MATH 136 & 235 Course Notes PDF

Preview

Summary

ContentsA Note to Students - READ THIS!.................. iiiHow To Make The Most Of This Book: SQ3R............ viii 1 Vectors in Euclidean Space Acknowledgments ix 1 Vector Addition and Scalar Multiplication 1 Bases 1 Subspaces 1 Dot Product, Cross Product, and Scalar Equations 1 Projections 2 Sys...

Description

Section 3.2 Linear Mappings

91

Our second interpretation of matrix vector multiplication says that a matrix times a vector gives us a linear combination of the columns of the matrix. Using this in reverse, we see that   x h i  .1  L(~x) = x1 L(~e1 ) + · · · + xn L(~en ) = L(~e1 ) · · · L(~en )  ..    x n

We have proven the following theorem.

THEOREM 2

Every linear mapping L : Rn → Rm can be represented as a matrix mapping with matrix whose i-th column is the image of the i-th standard basis vector of Rn under L for all 1 ≤ i ≤ n. That is, L(~x ) = [L]~x where h i [L] = L(~e1 ) · · · L(~en )

DEFINITION

h Let L : Rn → Rm be a linear mapping. The matrix [L] = L(~e1 ) · · · the standard matrix of L. It satisfies L(~x ) = [L]~x .

Standard Matrix

i L(~en ) is called

REMARK We are calling this the standard matrix of L since it is based on the standard basis. In Chapters 6 and 8, we will see that we can find the matrix representation of a linear mapping with respect to different bases.

EXAMPLE 5

Determine the standard matrix of L(x1 , x2 , x3 ) = (3 x1 − x2 , 2x1 + 2x3 ). Solution: By definition, the columns of the standard matrix [L] are the images of the standard basis vectors of R3 under L. We have L(1, 0, 0) = (3, 2) L(0, 1, 0) = (−1, 0) L(0, 0, 1) = (0, 2) Thus, " # 3 −1 0 [L] = L(~e1 ) L(~e2 ) L(~e3 ) = 2 0 2 h

i

92

Chapter 3 Matrices and Linear Mappings

REMARK Be careful to remember that when we write L(1, 0, 0) = (3, 2), we really mean   " # 1  3 so that L(~e1 ) is in fact a column of [L]. L(~e1 ) = L 0  = 2   0

EXAMPLE 6

"# 1 Let ~a = . Determine the standard matrix of proj~a : R2 → R2 . 2 Solution: We need to find the projection of the standard basis vectors of R2 onto ~a. " # " # 1 1 ~e1 · ~a 1/5 ~a = proj~a (~e1 ) = = 2/5 k~a k2 5 2 " # " # ~e2 · ~a 2 1 2/5 = proj~a (~e2 ) = ~a = 2 4/5 k~a k 5 2 h i "1/5 2/5# Hence, [proj~a ] = proj~a (~e1 ) proj~a (~e2 ) = . 2/5 4/5

EXERCISE 2

Determine the standard matrix of the linear mapping L( x1 , x2 , x3 ) = ( x1 + 4 x2 , x1 − x2 + 2 x3 )

We now look at two important examples of linear mappings that will be used later in the book. Rotations in R2 Let Rθ : R2 → R2 denote the function that rotates a vector ~x ∈ R2 about the origin through an angle θ. Using basic trigonometric identities, we can show that Rθ (x1 , x2 ) = ( x1 cos θ − x2 sin θ, x1 sin θ + x2 cos θ) and that Rθ is linear. Using this we get that the standard matrix of Rθ is " # cos θ − sin θ [Rθ ] = sin θ cos θ

x2 Rθ (~x)

θ

~x x1

Section 3.2 Linear Mappings

EXAMPLE 7

93

Determine the standard matrix of Rπ/4 . Solution: We have √ # " # " √ cos π/4 − sin π/4 1/ √2 −1/√ 2 [Rπ/4 ] = = sin π/4 cos π/4 1/ 2 1/ 2 We call Rθ a rotation in R2 and we call its standard matrix [Rθ ] a rotation matrix. The following theorem shows some important properties of a rotation matrix.

