y must be the matrix . −1 2 42.6 What is Y X ?

DEFINITION

YX =

43

•

1 0

˜ 0 . 1

Orientation of a Basis ⃗ 1 , . . . , ⃗b n } is right-handed or positively oriented if it can The ordered basis B = {b be continuously transformed to the standard basis (with ⃗b i ↦→ ⃗e i ) while remaining linearly independent throughout the transformation. Otherwise, B is called lefthanded or negatively oriented.

Let {e⃗1 , ⃗e2 } be the standard basis for R2 and let u ⃗θ be a unit vector. Let θ be the angle between u ⃗θ and ⃗e1 measured counter-clockwise starting at ⃗e1 . 43.1 For which θ is {e⃗1 , u ⃗θ } a linearly independent set? Every θ that is not a multiple of π. 43.2 For which θ can {e⃗1 , u ⃗θ } be continuously transformed into {e⃗1 , ⃗e2 } and remain linearly

independent the whole time? Every θ ∈ (0, π). For θ ∈ (π, 2π), a continuous transformation of u ⃗θ to ⃗e2 would have to cross the x-axis, at which point {e⃗1 , u ⃗θ } would cease to be linearly independent. 43.3 For which θ is {e⃗1 , u ⃗θ } right-handed? Left-handed?

It is right-handed for θ ∈ (0, π), and left handed for θ ∈ (π, 2π). 43.4 For which θ is {u ⃗θ , ⃗e1 } (in that order) right-handed? Left-handed?

It is right-handed for θ ∈ (π, 2π), and left handed for θ ∈ (0, π). 43.5 Is {2e⃗1 , 3e⃗2 } right-handed or left-handed? What about {2e⃗1 , −3e⃗2 }?

{2e⃗1 , 3e⃗2 } is right-handed and {2e⃗1 , −3e⃗2 } is left-handed.

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© Jason Siefken, 2015–2021

Visually understand orientation. The goal of this problem is to ■ Determine the orientation of a basis from a picture. ■ Recognize the order of vectors in a basis relates to the orientation of that basis.

Italicizing N 44

The citizens of Oronto want to erect a sign welcoming visitors to the city. They’ve commissioned letters to be built, but at the last council meeting, they decided they wanted italicised letters instead of regular ones. Can you help them?

Suppose that the “N” on the left is written in regular 12-point font. Find a matrix A that will transform the “N” into the letter on the right which is written in an italic 16-point font. Work with your group to write out your solution and approach. Make a list of any assumptions you notice your group making or any questions for further pursuit.

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© IOLA Team iola. math. vt. edu

Beyond the N 45

Some council members were wondering placed in other locations in the • how letters ˜ 1 1/3 plane would be transformed under A = . If other letters are placed around the 0 4/3 “N,” the council members argued over four different possible results for the transformed letters. Which choice below, if any, is correct, and why? If none of the four options are correct, what would the correct option be, and why? Original

(A)

(0, 0)

(B)

(0, 0)

(C)

(D)

(0, 0)

33

(0, 0)

© IOLA Team iola. math. vt. edu

(0, 0)

Linear Transformations 46

R : R2 → R2 is the transformation that rotates vectors counter-clockwise by 90◦ . • ˜ • ˜ • ˜ • ˜ • ˜ • ˜ 1 0 1 0 0 −1 46.1 Compute R and R . R = and R = . 0 1 0 1 1 0 • ˜ • ˜ • ˜ 1 1 0 46.2 Compute R . How does this relate to R and R ? 1 0 1 • ˜ • ˜ • ˜ • ˜ 1 −1 1 0 R = =R +R . 1 1 0 1  • ˜ • ˜‹ 1 0 46.3 What is R a +b ? 0 1  • ˜ • ˜‹ • ˜ • ˜ 1 0 0 −1 R a +b =a +b . 0 1 1 0

Apply geometric transformations to vectors. The goal of this problem is to ■ Given a transformation described in words, compute the result of the transformation applied to particular vectors. ■ Use linear combinations to compute the result of rotations applied to unknown vectors. ■ Distinguish between a general transformation and a matrix transformation.

Rotating a vector and then multiplying by a scalar gives the same result as multiplying first then rotating. Similarly, adding two vectors and then rotating their sum gives the same result as rotating them and then adding. 46.4 Write down a matrix R so that Rv ⃗ is ⃗v rotated counter-clockwise by 90◦ .

DEFINITION

R=

47

•

0 1

˜ −1 is such a matrix. 0

Linear Transformation Let V and W be subspaces. A function T : V → W is called a linear transformation if T (u ⃗ + ⃗v ) = T (u ⃗) + T (v⃗) and T (αv⃗) = αT (v⃗) for all vectors u ⃗, ⃗v ∈ V and all scalars α.

Apply the definition of a linear transformation to examples.

47.1 Classify the following as linear transformations or not.

(a) R from before (rotation counter-clockwise by 90◦ ). A linear transformation. We proved this in the previous problem. • ˜ • 2˜ x x 2 2 (b) W : R → R where W = . y y  • ˜‹ • ˜ • ˜ 1 4 2 Not a linear transformation, since for example W 2 = ̸= = 0 0 0 • ˜ 1 2W . 0 • ˜ • ˜ x x +2 2 2 (c) T : R → R where T = . y y  • ˜‹ • ˜ • ˜ 1 4 1 Not a linear transformation, since for example T 2 = ̸= 2T . 0 0 0 • ˜ • ˜ • ˜ x x 2 2 2 (d) P : R → R where P = v compu⃗ and u ⃗= . y y 3 A linear transformation. We found a general formula for v compu⃗ in a previous exercise: v compu⃗ ⃗x =

u ⃗ · ⃗x u ⃗ · ⃗x u ⃗= u ⃗. u ⃗·u ⃗ 13

For any two vectors ⃗x and ⃗y , we have u ⃗ · (x⃗ + ⃗y ) u ⃗ 13 u ⃗ · ⃗y u ⃗ · ⃗x = u ⃗+ u ⃗ 13 13 = v compu⃗ ⃗x + v compu⃗ ⃗y .

v compu⃗ (x⃗ + ⃗y ) =

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© Jason Siefken, 2015–2021

The goal of this problem is to ■ Distinguish between a linear transformation and a non-linear transformation. ■ Provide a proof of whether a transformation is linear or not.

For any ⃗x and scalar α, we have

DEFINITION

v compu⃗ (αx⃗ ) =

48

u ⃗ · (αx⃗ ) α(u ⃗ · ⃗x ) u ⃗= u ⃗ = α v compu⃗ ⃗x . 13 13

Image of a Set Let L : Rn → Rm be a transformation and let X ⊆ Rn be a set. The image of the set X under L, denoted L(X ), is the set L(X ) = { ⃗y ∈ Rm : ⃗y = L(x⃗ ) for some ⃗x ∈ X }.

§• ˜ ª x Let S = : 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 ⊆ R2 be the filled-in unit square and let y ⃗, ⃗e1 , ⃗e2 , ⃗e1 + ⃗e2 } ⊆ R2 be the corners of the unit square. C = {0 48.1 Find R(C), W (C), and T (C) (where R, W , and T are from the previous question).

⃗, ⃗e2 , −e⃗1 , −e⃗1 + ⃗e2 }. R(C) = {0 W (C) = C. §• ˜ • ˜ • ˜ • ˜ª 2 3 2 3 T (C) = , , , . 0 0 1 1

The goal of this problem is to ■ Compute images of sets under transformations. ■ Develop geometric intuition for transformations of Rn in terms of inputs and outputs. ■ Relate images to graphical problems like italicising N .

48.2 Draw R(S), T (S), and P (S) (where R, T , and P are from the previous question).

P (S)

R(S)

Work with Images.

T (S)

48.3 Let ℓ = {all convex combinations of a ⃗ and ⃗b} be a line segment with endpoints a ⃗ and

⃗b and let A be a linear transformation. Must A(ℓ) be a line segment? What are its endpoints? ⃗ ). A(ℓ) must be a line segment, with endpoints A(a ⃗ ) and A(b

For any scalars α1 and α2 , by the linearity of A we have: A(α1 a ⃗ + α2 ⃗b) = α1 A(a ⃗) + ⃗ α2 A(b). If α1 + α2 = 1, then the linear combination on the right is also convex, and so A(ℓ) ⃗ ). This is precisely the straight is the set of convex combinations of A(a ⃗ ) and A(b ⃗ line segment joining A(a ⃗ ) and A(b). ⃗ Note that if A(a ⃗ ) = A(b) (for example, if A is the zero transformation), then A(ℓ) will consist of the single point, which we think of as a “degenerate” line segment in this situation. 48.4 Explain how images of sets relate to the Italicizing N task.

The task asked us to find a linear transformation such that the image of the regular “N” is the italicized “N”. By the previous exercise, we now know it suffices to find a linear transformation that sends the four endpoints of line segments on the regular “N” to the corresponding four endpoints on the italicized “N”.

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© Jason Siefken, 2015–2021

Pat and Jamie 49

Suppose that the “N” on the left is written in regular 12-point font. Find a matrix A that will transform the “N” into the letter on the right which is written in an italic 16-point font.

Two students—Pat and Jamie—explained their approach to italicizing the N as follows:

Decompose a transformation into a composition of simpler transformations.

In order to find the matrix A, we are going to find a matrix that makes the “N” taller, find a matrix that italicizes the taller “N,” and a combination of those two matrices will give the desired matrix A.

The goal of this problem is to ■ Decompose a transformation into simpler ones. ■ Produce examples showing matrix multiplication is not commutative.

1. Do you think Pat and Jamie’s approach allowed them to find A? If so, do you think they found the same matrix that you did during Italicising N? 2. Try Pat and Jamie’s approach. Either (a) come up with a matrix A using their approach, or (b) explain why their approach does not work.

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© IOLA Team iola. math. vt. edu

50

• ˜ 2 Define P : R → R to be projection onto span{u ⃗} where u ⃗= , and let R : R2 → R2 3 be rotation counter-clockwise by 90◦ . 2

2

50.1 Find a matrix P so that P ⃗ x = P (x⃗ ) for all ⃗x ∈ R2 .

P=

1 13

•

4 6

Connect function composition and matrix multiplication. The goal of this problem is to ■ Distinguish between matrices and linear transformations. ■ Explain the relationship between matrix multiplication and composition of linear transformations.

