Textbook Exercise 7 Model Answers PDF

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Answers/Solutions of Exercise 7 (

) 1 1 1. (a) T1 is a linear transformation with standard matrix . −1 1 (b) T2 is not a linear transformation.   1 1 (c) T3 is a linear transformation with standard matrix  0 0. 0 0

(d) T4 is not a linear transformation.

( (e) T5 is a linear transformation with standard matrix y1 y2 · · ·

(f) T6 is not a linear transformation. 2. (a) There is enough information.      x + 2y x  3x + 2y + 4z    T  y  =    −y + z  z x + 4y + 6z  1 2  3 2 The standard matrix is   0 −1 1 4

) yn .

  x for y  ∈ R3 . z  0  4 . 1 6

(b) The information is not enough because the two vectors do not form a basis for R2 . (c) There is enough information. ) ( ) (( )) ( x x−y x ∈ R2 . for = T y x+y y ) ( 1 1 . The standard matrix is 1 −1

(d) There is enough information.      ( ) x x 1 x + 17y − 8z      T = y for y  ∈ R3 . 5 x + 22y − 8z z z ) ( 17 −8 The standard matrix is

1 5 1 5

5 22 5

1

5 −8 5

.

(e) There is enough information.     y + z x x        = for y  ∈ R3 . T y z z

x+z

z   0 1 1 The standard matrix is  0 0 1. 1 0 1

(f) The information is not enough because the three vectors do not form a basis for R3 .   (( )) 2x x =  2y . 3. (a) (S ◦ T ) y x+y

T ◦ S is not defined.      −2x − y + 3z x (b) (S ◦ T )  y  =  −x − y + 3z . z −3x − 2y + 6z (( )) ( ) x 2x + 2y (T ◦ S) . = 2x + y y

4. (⇒) It is a particular case of Theorem 7.1.4.2. (⇐) Suppose T (cu + dv) = c T (u) + d T (v) for all u, v ∈ Rn and c, d ∈ R.

(∗)

Let {e1 , e2 , . . . , en } be the standard basis for Rn and let A be the m × n ) ( matrix T (e1 ) T (e2 ) · · · T (en ) .

For any u = (u1 , u2 , . . . , un )T ∈ Rn , u = u1 e1 + u2 e2 + · · · + un en . By applying (∗) repeatedly, we have T (u) = u1 T (e1 ) + u2 T (e2 ) + · · · + un T (en )   u1  ( )  u2   = T (e1 ) T (e2 ) · · · T (en )  ..  . un

= Au.

Thus T is a linear transformation. 2

5. (a) For any u ∈ Rn , (T1 + T2 )(u) = T1 (u) + T2 (u) = Au + Bu = ( A + B)u. So T1 + T2 is a linear transformation and the standard matrix for T1 + T2 is A + B . (b) For any u ∈ Rn , (λT )(u) = λT (u) = λAu = (λA)u. So λT is a linear transformation and the standard matrix for λT is λA. 6. (a) (i) T is invertible and the inverse of T is T itself. (ii) T is not invertible. Assume there exists an inverse S : R2 → R2 . Then (1, 0)T = S ◦ T ((1, 0)T ) = S((1, 0)T ) = S ◦ T ((0, 1)T ) = (0, 1)T , a contradiction. (b) A−1 . 7. (a) Note that (n · x)n = nnT x where LHS is the scalar n · x multiplied to the vector n while all operations on RHS are matrix multiplications. (To verify the equation, let n = (a1 , . . . , an )T and x = (x1 , . . . , xn )T and then check that both sides give us the same vector.) For any x ∈ Rn , P (x) = x − (n · x)n = Ix − nnT x = (I − nnT )x. So P is a linear transformation and the standard matrix for P is I − nnT . (b) Since for all x ∈ Rn , (P ◦ P )(x) = P (P (x)) = P (x − (n · x)n) = x − (n · x)n − {n · [x − (n · x)n]}n = x − (n · x)n − {(n · x) − (n · x)(n · n)}n = x − ( n · x) n = P ( x) , P ◦ P = P. Alternatively, since n is a unit vector, nT n = n · n = 1. Thus (I − nnT )2 = (I − nnT )(I − nnT ) = I − 2nnT + nnT nnT = I − nnT . By Theorem 7.1.11, P ◦ P = P . 8. (a) Suppose T is not the zero transformation. So there exists x ∈ Rn such that T (x) = 0. Define u = T (x). Then u is a nonzero vector and T (u) = T (T (x)) = (T ◦ T )(x) = T (x) = u. (b) Suppose T is not the identity transformation. So there exists y ∈ Rn such that T (y) = y. Define v = T (y) − y. Then v is a nonzero vector and T (v) = T (T (y) − y) = (T ◦ T )(y) − T (y) = T (y) − T (y) = 0. 3