THEOREM 3

If Rθ : R2 → R2 is a rotation with rotation matrix A = [Rθ ], then the columns of A are orthogonal unit vectors. # # " " cos θ − sin θ Proof: The columns of A are ~a1 = and ~a2 = . Hence, sin θ cos θ ~a1 · ~a2 = − cos θ sin θ + cos θ sin θ = 0 k~a1 k2 = cos2 θ + sin2 θ = 1

k~a2 k2 = cos2 θ + sin2 θ = 1 

Reflections Let refl P : Rn → Rn denote the mapping that sends a x2 vector ~x to its mirror image in the hyperplane P with normal vector ~n. The figure shows a reflection over a line in R2 with normal vector ~n. From the figure, we see that the reflection of ~x over the line with normal ~n vector ~n is given by refl P (~x) = ~x − 2 proj~n (~x)

~x

θ θ

We extend this formula directly to the n-dimensional case.

EXAMPLE 8

Prove that refl P : Rn → Rn is a linear mapping. Solution: Since proj~n is a linear mapping we get refl P (b~x + c~y) = (b~x + c~y) − 2 proj~n (b~x + c~y) = b~x + c~y − 2(b proj~n (~x) + c proj~n (~y)) = b(~x − 2 proj~n (~x)) + c(~y − 2 proj~n (~y)) = b reflP (~x) + c reflP (~y) for all ~x, ~y ∈ Rn and b, c ∈ R as required.

proj~n (~x)

reflP (~x) x1

94

Chapter 3 Matrices and Linear Mappings

EXAMPLE 9

1      1  1 2 .  4    Find the reflection of ~x = 1 over the hyperplane P in R with normal vector ~n = 2   3 0

Solution: We have

      1  −1/9   1   ~x · ~n  1 10 2 −11/9 refl P (~x) = ~x − 2 proj~n (~x) = ~x − 2 2 ~n =   − 2 =  −11/9  1 9 k~n k     0  3  3

EXAMPLE 10

   1    Determine the standard matrix of refl P : R3 → R3 where ~n =  2  .   −3

Solution: We have

Hence,

 1 refl P (~e1 ) = 0 −  0  0   refl P (~e2 ) = 1 −  0  0   refl P (~e3 ) = 0 +  1

     1   6/7   2     2  =  −2/7 14 −3  3/7       1   −2/7     4     2  =  3/7  14 −3  6/7      3/7  1 6      2  =  6/7  14 −3  −2/7

   6/7 −2/7 3/7  [reflP ] = −2/7 3/7 6/7    3/7 6/7 −2/7

Section 3.2 Linear Mappings

95

Section 3.2 Problems    −2 3    3 0  1. Let A =   and let L(~x ) = A~x be the  1 5  4 −6 corresponding matrix mapping.

(a) Determine the domain and codomain of L. (b) Determine L(2, −5) and L(−3, 4). (c) Determine [L].    1 2 −3 0    2. Let B =  2 −1 0 3  and let f (~x) = B~x   1 0 2 −1 be the corresponding matrix mapping. (a) Determine the domain and codomain of f . (b) Determine f (2, −2, 3, 1) and f (−3, 1, 4, 2). (c) Determine [ f ]. 3. Determine which of the following are linear. Find the standard matrix of each linear mapping. L(x1 , x2 , x3 ) = (x1 + x2 , 0) L(x1 , x2 , x3 ) = (x1 − x2 , x2 + x3 ) L(x1 , x2 , x3 ) = (1, x1 , x2 + x3 ) L(x1 , x2 ) = (x12, x22) L(x1 , x2 , x3 ) = (0, 0, 0) L(x1 , x2 , x3 , x4 ) = (x4 − x1 , 2x2 + 3x3 ) " # " # 3 2 4. Let ~a = 2 and ~x = −1 . (a) (b) (c) (d) (e) (f)

(a) Find [proj~a ]. (b) Use the standard matrix you found in (a) to find proj~a (~x). (c) Check your answer in (b) by computing proj~a (~x) directly.    3  5. Let P be a plane with normal vector ~n =  1  .   −1 3 Find the standard matrix of refl P : R → R3 .