˜ 6 is such a matrix. 9 •

˜ a b The matrix P corresponding to P is a 2 × 2 matrix, so suppose P = for c d 2 some a, b, c, d ∈ R. Then we know that if {e⃗1 , ⃗e2 } is the standard basis for R , • ˜ • ˜ a b P(e⃗1 ) = and P(e⃗2 ) = . c d ⃗x · u ⃗ u ⃗. Therefore, the first column u ⃗·u ⃗ • ˜ • ˜ 2 1 4 a = P (e⃗1 ) = u ⃗= c 13 13 6

We know from an earlier exercise that P (x⃗ ) = of P is

and the second column of P is • ˜ • ˜ 3 1 6 b = P (e⃗2 ) = u ⃗= . d 13 13 9 50.2 Find a matrix R so that Rx ⃗ = R(x⃗ ) for all ⃗x ∈ R2 .

R=

•

˜ −1 is such a matrix. 0

0 1

Using the same reasoning as the previous part, we can compute • ˜ • ˜ 0 −1 R(e⃗1 ) = ⃗e2 = and R(e⃗2 ) = −e⃗1 = . 1 0 Therefore, the matrix R for R is the matrix with the two vectors above as its respective columns. 50.3 Write down matrices A and B for P ◦ R and R ◦ P .

A=

1 13

•

6 9

˜ −4 and B = −6

1 13

•

−6 4

˜ −9 are two such matrices. 6

Using the same reasoning as above, we can compute • ˜ • ˜ 1 6 1 −4 (P ◦R)(e⃗1 ) = P (R(e⃗1 )) = P (e⃗2 ) = and (P ◦R)(e⃗2 ) = P (R(e⃗2 )) = P (−e⃗1 ) = . 13 9 13 −6 Therefore, the matrix A for P ◦ R is the matrix with the two vectors above as its respective columns. Similarly, for R ◦ P , we can compute: • ˜‹ • ˜ 1 4 1 −6 (R ◦ P )(e⃗1 ) = R(P (e⃗1 )) = R = 4 13 6 13  • ˜‹ • ˜ 1 6 1 −9 (R ◦ P )(e⃗2 ) = R(P (e⃗2 )) = R = . 6 13 9 13 

Therefore, the matrix B for R ◦ P is the matrix with these two vectors as its respective columns. 50.4 How do the matrices A and B relate to the matrices P and R?

A = PR and B = RP. We can compute these matrix products to see this, but from the previous parts, we know that for any vector ⃗x Ax⃗ = (P ◦ R)(x⃗ ) = P (R(x⃗ )) = P (Rx⃗ ) = PRx⃗ 37

© Jason Siefken, 2015–2021

and B ⃗x = (R ◦ P )(x⃗ ) = R(P (x⃗ ) = R(P ⃗x ) = RP ⃗x .

DEFINITION

DEFINITION

Using ⃗x = ⃗e1 shows that first column of A must equal the first column of PR, and using ⃗x = ⃗e2 shows that the second column of A must equal the second column of PR, and therefore A = PR. For the same reason, we must also have B = RP.

51

Range The range (or image) of a linear transformation T : V → W is the set of vectors that T can output. That is, range(T ) = { ⃗y ∈ W : ⃗y = T ⃗x for some ⃗x ∈ V }.

Null Space The null space (or kernel) of a linear transformation T : V → W is the set of vectors that get mapped to the zero vector under T . That is, ⃗}. null(T ) = {x⃗ ∈ V : T ⃗ x =0

Let P : R2 → R2 be projection onto span{u ⃗} where u ⃗=

• ˜ 2 (like before). 3

51.1 What is the range of P ?

range(P ) = span{u ⃗}. P (x⃗ ) is by definition the vector in span{u ⃗} that is closest to ⃗x , so in particular P (x⃗ ) ∈ span{u ⃗} for all ⃗x ∈ R2 . Therefore range(P ) ⊆ span{u ⃗}.

Understanding ranges and null spaces. The goal of this problem is to ■ Read and apply the definition of range and null space. ■ Geometrically visualize the range and null space of a projection.

On the other hand, P (αu ⃗) = αP (u ⃗) = αu ⃗ for any scalar α, and so range(P ) = span{u ⃗}. 51.2 What is the null space of P ?

§• ˜ª 3 null(P ) = span . −2 ⃗ if and only if ⃗x is on the line orthogonal to span{u A vector ⃗x projects to 0 ⃗} passing through the origin.

52

Practicing proofs.

Let T : Rn → Rm be an arbitrary linear transformation. 52.1 Show that the null space of T is a subspace.

(i) Let u ⃗, ⃗v ∈ null(T ). Applying the linearity of T we see T (u ⃗ +v⃗) = T (u ⃗)+T (v⃗) = ⃗+0 ⃗=0 ⃗, and so u 0 ⃗ + ⃗v ∈ null(T ). (ii) Let u ⃗ ∈ null(T ) and let α be any scalar. Again using the linearity of T we see ⃗=0 ⃗, and so αu T (αu ⃗) = αT (u ⃗) = α0 ⃗ ∈ null(T ).

The goal of this problem is to ■ Practice proving an abstract set (the range or the null space) is a subspace.

52.2 Show that the range of T is a subspace.

Since range(T ) = T (Rn ) and Rn is non-empty, we know that range(T ) is nonempty. Therefore, to show that range(T ) is a subspace, we need to show (i) that it’s closed under vector addition, and (ii) that it is closed under scalar multiplication.

DEFINITION

(i) Let ⃗x , ⃗y ∈ range(T ). By definition, there exist u ⃗, ⃗v ∈ Rn such that T (u ⃗) = ⃗x and T (v⃗) = ⃗y . Since T is linear, ⃗x + ⃗y = T (u ⃗) + T (v⃗) = T (u ⃗ + ⃗v ), and so ⃗x + ⃗y ∈ range(T ). (ii) Let ⃗x ∈ range(T ) and let α be a scalar. By definition, there exists u ⃗ ∈ Rn such that T (u ⃗) = ⃗x , and so by the linearity of T , αx⃗ = αT (u ⃗) = T (αu ⃗). Therefore αx⃗ ∈ range(T ).

Induced Transformation Let M be an n × m matrix. We say M induces a linear transformation T M : Rm → Rn defined by [T M ⃗v ]E ′ = M [v⃗]E , where E is the standard basis for Rm and E ′ is the standard basis for Rn . 38

© Jason Siefken, 2015–2021

Formalizing the connection between matrices and linear transformations. The goal of this problem is to ■ Distinguish between linear transformations and matrices. ■ Explain how to relate matrices and linear transformations. ■ Practice using formal language and notation, avoiding category errors.

•

a Let M = c

53

˜ b , let ⃗v = ⃗e1 + ⃗e2 ∈ R2 , and let T M be the transformation induced by M . d

53.1 What is the difference between “M ⃗ v ” and “M [v⃗]E ”?

“M ⃗v ” is ambiguous notation, as it is only defined if ⃗v is a specific list of numbers. There are infinitely many different bases of R2 and so ⃗v has infinitely many different representations as a list of numbers (a different basis produces a different • ˜ 1 representation). For example, let B = {v⃗, ⃗e1 }. We now have [v⃗]E = and 1 • ˜ • ˜ • ˜ 1 1 1 [v⃗]B = . There’s no way to decide whether M ⃗v should mean M or M 0 1 0 or any of the infinitely many other options. “M [v⃗]E ” is unambiguous, as [v⃗]E is an explicit representation of ⃗v in a particular basis. 53.2 What is [T M ⃗e1 ]E ?

It is the first column of M . • ˜ 1 By definition, [T M ⃗e1 ]E = M [e⃗1 ]E = M , which equals the first column of M . 0 53.3 Can you relate the columns of M to the range of T M ?

The range of T M equals the span of the columns of M . By the previous part, the first column of M is in the range of T M . By a similar argument, the second column of M is also in the range of T M , since it equals [T M ⃗e2 ]E . Therefore the span of the columns of M is a subset of the range of T M . • ˜ • ˜ a b On the other hand, if ⃗v 1 = and ⃗v 2 = are the columns of M and ⃗x = c d α1 ⃗v 1 + α2 ⃗v 2 is an element of span{v⃗1 , ⃗v 2 }, then [x⃗ ]E = α1

• ˜ • ˜ a b + α2 = α1 [T M ⃗e1 ]E + α2 [T M ⃗e2 ]E = [T M (α1 ⃗e1 + α2 ⃗e2 )]E . c d

DEFINITION

Therefore ⃗x is in the range of T M .

Fundamental Subspaces Associated with any matrix M are three fundamental subspaces: the row space of M , denoted row(M ), is the span of the rows of M ; the column space of M , denoted col(M ), is the span of the columns of M ; and the null space of M , denoted null(M ), ⃗. is the set of solutions to M ⃗ x =0 Fundamental subspaces of a matrix.

54

•

1 Consider A = 0

0 1

˜ 0 . 0

The goal of this problem is to ■ Compute row and column spaces of a matrix.

54.1 Describe the row space of A.

⎧⎡ ⎤ ⎡ ⎤⎫ 0 ⎬ ⎨ 1 row(A) = span ⎣0⎦ , ⎣1⎦ , which is the x y-plane in R3 . ⎩ 0 0 ⎭ 54.2 Describe the column space of A.

§• ˜ • ˜ • ˜ª §• ˜ • ˜ª 1 0 0 1 0 col(A) = span , , = span , = R2 . 0 1 0 0 1 54.3 Is the row space of A the same as the column space of A?

No. Although they are both two dimensional spaces, row(A) is a subspace of R3 and all vectors in it have three coordinates (with the third always being zero), while col(A) is a subspace of R2 and all vectors in it have two coordinates. Therefore, these two spaces are different. 54.4 Describe the set of all vectors orthogonal to the rows of A.

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© Jason Siefken, 2015–2021

■ Recognize that row and column spaces may be unrelated. ■ Geometrically relate the row space to the null space. ■ Connect the fundamental subspaces of a matrix to the range and null space of a transformation.