(c) Let A be the standard matrix for T . If T is not the zero transformation and the identity transformation, then by (a) and (b), 1 and 0 are the eigenvalues of A. So by the result of Question 6.4, ( ) ) ( ) ( 0 0 r s 1 0 A= where st = r (1 − r ). or , t 1−r 0 1 0 0 9. (a) Similar to Question 7.7, for any x ∈ Rn , F (x) = x − 2(n · x)n = Ix − 2nnT x = (I − 2nnT )x. So F is a linear transformation and the standard matrix for F is I − 2nnT . (b) Since for all x ∈ Rn , (F ◦ F )(x) = F (F (x)) = F (x − 2(n · x)n) = x − 2(n · x)n − 2{n · [x − 2(n · x)n]}n = x − 2(n · x)n − 2{(n · x) − 2(n · x)(n · n)}n = x − 2(n · x)n − 2{−(n · x)} = x, F ◦ F is the identity transformation. Alternatively, (I − 2nnT )2 = (I − 2nnT )(I − 2nnT ) = I − 4nnT + 4nnT nnT = I. By Theorem 7.1.11, F ◦ F is the identity transformation. (c) Note that (I − 2nnT )T = I − 2(nnT )T = I − 2nnT . Thus (I − 2nnT )(I − 2nnT )T = (I − 2nnT )2 = I by (b). The standard matrix is an orthogonal matrix. 10. (a) By Theorem 7.1.4.2, T (u + v) · T (u + v) = (T (u) + T (v)) · (T (u) + T (v)) = T (u) · T (u) + 2(T (u) · T (v)) + T (v) · T (v) = ||T (u)||2 + ||T (v )||2 + 2(T (u) · T (v)) = ||u||2 + ||v ||2 + 2(T (u) · T (v)).

(1)

On the other hand, T (u + v) · T (u + v) = ||T (u + v )||2 = ||u + v||2 = (u + v) · (u + v ) = u · u + 2(u · v) + v · v = ||u||2 + ||v ||2 + 2(u · v). Thus (1) and (2) imply T (u) · T (v) = u · v . 4

(2)

(b) (⇐) Suppose A is an orthogonal matrix of order n. Then by Question 5.32, for all u ∈ Rn , ||T (u)|| = ||Au|| = ||u||. So T is an isometry. (⇒) Suppose T is an isometry on Rn . Let {e1 , e2 , . . . , en } be the standard basis for Rn . Then (Aei) · (Aej ) = (Aei)T Aej = eiT AT Aej = the (i, j)-entry of AT A.

(3)

{

(4)

On the other hand, by (a), (Aei) · (Aej ) = T (ei) · T (ej ) = ei · ej =

1 if i = j 0 if i = j.

By (3) and (4), AT A = I . By Remark 5.4.4, A is an orthogonal matrix. (c) All isometries on R2 are of the form ) (( )) ( x cos(θ) + δy sin(θ ) x = T x sin(θ ) − δy cos(θ ) y

( ) x ∈ R2 for y

where δ = ±1 and 0 ≤ θ < 2π . (

) 2 1 0 . 1 −1 1 ) ) Gauss-Jordan ( ( 1 0 13 2 1 0 −→ 1 −1 1 0 1 − 23 Elimination

11. The standard matrix of T is

(a) {(2, 1)T , (1, −1)T } is a basis for R(T ). (For this example, any two linearly independent vectors in R2 is a basis for R(T ). Why?) (b) {(− 31 , 32, 1)T } is a basis for Ker(T ). (c) rank(T ) + nullity(T ) = dim(R(T )) + dim(Ker(T )) = 2 + 1 = 3 = dim(R3 ). (d) For example, {(− 13 , 32, 1)T , (0, 1, 0)T , (0, 0, 1)T } is a basis for R3 .     Gauss-Jordan 1 0 1 2 3 −1 2 7  0 1 1 −1 −→ 12. 1 2 3 0  0 1 1 −1 Elimination 0 0 0 0 5

(a) {(3, 1, 0)T , (−1, 2, 1)T } is a basis for R(T ). (b) {(−1, −1, 1, 0)T , (−2, 1, 0, 1)T } is a basis for Ker(T ). (c) rank(T ) + nullity(T ) = dim(R(T )) + dim(Ker(T )) = 2 + 2 = 4 = dim(R4 ). 13. (a) 2. (b) 2. (c) 2. 14. (a) {0}. (b) Rn . 15. (a) Let {v1 , v2 , . . . , vk } be an orthonormal basis for V . By Theorem 5.2.15, P (u) = (u · v1 )v1 + (u · v2 )v2 + · · · + (u · vk )vk = v1 v1 T u + v2 v2T u + · · · + vk vkT u = (v1 v1T + v2 v2T + · · · + vk vkT )u Note that v1 v1T + v2 v2 T + · · · + vk vk T is an n × n matrix. So P is a linear transformation. (b) Ker(P ) = span{(a, b, c)} and R(P ) = V . 16. (⇒) Let u, v ∈ Rn such that T (u) = T (v). Then T (u − v ) = T (u) − T (v) = 0 and hence u − v ∈ Ker(T ). Since Ker(T ) = {0}, u − v = 0, i.e. u = v . Thus T is one-to-one. (⇐) By Theorem 7.1.4.1, T (0) = 0. Since T is one-to-one, for all v ∈ Rn , if v = 0, T (v) = T (0) = 0. Thus Ker(T ) = {0}. 17. (a) Let u ∈ Ker(S), i.e. S(u) = 0. Then T ◦ S(u) = T (S(u)) = T (0) = 0 and hence u ∈ Ker(T ◦ S ). Thus Ker(S) ⊆ Ker(T ◦ S ). (b) Let v ∈ R(T ◦ S), i.e. there exists u ∈ Rn such that v = T ◦ S(u). Put w = S(u) ∈ Rm . Then v = T (S (u)) = T (w). This means that v ∈ R(T ). Thus R(T ◦ S) ⊆ R(T ).

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