6. Let ~a ∈ Rn . Prove that perp~a : Rn → Rn is a linear mapping. 7. Let ~v ∈ Rn and define DOT~v : Rn → R by DOT~v (~x) = ~x · ~v. Prove that DOT~v is linear and find it’s standard matrix. 8. Prove that if L : Rn → Rm is linear, then L(~0) = ~0. 9. Let L be with standard matrix # " a linear mapping 1 3 2 1 [L] = −1 3 0 5 . (a) What is the domain of L? (b) What is the codomain of L? (c) What is L(~x) for any ~x in the domain of L.

10. Consider the rotation Rπ/3 : R2 → R2 .

(a) Find the standard matrix of Rπ/3 . (b) Use the√ standard matrix to determine √ Rπ/3 (1/ 2, 1/ 2).

11. (a) Invent a linear mapping L : R2 → R3 such that L(1, 0) = (3, −1, 4) and L(0, 1) = (1, −5, 9). (b) Invent a non-linear mapping L : R2 → R3 such that L(1, 0) = (3, −1, 4) and L(0, 1) = (1, −5, 9). 12. Invent a linear mapping L : R2 → R2 such that L(1, 1) = (2, 3) and L(1, −1) = (3, 1).

13. (a) Let L : Rn → Rm be a linear mapping. Prove that if {L(~v1 ), . . . , L(~vk )} is a linearly independent set in Rm , then {~v1 , . . . ,~vk } is linearly independent. (b) Give an example of a linear mapping L : Rn → Rm where {~v1 , . . . ,~vk } is linearly independent in Rn , but {L(~v1 ), . . . , L(~vk )} is linearly dependent. 14. Prove that if ~u is a unit vector, then [proj~u ] = ~u~uT

96

Chapter 3 Matrices and Linear Mappings

3.3

Special Subspaces

When working with a function we are often interested in the subset of the codomain which is the set of all possible images under the function, called the range of the function. In this section we will not only look at the range of a linear mapping, but a special subset of the domain as well. We will then use the connection between linear mappings and their standard matrices to derive four important subspaces of a matrix. Range

DEFINITION Range

EXAMPLE 1

Let L : Rn → Rm be a linear mapping. The range of L is defined by   Range(L) = L(~x) ∈ Rm | ~x ∈ Rn Let L : R2 → R3 be defined by L(x1 , x2 ) = (x1 + which of  x 2 , x1 −x2, x2 ). Determine   0 2 3         the following vectors is in the range of L: ~y1 =  0 , ~y2 =  3, ~y3 = −1.       0 2 1

Solution: Observe that L(0, 0) = (0 + 0, 0 − 0, 0) = (0, 0, 0), so ~y1 ∈ Range(L). " # x For ~y2 , we need to find if there exist ~x = x1 such that 2 (2, 3, 1) = L(x1 , x2 ) = (x1 + x2 , x1 − x2 , x2 ) Comparing corresponding entires gives us the system of linear of equations x1 + x2 = 2 x1 − x2 = 3 x2 = 1 It is easy to see that this system is inconsistent. Thus, ~y2 < Range(L). For ~y3 , we need to find if there exist x1 and x2 such that (3, −1, 2) = L(x1 , x2 ) = (x1 + x2 , x1 − x2 , x2 ) That is, we need to solve the system of equations x1 + x2 = 3 x1 − x2 = −1 x2 = 2

We find that the system is consistent with unique solution is x1 = 1, x2 = 2. Hence, L(1, 2) = (3, −1, 2), so ~y3 ∈ Range(L).