The z-axis in R3 . ⎡ ⎤ x A vector ⃗x = ⎣ y ⎦ is orthogonal to the rows of A if and only if its dot product z with both rows is zero. That is ⎡ ⎤ ⎡ ⎤ 1 0 ⃗x · ⎣0⎦ = x = 0 and ⃗x · ⎣1⎦ = y = 0. 0 0 ⎡ ⎤ 0 ⎣ ⃗x satisfies these equations if and only if ⃗x = 0⎦ for some real number t, or in t other words if ⃗x is on the z-axis. 54.5 Describe the null space of A.

The z-axis in R3 . ⎡ ⎤ x A vector ⃗x = ⎣ y ⎦ is in null(A) if and only if z Ax⃗ =

• ˜ x ⃗. =0 y

These are the same conditions as in the previous part, so the set of vectors satisfying this is the z-axis. 54.6 Describe the range and null space of TA, the transformation induced by A.

range(TA) = col(A) = R2 and null(TA) = null(A), which is the z-axis in R3 . By Problem 53.3, the range of an induced transformation equals the span of the columns of the matrix. In other words, range(TA) = col(A). ⃗. In other words, Next, by definition ⃗v ∈ null(TA) when [TA ⃗v ]E = A[v⃗]E = 0 ⃗v ∈ null(TA) if and only if [v⃗]E ∈ null(A). We know from the previous part that null(A) is the z-axis in R3 .

55 B=

•

1 1

2 1

3 1

˜

• 1 C = rref(B) = 0

0 1

˜ −1 2

55.1 How does the row space of B relate to the row space of C?

They are equal. Row operations replace rows with linear combinations of rows. Therefore, since C is the matrix B after the application of some row operations, row(C) ⊆ row(B). Since row operations are all reversible, we also know that B can be obtained from C by applying row operations, so row(B) ⊆ row(C). Therefore, row(B) = row(C). 55.2 How does the null space of B relate to the null space of C?

They are equal. A vector is in null(B) or null(C) if and only if it is orthogonal to all vectors in row(B) or all vectors in row(C), respectively. But row(B) = row(C) by the previous part, so their null spaces must also be equal. 55.3 Compute the null space of B.

⎧⎡ ⎤⎫ 1 ⎬ ⎨ null(C) = span ⎣−2⎦ . ⎩ 1 ⎭ We compute null(C), since it equals null(B) by the previous part. 40

© Jason Siefken, 2015–2021

Fundamental subspaces and row reduction. The goal of this problem is to ■ Explain why row reduction doesn’t change the row space or the null space.

⎡ ⎤ • ˜ x x −z ⎣ ⎦ ⃗. The complete solution ⃗x = y is in null(C) if and only if C ⃗x = =0 y + 2z z to this matrix equation is ⎧⎡ ⎫ ⎧⎡ ⎤⎫ ⎤ t 1 ⎬ ⎨ ⎬ ⎨ null(C) = ⎣−2t ⎦ ∈ R3 : t ∈ R = span ⎣−2⎦ . ⎩ ⎭ ⎩ t 1 ⎭

56 P=

•

0 1

0 2

˜

Q = rref(P) =

•

1 0

2 0

˜

Fundamental subspaces and row reduction. The goal of this problem is to ■ Recognize that row reduction may change the column space of a matrix.

56.1 How does the column space of P relate to the column space of Q?

They are not equal, but have the same dimension. 56.2 Describe the column space of P and the column space of Q.

DEFINITION

§• ˜ • ˜ª §• ˜ª 0 0 0 col(P) = span , = span , which is the y-axis in R2 . 1 2 1 §• ˜ • ˜ª §• ˜ª 1 2 1 col(Q) = span , = span , which is the x-axis in R2 . 0 0 0

Rank For a linear transformation T : Rn → Rm , the rank of T , denoted rank(T ), is the dimension of the range of T . For an m × n matrix M , the rank of M , denoted rank(M ), is the dimension of the column space of M . Rank of linear transformations.

57

• ˜ 2 Let P : R → R be projection onto span{u ⃗} where u ⃗= , and let R : R2 → R2 be 3 rotation counter-clockwise by 90◦ . 2

2

57.1 Describe range(P ) and range(R).

■ Use geometric intuition to compute the rank of a linear transformation.

range(P ) = span{u ⃗}, and range(R) = R2 . For P , by the definition of projection P (x⃗ ) is the vector in span{u ⃗} that is closest to ⃗x , so in particular P (x⃗ ) ∈ span{u ⃗} for all ⃗x ∈ R2 . Therefore range(P ) ⊆ span{u ⃗}. On the other hand, P (αu ⃗) = αP (u ⃗) = αu ⃗ for any scalar α, and so range(P ) = span{u ⃗}. For Q, we have that any vector ⃗x ∈ R2 , ⃗x = Q( ⃗y ), where ⃗y is the rotation of ⃗x clockwise by 90◦ . Therefore range(Q) = R2 . 57.2 What is the rank of P and the rank of R?

rank(P ) = 1 and rank(R) = 2. By the previous part, we know range(P ) is 1-dimensional and range(Q) is 2dimensional. 57.3 Let P and R be the matrices corresponding to P and R. What is the rank of P and the

rank of R? rank(P) = 1 and rank(R) = 2. • ˜ • ˜ 0 −1 1 4 6 By Problem 50, P = 13 and R = are the matrices corresponding 6 9 1 0 to P and R. Then we compute:   • ˜ 1 0 1 32 rref(P) = and rref(R) = . 0 1 0 0 These matrices have 1 and 2 pivots, respectively. 41

The goal of this problem is to ■ Apply the definition of rank to compute the rank of a linear transformation.

© Jason Siefken, 2015–2021

■ Relate the rank of a linear transformation to the rank of its matrix.

57.4 Make a conjecture about how the rank of a transformation and the rank of its corre-

sponding matrix relate. Can you justify your claim? They are equal. By Problem 53.3, the range of a transformation is equal to the column space of its corresponding matrix, and therefore the dimensions of these two spaces are equal. In other words, the rank of a transformation is equal to the dimension of the column space of its corresponding matrix. We already know that the dimension of the column space of a matrix is equal to the number of pivots in its reduced row echelon form, and that is by definition the rank of the matrix.

58

• 1 58.1 Determine the rank of (a) 2

˜

• 1 1 (b) 2 3

˜

•

2 1 (c) 4 0

1 0

⎡ ⎤ ⎡ 3 1 0 (d) ⎣3⎦ (e) ⎣0 1 2 0 ˜

0 1 0

⎤ 1 0 ⎦. 1

Rank of matrices. The goal of this problem is to ■ Use the definition of rank to compute the rank of matrices.

For each part, we compute the reduced row echelon form of the matrix and count the number of pivots. • ˜‹ • ˜‹ • ˜ 1 1 1 1 1 1 (a) rank = 1, since rref = has one pivot. 2 2 2 2 0 0 • ˜‹ • ˜‹ • ˜ 1 2 1 2 1 0 (b) rank = 2, since rref = has two pivots 3 4 3 4 0 1 • ˜‹ 1 1 0 (c) rank = 2. This matrix is already in reduced row echelon form, 0 0 1 and has two pivots. ⎛⎡ ⎤ ⎞ ⎛⎡ ⎤ ⎞ ⎡ ⎤ 3 3 1 (d) rank ⎝⎣3⎦⎠ = 1, since rref ⎝⎣3⎦⎠ = ⎣0⎦ has one pivot. 2 2 0 ⎛⎡ ⎤⎞ ⎛⎡ ⎤⎞ ⎡ ⎤ 1 0 1 1 0 1 1 0 0 (e) rank ⎝⎣0 1 0⎦⎠ = 3, since rref ⎝⎣0 1 0⎦⎠ = ⎣0 1 0⎦ has three 0 0 1 0 0 1 0 0 1 pivots. Connect rank to existing concepts. The goal of this problem is to ■ Connect rank(A) to the number of ⃗. solutions to Ax⃗ = 0 ■ Connect rank(A) to linear independence or dependence of the columns of A.

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© Jason Siefken, 2015–2021

59

Consider the homogeneous system x x −x

+2 y +2 y −2 y

+z +3z +z

=0 =0 =0 ⎡

1 and the non-augmented matrix of coefficients A = ⎣ 1 −1

(1)

2 2 −2

⎤ 1 3 ⎦. 1

59.1 What is rank(A)?

⎡

1 rank(A) = 2, since rref(A) = ⎣0 0

2 0 0

⎤ 0 1⎦ as two pivots. 0

59.2 Give the general solution to system (1).

⎡

⎤ −2 ⃗x is a solution to the system if ⃗x = t ⎣ 1⎦ for some real number t. 0 ⎡ ⎤ ⎡ ⎤⎡ ⎤ x 1 2 0 x If ⃗x = ⎣ y ⎦ is a solution to the system, then we must have ⎣0 0 1⎦ ⎣ y ⎦ = z 0 0 0 z ⎡ ⎤ ⎡ ⎤ x + 2y 0 ⎣ z ⎦ = ⎣0⎦, from which it follows that z = 0 and x = −2 y. In other words, 0 0 ⎡ ⎤ −2 any scalar multiple of ⎣ 1⎦ is a solution. 0 59.3 Are the column vectors of A linearly independent?

No. The second column is two times the first column. 59.4 Give a non-homogeneous system with the same coefficients as (1) that has

(a) infinitely many solutions (b) no solutions. (a) x x −x (b) x x −x

+2 y +2 y −2 y +2 y +2 y −2 y

+z +3z +z +z +3z +z

=1 =1 = −1 =0 =0 =1 Connect the rank of a matrix to the linear independence/dependence of its columns. The goal of this problem is to ■ Determine the linear independence/dependence of the columns of a matrix based on its size and rank.

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© Jason Siefken, 2015–2021

60

60.1 The rank of a 3 × 4 matrix A is 3. Are the column vectors of A linearly independent?

No. A 3 × 4 matrix has four columns, each of which are vectors in R3 . It is not possible for four different vectors in R3 to be linearly independent. 60.2 The rank of a 4 × 3 matrix B is 3. Are the column vectors of B linearly independent?

THEOREM

Yes. Since rank(B) = 3, there are three pivots in rref(B). Pivot positions in rref(B) indicate a maximal linearly independent subset of the columns of B. Since there are three columns in B and three pivots, the three columns of B must be linearly independent.

61

Rank-nullity Theorem The nullity of a matrix is the dimension of the null space. The rank-nullity theorem for a matrix A states rank(A) + nullity(A) = # of columns in A.

61.1 Is there a version of the rank-nullity theorem that applies to linear transformations

instead of matrices? If so, state it. Yes. If T : V → W is a linear transformation, then rank(T ) + dim(null(T )) = dim(V ).

Relate linear transformation concepts with matrix concepts. The goal of this problem is to ■ Rephrase the rank-nullity theorem as stated for matrices as the rank-nullity theorem for linear transformations.

If A is the matrix corresponding to T , then rank(T ) = rank(A) by Problem 57.4. null(T ) = null(A) by Problem 54.6, since T = TA, and so dim(null(T )) = nullity(A). Finally, the number of columns of A is equal to the dimension of the domain of T . Apply the rank-nullity theorem. The goal of this problem is to ■ Apply the rank-nullity theorem to compute the rank or nullity of unknown matrices.

62

The vectors u ⃗, ⃗v ∈ R9 are linearly independent and w ⃗ = 2u ⃗ − ⃗v . Define A = [u ⃗|v⃗|w ⃗ ]. 62.1 What is the rank and nullity of A?

rank(A) = 2 and nullity(A) = 1. We know that rank(A) equals the number of pivots in rref(A), which in turn equals the dimension of col(A). Since A has two linearly independent columns, dim(col(A)) = 2. Again, the rank-nullity theorem then says that 2+nullity(A) = 3, and so nullity(A) = 1 62.2 What is the rank and nullity of AT ?

rank(AT ) = 2 and nullity(AT ) = 7. AT is the matrix with rows u ⃗, ⃗v , and w ⃗ . Since w ⃗ = 2u ⃗ − ⃗v , the third row of AT can be reduced to a row of zeros by the row operation R3 ↦→ R3 − 2R1 + R2 . Neither of the first two rows can be reduced to rows of zeros since they are linearly independent. Therefore rref(AT ) has two pivots, meaning rank(AT ) = 2. The rank-nullity theorem then says that 2+nullity(AT ) = 9, and so nullity(AT ) = 7

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© Jason Siefken, 2015–2021

Getting back N 63

“We’ve made a terrible mistake,” a council member says. “Can we go back to the regular N?” Recall the original Italicising N task.

Suppose that the “N” on the left is written in regular 12-point font. Find a matrix A that will transform the “N” into the letter on the right which is written in an italic 16-point font.

Pat and Jamie explained their approach to the Italicizing N task as follows: In order to find the matrix A, we are going to find a matrix that makes the “N” taller, find a matrix that italicizes the taller “N,” and a combination of those two matrices will give the desired matrix A. The Oronto city council has asked you to unitalicise the N. Your new task is to find a matrix C that transforms the “N” on the right to the “N” on the left. 1. Use any method you like to find C. 2. Use a method similar to Pat and Jamie’s method, only use it to find C instead of A.

45

© IOLA Team iola. math. vt. edu

Inverses Elementary matrices.

64.1 Apply the row operation R3 ↦→ R3 + 2R1 to the 3 × 3 identity matrix and call the result

E1 . ⎡

1 ⎣0 0

⎤ ⎡ 0 1 R3 ↦→R3 +2R1 0⎦ −−−−−−−→ ⎣0 1 2

0 1 0

■ Relate elementary matrices to row reduction.

⎤ 0 0⎦ = E1 . 1

0 1 0

The goal of this problem is to ■ Define elementary matrices.

■ Use the “reversibility” of elementary row operations to create inverses to elementary matrices.

64.2 Apply the row operation R3 ↦→ R3 − 2R1 to the 3 × 3 identity matrix and call the result

E2 . ⎡

DEF

1 ⎣0 0

⎤ ⎡ 0 1 R3 ↦→R3 −2R1 0⎦ −−−−−−−→ ⎣ 0 1 −2

0 1 0

0 1 0

⎤ 0 0⎦ = E2 . 1

Elementary Matrix A matrix is called an elementary matrix if it is an identity matrix with a single elementary row operation applied.

⎡

1 A = ⎣4 7

2 5 8

⎤ 3 6⎦ 9

64.3 Compute E1 A and E2 A. How do the resulting matrices relate to row operations?

⎡

1 E1 A = ⎣4 9

2 5 12

⎤ ⎡ 3 1 6⎦ and E2 A = ⎣4 15 5

2 5 4

⎤ 3 6 ⎦. 3

E1 A is the result applying the row operation R3 ↦→ R3 + 2R1 to A, and similarly E2 A is the result of applying the row operation R3 ↦→ R3 − 2R1 to A. 64.4 Without computing, what should the result of applying the row operation R3 ↦→ R3 − 2R1

to E1 be? Compute and verify. It should be the identity matrix, since the row operation R3 ↦→ R3 − 2R1 should undo the operation R3 ↦→ R3 + 2R1 . 64.5 Without computing, what should E2 E1 be? What about E1 E2 ? Now compute and verify.

They should both be the identity matrix. The solution to part 3 above lead us to believe that applying E1 to a matrix has the effect of applying the row operation R3 ↦→ R3 + 2R1 to it. Applying that row operation to E2 would produce the identity matrix, so we expect that E1 E2 should equal the identity matrix. Similar reasoning leads us to believe that E2 E1 should also equal the identity matrix. Indeed, we can compute ⎡

1 ⎣ E1 E2 = 0 2

0 1 0

⎤⎡ 0 1 0⎦ ⎣ 0 1 −2

0 1 0

⎤ ⎡ 0 1 0 ⎦ = ⎣0 1 0

0 1 0

⎤ ⎡ 0 1 0⎦ = ⎣ 0 1 −2

0 1 0

⎤⎡ 0 1 0 ⎦ ⎣0 1 2

0 1 0

⎤ 0 0⎦ = E2 E1 . 1 Apply the definition of inverse matrix.

DEF

64

Matrix Inverse The inverse of a matrix A is a matrix B such that AB = I and BA = I. In this case, B is called the inverse of A and is notated A−1 . 46

© Jason Siefken, 2015–2021

The goal of this problem is to ■ Use the definition of inverse matrix to identify whether two matrices are inverses of each other.

65

Consider the matrices ⎡

1 A=⎣ 0 −3 ⎡

2 1 −6

⎡

B=

⎤ 0 0⎦ 1

−2 1 0

1 D = ⎣0 3

⎤ 0 0⎦ 1

•

1 0

⎡

1 E = ⎣0 0

0 1

0 0 ⎤ 0 0⎦ 1

0 2 1

⎤ 0 1⎦ 0

1 C = ⎣0 0

˜

⎡

1 F = ⎣0 0

⎤ 0 0⎦ 1

0 1 0

65.1 Which pairs of matrices above are inverses of each other?

A and D are inverses of each other, and F is its own inverse. Compute inverses.

66 B=

•

1 0

4 2

The goal of this problem is to ■ Use elementary matrices to compute matrix inverses.

˜

66.1 Use two row operations to reduce B to I2×2 and write an elementary matrix E1 corre-

■ Decompose an invertible matrix into the product of elementary matrices.

sponding to the first operation and E2 corresponding to the second. • ˜ • ˜ • ˜ 1 4 R2 ↦→ 12 R2 1 4 R1 ↦→R1 −4R2 1 0 −−−−→ −−−−−−−→ . 0 2 0 1 0 1   • ˜ 1 0 1 −4 The two elementary matrices are E1 = and E2 = . 0 1 0 12 66.2 What is E2 E1 B?

•

1 E2 E1 B = 0

−4 1

˜ 1 0

0

•

1 2

˜ • 4 1 = 2 0

1 0

˜ 0 . 1

66.3 Find B −1 .

B −1 = E2 E1 =



1 0

 −2 1 2

=

1 2

•

2 0

˜ −4 . 1

By the previous part we already know that (E2 E1 )B = I. We can also check that B(E2 E1 ) = I, meaning E2 E1 is the inverse of B. 66.4 Can you outline a procedure for finding the inverse of a matrix using elementary matrices?

Suppose A is a matrix that can be row reduced to the identity. Let E1 , E2 , . . . , En be the elementary matrices corresponding to the sequence of row operations that reduces A to I. Then as we have seen, we have En En−1 · · · E2 E1 A = I. Thus En En−1 · · · E2 E1 is the inverse of A. Solve systems with inverses.

67 ⎡

1 ⎣ A= 2 1

2 2 3

⎤ −1 4⎦ −3

⎡ ⎤ 1 ⃗b = ⎣2⎦ 3

⎡

⃗] C = [A|b

9 −1 ⎣ −5 A = −2

−3/2 1 1/2

⎤ −5 3⎦ 1

67.1 What is A−1 A?

A−1 A = I. This is true by the definition of an inverse, but we can also verify it by hand. 67.2 What is rref(A)?

⎡

1 rref(A) = ⎣0 0

0 1 0

⎤ 0 0⎦ = I. 1

67.3 What is rref(C)? (Hint, there is no need to actually do row reduction!)

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© Jason Siefken, 2015–2021

The goal of this problem is to ■ Relate inverse matrices to the previous methods for solving equations, row reduction. ■ Symbolically write the solution to a matrix equation using inverses.

⎡ 1  |︁ −1  rref(C) = I |︁A ⃗b = ⎣ 0 0

0 1 0

0 0 1

⎤ −9 6 ⎦. 2

We know that the reduced row echelon form of C must be of the form [I|c⃗] for some ⃗c , and we know that multiplying on the left by A−1 is equivalent to applying the sequence of row operations that reduces A to rref(A) = I. So the same sequence of row ⎡ operations applied to ⃗b, the last column of C, will produce ⎤ −9 the vector ⃗c = A−1 ⃗b = ⎣ 6⎦. 2 67.4 Solve the system Ax ⃗ = ⃗b.