Section 3.3 Special Subspaces

EXAMPLE 2

97

Find a basis for the range of L(x1 , x2 , x3 ) = (3 x1 − x2 , 2x1 + 2x3 ). Solution: We see that every vector L(~x ) has the form " " # " # "# # 3x1 − x2 0 −1 3 + x3 + x2 = x1 2 0 2 2 x1 + 2 x3 Hence,

(" # " # " #) (" # " #) 3 −1 0 −1 0 Range(L) = Span , = Span , , 2 0 2 0 2 " # " # " # (" # " #) 3 −1 0 −1 0 since = (−3) + . Thus, B = , spans Range(L) and is clearly 2 0 2 0 2 linearly independent, so it is a basis for Range(L).

EXAMPLE 3

"

# 2 Let ~a = . Find a basis for the range of proj~a : R2 → R2 . −1 Solution: Let ~y ∈ Range(proj~a ). Then, by definition, there exists ~x ∈ R2 such that ~y = proj~a (~x) =

~x · ~a ~a ∈ Span{~a} k~a k2

So, "Range(proj a}. On the other hand, pick any c~a ∈ Span{~a}. Taking ~a ) ⊆ Span{~ # 0 ~y = −5c gives 0 + (−5c)(−1) ~y · ~a ~a = ~a = c~a proj~a (~y) = 2 k~a k 5 Thus, Span{~a} ⊆ Range(proj~a ). Hence, Range(proj~a ) = Span{~a}. Since {~a} is also clearly linearly independent, it is a basis for Range(proj~a ).

EXERCISE 1

Find a basis for the range of L(x1 , x2 , x3 ) = (x1 − x2 , x1 − 2x3 ). Observe in the examples above that the range of a linear mapping is a subspace of the codomain. To prove that this is always the case, we will use the following lemma.

LEMMA 1

THEOREM 2

~ = ~0. If L : Rn → Rm is a linear mapping, then L(0) If L : Rn → Rm is a linear mapping, then Range(L) is a subspace of Rm . In many cases we are interested in whether the range of a linear mapping L equals its codomain. That is, for every vector ~y in the codomain, can we find a vector ~x in the domain such that L(~x ) = ~y.

98

Chapter 3 Matrices and Linear Mappings

REMARK A function whose range equals its codomain is said to be onto. We will look at onto mappings in Chapter 8.

EXAMPLE 4

Let L : R3 → R3 be defined by L(x1 , x2 , x3 ) = (x1 + x2 + x3 , 2x1 − 2x3 , x2 + 3x3 ). Prove that Range(L) = R3 . Solution: Since Range(L) is a subset of R3 by definition, to prove that Range(L) = R3 we only need to show that if we pick any vector ~y ∈ R3 , then we can find a vector    y1   ~x ∈ R3 such that L(~x ) = ~y. Let ~y =  y2 ∈ R3 . Consider   y3 (y1 , y2 , y3 ) = L(x1 , x2 , x3 ) = (x1 + x2 + x3 , 2x1 − 2x3 , x2 + 3x3 )

This gives the system of linear equations x1 + x2 + x3 = y1 2 x1 − 2 x3 = y2 x2 + 3 x3 = y3 Row reducing the corresponding augmented matrix gives     1 1 1 y1   1 0 0 −y1 + y2 + y3  3    2 0 −2 y2  ∼  0 1 0 3y1 − 2 1y2 − 2y3 0 1 3 y3 0 0 1 −y1 + 2 y2 + y3

   

Therefore, Range(L) = R3 since the system is consistent for all y1 , y2 , and y3 . In particular, taking x1 = −y1 + y2 + y3 , x2 = 3y1 − 23y2 − 2y3 , x3 = −y1 + 21 y2 + y3 we get L( x1 , x2 , x3 ) = (y1 , y2 , y3 )

EXAMPLE 5

Let A be an n × n matrix and let L : Rn → Rn be defined by L(~x ) = A~x. Prove that Range(L) = Rn if and only if rank A = n. Solution: If Range(L) = Rn , then for every ~y ∈ Rn there exists a vector ~x such that ~y = L(~x ) = A~x. Hence, the system of equations A~x = ~y with coefficient matrix A is consistent for every ~y ∈ Rn . The System-Rank Theorem then implies that rank(A) = n. On the other hand, if rank A = n, then by the System-Rank Theorem L(~x ) = A~x = ~y is consistent for every ~y ∈ Rn , so Range(L) = Rn .