⎡

⎤ −9 The system has one solution: ⃗x = A−1 ⃗b = ⎣ 6⎦. 2 We can read this solution from the reduced row echelon form of the augmented matrix C representing this system. We can also multiply both sides of the equation on the left by A−1 : Ax⃗ = ⃗x =⇒ A−1 Ax⃗ = A−1 ⃗b =⇒ ⃗x = A−1 ⃗b.

68

Inverses and composition.

68.1 For two square matrices X , Y , should (X Y )−1 = X −1 Y −1 ?

No. By the definition of an inverse we need (X Y )−1 (X Y ) = I, so that multiplying by (X Y )−1 undoes multiplication by X Y . To do this, we must first undo multiplication by X , then undo multiplication by Y . That is, we must first multiply by X −1 then multiply by Y −1 .

The goal of this problem is to ■ Create a correct formula for (X Y )−1 and explain it algebraically or in terms of function composition. ■ Relate invertibility of a matrix and its induced transformation.

In other words, we expect that (X Y )−1 = Y −1 X −1 . We can then verify this by computing (X Y )(Y −1 X −1 ) = X Y Y −1 X −1 = X I X −1 = X X −1 = I and (Y −1 X −1 )(X Y ) = Y −1 X −1 X Y = Y −1 I Y = Y −1 Y = I. 68.2 If M is a matrix corresponding to a non-invertible linear transformation T , could M be

invertible? No. Suppose M −1 exists. Then M −1 M = M M −1 = I. Let S be the linear transformation induced by M −1 . Since M is the matrix for T we must have S ◦ T = T ◦ S = id. But then S would be the inverse of T , which is impossible.

More Change of Basis Inverses and change of basis.

69

• ˜ • ˜ ⃗ 1 , ⃗b2 } where ⃗b1 = 1 , ⃗b2 = 1 and let X = [b ⃗ 1 |b ⃗ 2 ] be the matrix whose Let B = {b 1 E −1 E ⃗ 1 ]E and [b ⃗ 2 ]E . columns are [b • ˜ 1 1 69.1 Write down X . 1 −1 69.2 Compute [e⃗1 ]B and [e⃗2 ]B .

[e⃗1 ]B =

1 2

• ˜ 1 and [e⃗2 ]B = 1

1 2

•

˜ 1 . −1

⃗ 1 + ⃗b2 ) and ⃗e2 = 1 (b ⃗ 1 − ⃗b2 ) This is because ⃗e1 = 12 (b 2 48

© Jason Siefken, 2015–2021

The goal of this problem is to ■ Use inverses to answer change-ofbasis questions. ■ Explain why the inverse of a changeof-basis matrix is another change of basis matrix.

69.3 Compute X [e⃗1 ]B and X [e⃗2 ]B . What do you notice?

X [e⃗1 ]B =

1 2

X [e⃗2 ]B =

1 2

•

1 1

•

1 1

˜• ˜ • ˜ • ˜ • ˜ 1 1 1 1 = 12 + 12 = 12 ⃗b1 + 12 ⃗b2 = and 1 1 −1 0 ˜• ˜ • ˜ • ˜ • ˜ 1 1 1 1 0 = 12 − 12 = 12 ⃗b1 − 12 ⃗b2 = −1 −1 1 −1 1 1 −1

We notice that multiplying by X turns the representations of these two vectors in the basis B into representations in the standard basis. 69.4 Find the matrix X −1 . How does X −1 relate to change of basis?

X −1 =

1 2

• 1 1

˜ 1 . −1

X −1 should undo what X does. In the previous part we saw that X takes vectors represented in B and represents them in the standard basis. So X −1 should do the reverse, and take vectors represented in the standard basis and represent them in the basis B.

70

⃗ 1 , ⃗b2 , . . . , ⃗b n } be Let E = {e⃗1 , ⃗e2 , . . . , ⃗e n } be the standard basis for Rn and let B = {b ⃗ 1 |b ⃗ 2 | · · · |b ⃗ n ] to be the matrix whose another basis for Rn . Define the matrix X = [b columns are the ⃗b i vectors written in the standard basis. Notice that X converts vectors from the B basis into the standard basis. In other words,

Inverses and change of basis in arbitrary dimensions. The goal of this problem is to ⃗ 1 | · · · |b ⃗ n ] as a change-of■ Recognize [b basis matrix. ■ Explain why changing basis is an invertible operation.

X [v⃗]B = [v⃗]E .

70.1 Should X −1 exist? Explain.

Yes. X converts vectors from the B basis to the standard basis, and this process can be undone. X −1 is the matrix that does this.

■ Explain how the representation of the standard basis vectors as columns of 0’s and one 1 is a result of representing a vector in its own basis and not something special about the standard basis.

70.2 Consider the equation

X −1 [v⃗]? = [v⃗]? . Can you fill in the “?” symbols so that the equation makes sense? X −1 [v⃗]E = [v⃗]B . As we said in the previous part X −1 should undo what X does, meaning it should convert vectors from the standard basis into the B basis. ⃗ 1 ]B ? How about [b ⃗ 2 ]B ? Can you generalize to [b ⃗ i ]B ? 70.3 What is [b ⎡ ⎤ ⎡ ⎤ 1 0 0 ⎢ ⎥ ⎢1⎥ ⎢ ⎥ ⎢ ⎥ 0 ⃗ 1 ]B = ⎢ ⎥, and [b ⃗ 2 ]B = ⎢0⎥, where each of these vectors have n coordinates. [b ⎢.⎥ ⎢.⎥ ⎣ .. ⎦ ⎣ .. ⎦ 0

0

⃗ i ]B should be the column vector with zeroes in all coordinates except In general, [b for a 1 in the i th coordinate. Representations of transformations.

71

• ˜ • ˜ • ˜ • ˜ 2 5 2 5 3 −5 Let ⃗c 1 = , ⃗c 2 = , C = {c⃗1 , ⃗c 2 }, and A = . Note that A−1 = and 1 E 3 E 1 3 −1 2 that A changes vectors from the C basis to the standard basis and A−1 changes vectors from the standard basis to the C basis. • ˜ • ˜ 1 0 71.1 Compute [c⃗1 ]C and [c⃗2 ]C . [c⃗1 ]C = and [c⃗2 ]C = . 0 1 Let T : R2 → R2 be the linear transformation that stretches in the ⃗c 1 direction by a factor of 2 and doesn’t stretch in the ⃗c 2 direction at all. • ˜ • ˜ • ˜ • ˜ • ˜ 2 5 2 4 5 71.2 Compute T and T . T = Tc⃗1 = 2c⃗1 = and T = Tc⃗2 = ⃗c 2 = 1 E 3 E 1 E 2 E 3 E • ˜ 5 . 3 E 49

© Jason Siefken, 2015–2021

The goal of this problem is to ■ Represent a transformation as a matrix in different bases. ■ Recognize that some bases give nicer matrix representations than others. ■ Connect the definition of similar matrices to change-of-basis.

• ˜ • ˜ 2 0 71.3 Compute [Tc⃗1 ]C and [Tc⃗2 ]C . [Tc⃗1 ]C = [2c⃗1 ]C = and [Tc⃗2 ]C = [c⃗2 ]C = . 0 1 • ˜ α 71.4 Compute the result of T and express the result in the C basis (i.e., as a vector of β C • ˜ ? the form ). ? C • ˜ • ˜ α 2α T = . β C β C • ˜ α If ⃗v is a vector such that [v⃗]C = , then ⃗v = αc⃗1 + β ⃗c 2 . Since T is linear, we β can then compute T ⃗v = T (αc⃗1 + β ⃗c 2 ) = αT (c⃗1 ) + β T (c⃗2 ) = 2αc⃗1 + β ⃗c 2 =

•

˜ 2α . β C

71.5 Find [T ]C , the matrix for T in the C basis.

[T ]C =

•

2 0

˜ 0 . 1

• ˜ • ˜ 1 2 From the results of the previous parts, we know that we must have [T ]C = 0 0 • ˜ • ˜ 0 0 and [T ]C = , so these must be the first and second columns of [T ]C , 1 1 respectively. 71.6 Find [T ]E , the matrix for T in the standard basis.

•

7 [T ]E = 3

˜ −10 −4

There are two methods to determine this. Method 1: Since ⃗e1 = 3c⃗1 − ⃗c 2 and ⃗e2 = −5c⃗1 + 2c⃗2 , we compute [T ⃗e1 ]E = [T (3c⃗1 − ⃗c 2 )]E = 3[T (c⃗1 )]E − [T (c⃗2 )]E = 3

• ˜ • ˜ • ˜ 4 5 7 − = 2 3 3

and [T ⃗e2 ]E = [T (−5c⃗1 +2c⃗2 )]E = −5[T (c⃗1 )]E +2[T (c⃗2 )]E = −5

• ˜ • ˜ • ˜ 4 5 −10 +2 = . 2 3 −4

These two vectors are the respective columns of [T ]E , as usual. Method 2: Since A changes vectors from the C basis to the standard basis and A−1 changes vectors from the standard basis to the C basis, we know [T ]E = A[T ]C A−1 . Using [T ]C from the previous part, we compute

DEFINITION

[T ]E =

•

2 1

5 3

˜•

2 0

0 1

˜•

3 −1

˜ • −5 7 = 2 3

˜ −10 . −4

Similar Matrices The matrices A and B are called similar matrices, denoted A ∼ B, if A and B represent the same linear transformation but in possibly different bases. Equivalently, A ∼ B if there is an invertible matrix X so that A = X BX −1 .

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DEFINITION

Determinants Unit n-cube The unit n-cube is the n-dimensional cube with sides given by the standard basis vectors and lower-left corner located at the origin. That is ¨ « n ∑︂ n Cn = ⃗x ∈ R : ⃗x = αi ⃗e i for some α1 , . . . , αn ∈ [0, 1] = [0, 1]n . i=1

The sides of the unit n-cube are always length 1 and its volume is always 1.

72

The picture shows what the linear transformation T does to the unit square (i.e., the unit 2-cube).

2

Volumes of images. The goal of this problem is to ■ Apply the definitions of unit n-cube and image of a set. ■ Use tools from outside of linear algebra class to compute the area of a polygon.

2

■ Be comfortable using the word “volume” in R2 .

1

1

1

2

1

2

• ˜ • ˜ • ˜ 1 0 1 72.1 What is T ,T ,T ? 0 1 1 • ˜   • ˜ 1 • ˜ 3 1 1 0 1 T = 1 ,T = 2 , and T = 23 ? 0 1 1 1 2 2 We can see first two directly in the picture. Using the linearity of T , we can compute T

• ˜ • ˜ • ˜‹ • ˜ • ˜   1 3 1 1 1 0 1 0 =T + =T +T = 1 + 2 = 23 . 1 0 1 0 1 1 2 2

72.2 Write down a matrix for T .

 The matrix for T in the standard basis is

1 2

1 1 2

1

 .

72.3 What is the volume of the image of the unit square (i.e., the volume of T (C2 ))? You may

use trigonometry.

DEF

The volume is 34 .

Determinant The determinant of a linear transformation T : Rn → Rn , denoted det(T ) or |T |, is the oriented volume of the image of the unit n-cube. The determinant of a square matrix is the determinant of its induced transformation. Apply the definition of determinant.

73

We know the following about the transformation A: A

• ˜ • ˜ 1 2 = 0 0

and

51

A

• ˜ • ˜ 0 1 = . 1 1

© Jason Siefken, 2015–2021

The goal of this problem is to ■ Compute a determinant from the definition. ■ Practice finding the area of a parallelogram.

73.1 Draw C2 and A(C2 ), the image of the unit square under A.

2

A(C2 )

1

1

2

3

73.2 Compute the area of A(C2 ). The area of this parallelogram is 2. 73.3 Compute det(A).

det(A) = 2. The parallelogram with sides

• ˜ • ˜ 2 0 and is positively oriented, so det(A) = +2. 0 1 Apply the definition of determinant. The goal of this problem is to ■ Compute a determinant from the definition by applying geometric reasoning.

74

Suppose R is a rotation counter-clockwise by 30◦ . 74.1 Draw C2 and R(C2 ).

2 R(C2 ) 1

1

2 Apply the definition of determinant.

74.2 Compute the area of R(C2 ).

The area is 1. R rotates the entire unit square, which does not change its area. 74.3 Compute det(R).

Since R preserves orientation, det(R) must be positive. Since R does not change the area of the unit square, det(R) = +1.

52

© Jason Siefken, 2015–2021

The goal of this problem is to ■ Compute a determinant from the definition when orientation is reversed.

75

We know the following about the transformation F : • ˜ • ˜ 1 0 F = 0 1

and

• ˜ • ˜ 0 1 F = . 1 0

75.1 What is det(F )?

det(F ) = −1. F does not change the area of the unit square, but reverses its orientation, so det(F ) = −1.

Relate determinants of transformations and matrices.

THM

Volume Theorem I For a square matrix M , det(M ) is the oriented volume of the parallelepiped (ndimensional parallelogram) given by the column vectors of M .

THM

The goal of this problem is to ■ Relate the image of the unit cube under a transformation T to the columns of T ’s matrix representation.

Volume Theorem II For a square matrix M , det(M ) is the oriented volume of the parallelepiped (ndimensional parallelogram) given by the row vectors of M .

■ Relate the determinant of a matrix and its transpose.

Determinants and areas.

76

76.1 Explain Volume Theorem I using the definition of determinant.

If T : Rn → Rn is a transformation with standard matrix M = [c⃗1 | · · · |c⃗n ], then T (e⃗i ) (represented in the standard basis) will be ci , the ith column of M . The image of the unit cube will be a parallelepiped with sides T (e⃗1 ) = ⃗c 1 , . . . , T (e⃗n ) = ⃗c n , and so det(T ) will be the oriented volume of the parallelepiped with sides given by ⃗c 1 , . . . , ⃗c n . 76.2 Based on Volume Theorems I and II, how should det(M ) and det(M T ) relate for a square

matrix M ? det(M ) = det(M T ). Since the transpose switches columns for rows, this is an immediate consequence of Volume Theorems I and II.

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The goal of this problem is to ■ Use determinants to compute areas/volumes of images of arbitrary sets. ■ See determinants as a “change of area/volume” factor. ■ Explain the multiplicative property of determinants in terms of area/volume changes.

77 1 R4

Let R = R1 ∪ R2 ∪ R3 ∪ R4 . You know the following about the linear transformations M , T , and S.

R3

• ˜ • ˜ x 2x M = y y R1

R2

T : R2 → R2 has determinant 2 S : R2 → R2 has determinant 3 1

77.1 Find the volumes (areas) of R1 , R2 , R3 , R4 , and R.

The volumes of R1 , R2 , R3 , and R4 are 14 . The volume of R is 1. 77.2 Compute the oriented volume of M (R1 ), M (R2 ), and M (R).

The oriented volumes of M (R1 ) and M (R2 ) are M (R) is 2.

1 2

and the oriented volume of

77.3 Do you have enough information to compute the oriented volume of T (R2 )? What about

the oriented volume of T (R + {e⃗2 })?

Yes. The oriented volume of T (R2 ) =

1 2

and the oriented volume of T (R+{e⃗2 }) = 2.

We don’t have enough information to determine what T (R2 ) looks like, but we do know (i) T (R2 ) will be a parallelogram, (ii) T (R) has oriented volume 2, and (iii) T (R) is made of four translated copies of T (R2 ). From
this we deduce that the oriented volume of T (R2 ) = 14 oriented volume of T (R) = ( 14 )(2). To find the oriented volume of T (R + {e⃗2 }), we use linearity to observe T (R + {e⃗2 }) = T (R) + T ({e⃗2 }) = T (R) + {T ⃗e2 }. This shows that T (R + {e⃗2 }) is just a translation of T (R) and therefore has the same oriented volume. 77.4 What is the oriented volume of S ◦ T (R)? What is det(S ◦ T )?

They are both equal to 6. S ◦ T (R) = S(T (R)). We already know T (R) has a volume of 2, and so S(T (R)) has a volume of 6, since S scales the volumes of all regions by 3. The oriented volume of S ◦ T (R) is the determinant of S ◦ T by definition. Determinants of elementary matrices.

78 • E f is I3×3 with the first two rows swapped.

The goal of this problem is to ■ Memorize the determinant of each type of elementary matrix.

• Em is I3×3 with the third row multiplied by 6.

■ Justify why the determinant of an elementary matrix of type “add a multiple of one row to another” is always 1.

• Ea is I3×3 with R1 ↦→ R1 + 2R2 applied. 78.1 What is det(E f )?

■ Outline a method to compute determinants of arbitrary matrices.

det(E f ) = −1. det(I3×3 ) = 1, and swapping one pair of rows of a matrix changes the sign of its determinant. 78.2 What is det(Em )?

det(Em ) = 6. Multiplying one row of a matrix by a constant multiplies its determinant by the same constant. 78.3 What is det(Ea )?

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det(Ea ) = 1. Adding a multiple of one row of a matrix to another row has no effect on its determinant. 78.4 What is det(E f Em )? det(E f Em ) = det(E f ) det(Em ) = (−1)(6) = −6. 78.5 What is det(4I3×3 )? det(4I3×3 ) = 43 = 64. 78.6 What is det(W ) where W = E f Ea E f Em Em ?

det(W ) = det(E f ) det(Ea ) det(E f ) det(Em ) det(Em ) = (−1)(1)(−1)(6)(6) = 36.

79

⎡

1 ⎢0 U =⎣ 0 0

2 3 0 0

1 −2 −1 0

Reasoning about determinants via elementary matrices.

⎤ 2 4⎥ 0⎦ 4

The goal of this problem is to ■ Develop a shortcut for computing determinants of triangular matrices. ■ Reason about the determinant of a matrix when given its reduced row echelon form.

79.1 What is det(U)?

det(U) = −12. 79.2 V is a square matrix and rref(V ) has a row of zeros. What is det(V )?

det(V ) = 0.

80

80.1 V is a square matrix whose columns are linearly dependent. What is det(V )?

det(V ) = 0. §• ˜ª −1 . What is det(P)? −1

80.2 P is projection onto span

Determinants of singular matrices. The goal of this problem is to ■ Reason geometrically about why a transformation that isn’t one-to-one has a zero determinant.

det(P) = 0. The image of the unit square under P is a line segment, which has zero volume.

81

Determinants and invertibility.

Suppose you know det(X ) = 4.

The goal of this problem is to ■ Produce the determinant of X −1 give the determinant of X .

81.1 What is det(X −1 )?

det(X −1 ) = 14 .

■ Explain why a transformation T is

We know that X X −1 = I. Therefore we must have that det(X X −1 ) = det(X ) det(X −1 ) = invertible if and only if det(T ) ̸= 0. det(I) = 1, and so det(X −1 ) = 14 . 81.2 Derive a relationship between det(Y ) and det(Y −1 ) for an arbitrary matrix Y .

det(Y −1 ) =

1 det(Y ) .

Using the same reasoning as the previous part, we know that Y Y −1 = I. Therefore we must have det(Y ) det(Y −1 ) = det(Y Y −1 ) = det(I) = 1, and so det(Y −1 ) = 1 det(Y ) . 81.3 Suppose Y is not invertible. What is det(Y )?

det(Y ) = 0. If Y is not invertible, it has linearly dependent columns. Therefore the parallelepiped formed by the columns of Y will be “flattened” and have zero volume. This is consistent with our previous findings. For a square matrix Y , det(Y −1 ) = 1 det(Y ) . This formula always works, except when det(Y ) = 0.

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The Green and the Black 82

The subway system of Oronto is laid out in a skewed grid. All tracks run parallel to one of the green lines shown. Compass directions are given by the black lines. While studying the subway map, you decide to pick two bases to help: the green basis G = {g ⃗ 1 , ⃗g 2 }, and the black basis B = {e⃗1 , ⃗e2 }. ⃗f

a ⃗ d⃗

⃗b

⃗e

⃗c

1. Write each point above in both the green and the black bases. 2. Find a change-of-basis matrix X that converts vectors from a green basis representation to a black basis representation. Find another matrix Y that converts vectors from a black basis representation to a green basis representation. 3. The city commission is considering renumbering all the stops along the y = −3x direction. You deduce that the commission’s proposal can be modeled by a linear transformation. Let T : R2 → R2 be the linear transformation that stretches in the y = −3x direction by a factor of 2 and leaves vectors in the y = x direction fixed. Describe what happens to the vectors u ⃗, ⃗v , and w ⃗ when T is applied given that [u ⃗ ]G =

• ˜ 6 1

[v⃗]G =

•

4 −3

˜

[w ⃗ ]B =

•

˜ −8 . −7

4. When working with the transformation T , which basis do you prefer vectors be represented in? What coordinate system would you propose the city commission use to describe their plans?