Section 3.3 Special Subspaces

99

Kernel We now look at a special subset of the domain of a linear mapping.

DEFINITION Kernel

Let L : Rn → Rm be a linear mapping. The kernel (nullspace) of L is the set of all vectors in the domain which are mapped to the zero vector in the codomain. That is, Ker(L) = {~x ∈ Rn | L(~x) = ~0}

EXAMPLE 6

Let L : R2 → R2 be defined by L(x1 , x2 ) = (2x1 − which of # 2 , −x1" +#x2 ). Determine " 2x " # 0 3 2 the following vectors is in the kernel of L: ~x1 = . , ~x = , ~x = 2 3 −1 0 2 Solution: We have L(~x1 ) = L(0, 0) = (0, 0) L(~x2 ) = L(2, 2) = (0, 0) L(~x3 ) = L(3, −1) = (8, −4) Hence, ~x1 , ~x2 ∈ Ker(L) while ~x3 < Ker(L).

EXAMPLE 7

Find a basis for the kernel of L(x1 , x2 , x3 ) = (3 x1 − x2 , 2x1 + 2x3 ).    x1 Solution: By definition, ~x =  x2 is in the kernel of L if   x3 (0, 0) = L(x1 , x2 , x3 ) = (3x1 − x2 , 2x1 + 2x3 )

Hence, ~x is in the kernel of L if and only if it satisfies 3x1 − x2 = 0 2 x1 + 2 x3 = 0 Row reducing the corresponding coefficient matrix gives # " # " 1 0 1 3 −1 0 ∼ 2 0 2 0 1 3 We find   system has   vector  equation  that  the solution space of the homogeneous     −1 −1    −1                    −3 −3 . Since ~x = t  −3 , t ∈ R. Thus, Ker(L) = Span  is also linearly                          1 1 1 independent, it is a basis for Ker(L).

100

EXAMPLE 8

Chapter 3 Matrices and Linear Mappings

Find a basis for the kernel of Rθ : R2 → R2 . Solution: Let’s first think about this geometrically. We are looking for all vectors in R2 such that rotating them by an angle of θ gives the zero vector. Since a rotation does not affect length, the only vector which could have image of the zero vector would be the zero vector. We verify our geometric intuition algebraically. Consider " # " # 0 cos θ − sin θ = Rθ (~x) = [Rθ ]~x = ~x 0 sin θ cos θ Row reducing the coefficient matrix of the corresponding homogeneous system gives # " # " 1 0 cos θ − sin θ ∼ sin θ cos θ 0 1 Hence, the system has the unique solution ~x = ~0. So, (" #) 0 Ker(Rθ ) = 0 Recall that we defined a basis for {~0} to be the empty set. Hence, the empty set is a basis for Ker(Rθ ).

EXERCISE 2

Find a basis for the kernel of L(x1 , x2 , x3 ) = (x1 − x2 , x1 − 2x3 ). In the examples, the kernel of a linear mapping was always a subspace of the appropriate Rn .

THEOREM 3

If L : Rn → Rm is a linear mapping, then Ker(L) is a subspace of Rn . Proof: We use the Subspace Test. We first observe that by definition Ker(L) is a subset of Rn . Also, we have that ~0 ∈ Ker(L) by Lemma 1. Let ~x, ~y ∈ Ker(L). Then, L(~x ) = ~0 and L(~y) = 0~ so L(~x + ~y) = L(~x ) + L(~y) = 0~ + ~0 = ~0 so ~x +~y ∈ Ker(L). Thus, Ker(L) is closed under addition. Similarly, for any c ∈ R we have L(c~x ) = cL(~x ) = c ~0 = ~0 so c~x ∈ Ker(L). Therefore, Ker(L) is also closed under scalar multiplication. Hence, Ker(L) is a subspace of Rn . 