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DEFINITION

Eigenvectors

83

Eigenvector Let X be a linear transformation or a matrix. An eigenvector for X is a non-zero vector ⃗ is an eigenvector for that doesn’t change directions when X is applied. That is, ⃗v ̸= 0 X if X ⃗v = λv⃗ for some scalar λ. We call λ the eigenvalue of X corresponding to the eigenvector ⃗v .

The picture shows what the linear transformation T does to the unit square (i.e., the unit 2-cube).

2

2

1

1

1

Apply the definition of eigenvector/value geometrically. The goal of this problem is to ■ Find eigenvectors/values from transformations defined geometrically. ■ Produce new eigenvectors from existing ones by scaling.

2

1

2

83.1 Give an eigenvector for T . What is the eigenvalue?

• ˜ 1 is an eigenvector for T , with corresponding eigenvalue 32 . 1 • ˜ 3 • ˜ 1 1 We can see from the image that T = 23 = 32 . 1 1 2 ⃗v =

83.2 Can you give another?

• ˜ 1 is also an eigenvector for T . 1  • ˜‹ • ˜ • ˜ • ˜ 1 1 1 1 For any scalar α, we have T α = αT = α 32 , meaning α is an 1 1 1 1 eigenvector for T with eigenvalue 32 α. • ˜ • ˜ • ˜ −1 −1 −1 More interestingly, since T = 12 , is an eigenvector with corre1 1 1 sponding eigenvalue 12 . Any scalar multiple of

84

Apply the definition of eigenvector/value algebraically.

For some matrix A, ⎡ ⎤ ⎡ ⎤ 3 2 A ⎣ 3⎦ = ⎣ 2⎦ 1 2/3

and

B = A − 23 I.

84.1 Give an eigenvector and a corresponding eigenvalue for A.

⎡ ⎤ 3 ⎣3⎦ is an eigenvector for A, with corresponding eigenvalue 2 . 3 1 ⎡ ⎤ 3 84.2 What is B ⎣3⎦? 1 ⎡ ⎤ ⎡ ⎤ 3 0 B ⎣3⎦ = ⎣0⎦ . 1 0 57 © Jason Siefken, 2015–2021

The goal of this problem is to ■ Identify numerically whether a vector is an eigenvector. ■ Use numerical evidence to compute an eigenvalue. ■ Reason about the matrix A− λI given λ is an eigenvalue.

We compute ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 3 3 3 3 2 2 0 2 2 B ⎣3⎦ = (A − I) ⎣3⎦ = A ⎣3⎦ − I ⎣3⎦ = ⎣ 2⎦ − ⎣ 2⎦ = ⎣0⎦ . 3 3 1 1 1 1 2/3 2/3 0 84.3 What is the dimension of null(B)?

The most we can say is that nullity(B) ≥ 1. ⎡ ⎤ 3 We know ⎣3⎦ ∈ null(B) by the previous part, and so the dimension of null(B) is 1 at least 1. It could be larger, but we do not have enough information to say for sure. 84.4 What is det(B)? det(B) = 0. Explore the matrix C − λI .

• −1 Let C = 1

85

The goal of this problem is to ■ Relate the matrix C − λI to the problem of finding eigenvectors/values.

˜

2 and Eλ = C − λI. 0

85.1 For what values of λ does Eλ have a non-trivial null space?

λ = −2 and λ = 1.

■ Relate the equation C ⃗x = λx⃗ to the null space of the matrix Eλ = C − λI .

Eλ has a non-trivial null space exactly when its determinant is zero. We compute:

■ Use the determinant to determine when a parameterized family of matrices is invertible or not.

• −1 − λ det(Eλ ) = det 1

■ Numerically compute eigenvalues/vectors without extra geometric information.

2 −λ

˜‹

= (−1−λ)(−λ)−(1)(2) = λ2 +λ−2 = (λ−1)(λ+2).

This equals zero exactly when λ = −2 or λ = 1. 85.2 What are the eigenvalues of C?

−2 and 1. ⃗. Thus The scalar λ is an eigenvalue of C if and only if C ⃗v = λv⃗ for some ⃗v ̸= 0 ⃗ = C ⃗v − λv⃗ = (C − λI)v⃗ = Eλ ⃗v , 0 and so Eλ has a non-trivial null space if and only if λ is an eigenvalue of C. 85.3 Find the eigenvectors of C.

•

˜ • ˜ −2 1 and (along with all non-zero scalar multiples of these). 1 1

We know from the previous part that finding an eigenvector with corresponding eigenvalue −2 amounts to finding the non-zero vectors in the null space of E−2 = • ˜ 1 2 . Computing, 1 2 §• ˜ª −2 null(E−2 ) = span . 1

DEFINITION

Similarly, the eigenvectors with corresponding eigenvalue 1 are the non-zero • ˜ −2 2 vectors in the null space of E1 = , and we compute that null(E1 ) = 1 −1 §• ˜ª 1 span . 1 Apply the definition of the characteristic polynomial.

Characteristic Polynomial For a matrix A, the characteristic polynomial of A is

The goal of this problem is to ■ Compute a characteristic polynomial by applying the definition.

char(A) = det(A − λI).

■ Relate the characteristic polynomial to eigenvalues.

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86

•

1 Let D = 3

˜ 2 . 0

86.1 Compute char(D).

char(D) = (−2 − λ)(3 − λ). We compute: det(D−λI) = det

• 1−λ 3

˜‹ 2 = (1−λ)(−λ)−(2)(3) = λ2 −λ−6 = (−2−λ)(3−λ). −λ

86.2 Find the eigenvalues of D.

The eigenvalues of D are −2 and 3. The eigenvalues of D are the roots of char(D).

87

Suppose char(E) = −λ(2 − λ)(−3 − λ) for some unknown 3 × 3 matrix E. 87.1 What are the eigenvalues of E?

Getting information from the characteristic polynomial. The goal of this problem is to ■ Use the characteristic polynomial to determine eigenvalues and invertibility of matrices.

0, 2, and −3. The eigenvalues of E are the roots of char(E).

■ Relate the characteristic polynomial to determinants.

87.2 Is E invertible?

No. Since 0 is an eigenvalue of E, there must be a non-zero vector ⃗v such that E ⃗v = ⃗. This means nullity(E) > 0, which implies E is not invertible. 0v⃗ = 0 87.3 What can you say about nullity(E), nullity(E − 3I), nullity(E + 3I)?

nullity(E) = 1, nullity(E − 3I) = 0, nullity(E + 3I) = 1 Notice that evaluating the characteristic polynomial at λ give the determinant of E − λI. From this, we can determine the invertibility of any matrix of the form E − λI. Since E is not invertible, nullity(E) ≥ 1. Since E −3I is invertible, nullity(E −3I) = 0, and since E + 3I is not invertible, nullity(E + 3I) ≥ 1. To pin down the nullities of E and E + 3I exactly takes more work. E has three distinct eigenvalues and so E must have three linearly independent eigenvectors ⃗v 0 , ⃗v 2 , ⃗v −3 . Further, span{v⃗ i } ⊆ null(E − λi I). Since ⃗v 0 , ⃗v 1 , ⃗v −3 ∈ R3 , it must be the case that null(E − λi I) is one dimensional for i ∈ {0, 2, −3}. Eigen bases.

88

Consider ⎡

1 A = ⎣0 1

0 1 1

⎤ 1 1⎦ 0

⎡ ⎤ 1 ⃗v 1 = ⎣1⎦ 1

⎡

⎤ 1 ⃗v 2 = ⎣ 1⎦ −2

⎡

⎤ −1 ⃗v 3 = ⎣ 1⎦ 0

and notice that ⃗v 1 , ⃗v 2 , ⃗v 3 are eigenvectors for A. Let TA be the transformation induced by A. 88.1 Find the eigenvalues of TA.

The eigenvalues of TA are 2, −1, and 1. We compute that TA ⃗v 1 = 2v⃗1 , TA ⃗v 2 = −v⃗2 , and TA ⃗v 3 = ⃗v 3 , so 2, −1, and 1 are eigenvalues of TA. By the last part of the previous problem, there are no other eigenvalues. 88.2 Find the characteristic polynomial of TA.

char(A) = (2 − λ)(1 − λ)(−1 − λ) = −(λ − 2)(λ + 1)(λ − 1). Sine we know the roots of the characteristic polynomial are the eigenvalues and we know char(TA) is a cubic, we can immediately write down char(TA) = (2 − λ)(1 − λ)(−1 − λ) without computing a determinant. 88.3 Compute TA w ⃗ where w = 2v⃗1 − ⃗v 2 .

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The goal of this problem is to ■ Compute eigenvalues when given eigenvectors. ■ Compute a characteristic polynomial without using a determinant when given eigenvalues. ■ Compute the result of a transformation when vectors are written in an eigen basis.

TA w ⃗ = 4v⃗1 + ⃗v 2 . Using the computations we did in the first part above, we find TA w ⃗ = TA(2v⃗1 − ⃗v 2 ) = 2TA ⃗v 1 − Av⃗2 = 2(2v⃗1 ) − (−v⃗2 ) = 4v⃗1 + ⃗v 2 . 88.4 Compute TAu ⃗ where u ⃗ = av⃗1 + bv⃗2 + c ⃗v 3 for unknown scalar coefficients a, b, c.