Section 3.3 Special Subspaces

101

The Four Fundamental Subspaces of a Matrix We now look at the relationship of the standard matrix of a linear mapping to its range and kernel. In doing so, we will derive four important subspaces of a matrix.

THEOREM 4

Let L : Rn → Rm be a linear mapping with standard matrix [L]. Then, ~x ∈ Ker(L) if and only if [L]~x = ~0. Proof: If ~x ∈ Ker(L), then 0~ = L(~x ) = [L]~x . ~ then ~0 = [L]~x = L(~x ), so ~x ∈ Ker(L). On the other hand, if [L]~x = 0,



We get an immediate corollary to this theorem.

COROLLARY 5

Let A ∈ Mm×n (R). The set {~x ∈ Rn | A~x = ~0} is a subspace of Rn . The corollary motivates the following definition.

DEFINITION Nullspace

EXAMPLE 9

EXAMPLE 10

Let A be an m × n matrix. The nullspace (kernel) of A is defined by ~ Null(A) = {~x ∈ Rn | A~x = 0}

"

# " # 1 2 −1 Let A = . Determine if ~x = is in the nullspace of A. −1 0 2 " # 3 Solution: We have A~x = . Since A~x , ~0, ~x < Null(A). 1 "

# 1 3 1 Let A = . Find a basis for the nullspace of A. −1 −2 2

Solution: We need to find all ~x such that A~x = ~0. We observe that this is a homogeneous system with coefficient matrix A. Row reducing gives # # " " 1 0 −8 1 3 1 ∼ 0 1 3 −1 −2 2 Thus, a vector equation for the solution set of the system is    8  ~x = t  −3 , t ∈ R   1      8        Hence,  −3 spans Null(A) and is linearly independent, so it is a basis for Null(A).   1  

102

Chapter 3 Matrices and Linear Mappings

EXAMPLE 11

  1 2 −1   Let A =  5 0 −4 . Show that Null(A) = {~0}.    −2 6 4 Solution: We need to show that the only solution to the homogeneous system A~x = ~ 0 is ~x = ~0. Row reducing the associated coefficient matrix we get    1 2 −1  1 0 0         5 0 −4 ∼  0 1 0 −2 6 4 0 0 1

Hence, the rank of the coefficient matrix equals the number of columns so by the System-Rank Theorem there is a unique solution. Therefore, since the system is homogeneous, the only solution is ~x = ~0. Thus, Null(A) = {0~}. The connection between the range of a linear mapping and its standard matrix follows easily from our second interpretation of matrix vector multiplication.

THEOREM 6

h i If L : Rn → Rm is a linear mapping with standard matrix [L] = A = ~a1 · · · ~an , then Range(L) = Span{~a1 , . . . ,~an}

Proof: Observe that for any ~x ∈ Rn we have    h i x.1 L(~x ) = A~x = ~a1 · · · ~an  ..  = x1~a1 + · · · + xn~an x  n

Thus, every vector in Range(L) is in Span{~a1 , . . . , ~an } and every vector in Span{~a1 , . . . , ~an} is in Range(L) as required.  Therefore, the range of a linear mapping is the subspace spanned by the columns of its standard matrix.

DEFINITION Column Space

EXAMPLE 12

Let A ∈ Mm×n (R). The column space of A is the subspace of Rm defined by Col(A) = {A~x ∈ Rm | ~x ∈ Rn }

  " (" # " # " #) # 1 −3 1 5 −3 1 5 −3   Let A = and , , and B = 0 2  , then Col(A) = Span 1 9 2 9 2 1   0 1      1 −3        Col(B) = Span  0 ,  2 .      0  1 

Section 3.3 Special Subspaces

EXERCISE 3

103

Let A be an m × n matrix. Prove that Col(A) = Rm if and only if rank(A) = m. Recall that taking the transpose of a matrix changes the columns of a matrix into ro...