TAu ⃗ = 2av⃗1 − bv⃗2 + c ⃗v 3 . Using the same reasoning as the previous part, we compute TAu ⃗ = aAv⃗1 + bAv⃗2 + cAv⃗3 = 2av⃗1 − bv⃗2 + cv⃗3 . Notice that V = {v⃗1 , ⃗v 2 , ⃗v 3 } is a basis for R3 . ⎡ ⎤ 1 88.5 If [x ⃗ ]V = ⎣3⎦ is ⃗x written in the V basis, compute TA ⃗x in the V basis. 4 ⎡ ⎤ 2 [TA ⃗x ]V = ⎣−3⎦. 4 ⎡ ⎤ 1 If [x⃗ ]V = ⎣3⎦, then ⃗x = ⃗v 1 + 3v⃗2 + 4v⃗3 . Using the previous part, we then have 4 ⎡ ⎤ 2 that TA ⃗x = 2v⃗1 − 3v⃗2 + 4v⃗3 , so [TA ⃗x ]V = ⎣−3⎦. 4

89

Recall from Problem 88 that ⎡ ⎤ 1 0 1 A = ⎣0 1 1 ⎦ 1 1 0

Diagonalizing matrices.

⎡ ⎤ 1 ⃗v 1 = ⎣1⎦ 1

⎡

⎤ 1 ⃗v 2 = ⎣ 1⎦ −2

⎡

⎤ −1 ⃗v 3 = ⎣ 1⎦ 0

and V = {v⃗1 , ⃗v 2 , ⃗v 3 }. Let TA be the transformation induced by A and let P = [v⃗1 |v⃗2 |v⃗3 ] be the matrix with columns ⃗v 1 , ⃗v 2 , and ⃗v 3 (written in the standard basis). 89.1 Describe in words what P and P −1 do in terms of change-of-basis.

P takes vectors in the V -basis representation and outputs vectors in the standard basis. P −1 does the opposite of what P does. That is, it takes vectors written in the standard basis and rewrites them in the V basis. 89.2 If you were asked to compute TA ⃗ y for some ⃗y ∈ R3 , which basis would you prefer to do

your computations in? Explain. Computing TA ⃗y in the V basis takes only three multiplications. Computing TA ⃗y in the E basis takes nine multiplications and three additions. So, the answer depends on how much you enjoy multiplying numbers. 89.3 Given a vector ⃗ y ∈ R3 written in the standard basis, is there a way to compute TA ⃗y

without using the matrix A? (You may use P and P −1 , just not A.) Explain.

Yes! Computing [TA]V [ ⃗y ]V is easy, since [TA]V just multiplies each coordinate of [ ⃗y ]V by a scalar. We know that P −1 [ ⃗y ]E = [ ⃗y ]V and that P[x⃗ ]V = [x⃗ ]E and so given any vector ⃗v represented by [v⃗]E in the standard basis, we have A[v⃗]E = P[TA]V P −1 [v⃗]E since P[TA]V P −1 [v⃗]E = P[TA]V [v⃗]V = P[TA ⃗v ]V = [TA ⃗v ]E = A[v⃗]E . 89.4 Can you find a matrix D so that

P DP −1 = A?

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The goal of this problem is to ■ Diagonalize a matrix. ■ Explain diagonalization in terms of change of basis. ■ Use diagonalization to compute large matrix powers.

⎡

2 D = ⎣0 0

0 −1 0

⎤ 0 0 ⎦. 1

D = [A]V , so by the previous part we have that that for any vector ⃗v A[v⃗]E = P DP −1 [v⃗]E . ⎡ ⎤ 1 89.5 [x ⃗ ]V = ⎣3⎦. Compute TA100 ⃗x . Express your answer in both the V basis and the standard 4 basis. ⎡ ⎤ 2100 TA100 ⃗x = ⎣ 3⎦ . 4 V

By the previous problem, we know how TA acts on vectors represented in the V basis: it multiplies the first coordinate by 2, the second by −1, and leaves the third coordinate unchanged. So we compute ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 2 22 23 100 ⎣ ⎦ 99 ⎣ 98 ⎣ ⎦ 97 ⎣ ⎦ 3 = TA −3 = TA 3 = TA −3⎦ = . . . TA 4 4 4 4

DEF

To express TA100 ⃗x in the standard basis, we use P. ⎡ ⎤ 2100 − 1 [TA100 ⃗x ]E = P[TA100 ⃗x ]V = ⎣2100 + 7⎦ . 2100 − 6

90

Diagonalizable A matrix is diagonalizable if it is similar to a diagonal matrix.

Let B be an n × n matrix and let TB be the induced transformation. Suppose TB has eigenvectors ⃗v 1 , . . . , ⃗v n which form a basis for Rn , and let λ1 , . . . , λn be the corresponding eigenvalues. 90.1 How do the eigenvalues and eigenvectors of B and TB relate?

The eigenvalues of B and TB are the same. The eigen vectors are also the same, except eigenvectors for B must be written in the standard basis. E.g., [v⃗1 ]E , . . . , [v⃗ n ]E are eigenvectors for B. 90.2 Is B diagonalizable (i.e., similar to a diagonal matrix)? If so, explain how to obtain its

diagonalized form.

Eigenvectors and diagonalization. The goal of this problem is to ■ Explain how the existence of a basis of eigenvectors implies diagonalizability. ■ Explain how if there isn’t a basis of eigenvectors a matrix is not diagonalizable. ■ Not confuse diagonalizability and invertibility.

Yes. V = {v⃗1 , . . . ⃗v n } is a basis consisting of eigenvectors for TB . By definition, B[v⃗ i ]E = λi [v⃗ i ]E for each i.

Let P be the matrix that takes vectors represented in the V basis and outputs their representations in the standard basis E . Then, for example, we should have that ⎡ ⎤ ⎡ ⎤ 1 λ1 0 ⎢ ⎥ ⎢0⎥ ⎥ = P −1 BP[v⃗ ] = P −1 B[v⃗ ] = λ P −1 [v⃗ ] = λ [v⃗ ] = ⎢ . ⎥ . P −1 BP ⎢ . 1 V 1 E 1 1 E 1 1 V ⎣.⎦ ⎣ . ⎦ . . 0 0 ⎡ ⎤ λ1 ⎢0⎥ th ⎥ Therefore, the first column of P −1 BP is ⎢ ⎣ .. ⎦. By similar reasoning, the i column . 0 −1 of P BP consists of all zeroes except for λi in the i th position. In other words, B is similar to the diagonal matrix D with λ1 , λ2 , . . . , λn along the diagonal, in that order. 61

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90.3 What if one of the eigenvalues of TB is zero? Would B be diagonalizable?

Yes. The argument in the previous part does not depend on any of the eigenvalues being non-zero. 90.4 What if the eigenvectors of TB did not form a basis for Rn . Would B be diagonalizable?

No. The argument we used in the first part definitely would not work. Consider the converse, and assume B is similar to diagonal matrix D. That is, suppose there is an invertible matrix P such that B = P DP −1 . Then, if ⃗v 1 is the first column of P and λ1 is the first entry on the diagonal of D, we would have ⎡ ⎤ ⎡ ⎤ 1 1 ⎢ 0⎥ ⎢ 0⎥ ⎥ ⎢ ⎥ B[v⃗1 ]E = P DP −1 [v⃗1 ]E = P D[v⃗1 ]V = P D ⎢ ⎣ .. ⎦ = λ1 P ⎣ .. ⎦ = λ1 [v⃗1 ]E , . . 0

0

DEFINITION

meaning that [v⃗1 ]E is an eigenvector of B. Similarly, all of the columns of P would be eigenvectors of B, with eigenvalues equal to the corresponding entry on the diagonal of D. Since P is invertible, its columns must be linearly independent, and therefore the n columns of P would form a basis of Rn consisting of eigenvectors of B.

91

Eigenspace Let A be an n × n matrix with eigenvalues λ1 , . . . , λm . The eigenspace of A corresponding to the eigenvalue λi is the null space of A − λi I. That is, it is the space spanned by all eigenvectors that have the eigenvalue λi . The geometric multiplicity of an eigenvalue λi is the dimension of the corresponding eigenspace. The algebraic multiplicity of λi is the number of times λi occurs as a root of the characteristic polynomial of A (i.e., the number of times x − λi occurs as a factor).

Let F =

•

3 0

˜ • 1 3 and G = 3 0

Non-diagonalizable matrices.

˜ 0 . 3

The goal of this problem is to ■ Explain why not all matrices are diagonalizable.

91.1 Is F diagonalizable? Why or why not?

■ Memorize an example of a nondiagonalizable matrix.

No. F is diagonalizable if and only if there is a basis of R2 consisting of eigenvectors of F , so we begin by computing all eigenvectors of F . char(F ) = (3 − λ)2 , so the only eigenvalue of F is 3, meaning that the eigenvectors of F are precisely the non-zero vectors in null(F − 3I). We check that null(F − 3I) = null

• 0 0

1 0

˜‹

= span

§• ˜ª 1 , 0

which is one-dimensional, and so there cannot be a basis of R2 consisting of eigenvectors of F . 91.2 Is G diagonalizable? Why or why not?

Yes. G is already diagonal (and is necessarily similar to itself). 91.3 What are the geometric and algebraic multiplicities of each eigenvalue of F ? What about

the multiplicities for each eigenvalue of G? The only eigenvalue for F is 3. Its geometric multiplicity is 1, and its algebraic multiplicity is 2. The only eigenvalue for G is 3. Its geometric and algebraic multiplicity is 2. 91.4 Suppose A is a matrix where the geometric multiplicity of one of its eigenvalues is smaller

than the algebraic multiplicity of the same eigenvalue. Is A diagonalizable? What if all the geometric and algebraic multiplicities match? 62

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If one of the geometric multiplicities is smaller than the corresponding algebraic multiplicity, A cannot be diagonalizable. Since the characteristic polynomial of an n × n matrix has degree n, it has at most n real roots. Since each root is an eigenvalue, we have ∑︂ algebraic multiplicities ≤ n. If one of the geometric multiplicities is smaller than the algebraic multiplicities, we have ∑︂ geometric multiplicities < n, and so there cannot be a basis for Rn consisting of eigenvectors. For the converse statement, we need the fundamental theorem of algebra: a degree n polynomial has exactly n complex roots, counting multiplicity. If we allow eigenvalues to be complex numbers, then ∑︂ algebraic multiplicities = n, and so if all geometric and algebraic multiplicities are equal, we have ∑︂ geometric multiplicities = n. Thus, there would be a basis of eigenvectors.

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