7 Since T is a linear transformation, c 1v 1 + c 2v 2 + :::+ c nv n = 0)T(c 1v 1 + c 2v 2 + :::+ c nv n) = 0)c 1T(v 1) + c 2T(v 2) + :::+ c nT(v n) = ...

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(1) Let T1 : R2 → R2 and T2 : R2 → R2 be two linear transformations satisfying rank(T1 ◦ T2 ) = rank(T2 ◦ T1 ) = 0, T1 (1, 1) = (1, −1) and T2 (1, 2) = (2, 1). Find T1 and T2 . Justify your answer. Solution. Since rank(T1 ◦ T2 ) = rank(T2 ◦ T1 ) = 0, T1 ◦ T2 = T2 ◦ T1 = 0. Therefore, 0 = T1 ◦ T2 (1, 2) = T1 (T2 (1, 2)) = T1 (2, 1) 0 = T2 ◦ T1 (1, 1) = T2 (T1 (1, 1)) = T2 (1, −1). Suppose that x x x x T1 = A1 and T2 = A2 y y y y for some 2 × 2 matrices A1 and A2 . Then 1 2 1 0 1 1 2 0 A1 = and A2 = . 1 1 −1 0 2 −1 1 0 So −1 −1 2 1 0 1 2 = A1 = 1 −2 −1 0 1 1 −1 2 0 1 1 2/3 2/3 A2 = = . 1 0 2 −1 1/3 1/3

Therefore, T1 (x, y) = (−x + 2y, x − 2y) 2 2 1 1 T2 (x, y) = ( x + y, x + y). 3 3 3 3 (2) Let P2 be the vector space of real polynomials in x of degree ≤ 2 and let T1 : P2 → P2 and T2 : P2 → P2 be the linear transformations given by T1 (f (x)) = f (x + 1) and T2 (f (x)) = xf 0 (x). 1

http://www.math.ualberta.ca/˜xichen/math22514f/hw7sol.pdf 1

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(a) Find R(T1 ), R(T2 ), R(T1 ◦ T2 ) and R(T2 ◦ T1 ). (b) Verify that rank(T1 ◦ T2 ) ≤ min(rank(T1 ), rank(T2 )) and rank(T2 ◦ T1 ) ≤ min(rank(T1 ), rank(T2 )). Solution. Since T1 ◦ T2 (f (x)) = T1 (T2 (f (x))) = T1 (xf 0 (x)) = (x + 1)f 0 (x + 1) T2 ◦ T1 (f (x)) = T2 (T1 (f (x))) = T2 (f (x + 1)) = xf 0 (x + 1), we obtain R(T1 ) = Span{T1 (1), T1 (x), T1 (x2 )} = Span{1, x + 1, (x + 1)2 } = P2 R(T2 ) = Span{T2 (1), T2 (x), T2 (x2 )} = Span{0, x, 2x2 } = Span{x, x2 } R(T1 ◦ T2 ) = Span{T1 ◦ T2 (1), T1 ◦ T2 (x), T1 ◦ T2 (x2 )} = Span{0, x + 1, 2(x + 1)2 } = Span{x + 1, (x + 1)2 } R(T2 ◦ T1 ) = Span{T1 ◦ T2 (1), T1 ◦ T2 (x), T1 ◦ T2 (x2 )} = Span{0, x, 2x(x + 1)} = Span{x, x2 }. So rank(T1 ) = 3 rank(T2 ) = 2 rank(T1 ◦ T2 ) = 2 rank(T2 ◦ T1 ) = 2 and hence rank(T1 ◦ T2 ) ≤ min(rank(T1 ), rank(T2 )) rank(T2 ◦ T1 ) ≤ min(rank(T1 ), rank(T2 )). (3) Which of the following statements are true and which are false? Justify your answer. (a) For any two linear transformations T1 : V → W and T2 : V → W , rank(T1 + T2 ) = rank(T1 ) + rank(T2 ).

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Proof. False. Let T1 : R2 → R2 and T2 : R2 → R2 be the maps given by T1 (x, y) = (2x, 0) and T2 (x, y) = (x, y). Then rank(T1 + T2 ) = 2 6= 3 = rank(T1 ) + rank(T2 ). (b) For any two linear transformations T1 : V → W and T2 : V → W , rank(T1 − T2 ) = rank(T1 ) − rank(T2 ). Proof. False. Using the same example in (a), we have rank(T1 − T2 ) = 2 6= −1 = rank(T1 ) − rank(T2 ). (c) For any two linear transformations T1 : V → W and T2 : U → V , rank(T1 ◦ T2 ) = rank(T1 ) rank(T2 ). Proof. False. Using the same example in (a), we have rank(T1 ◦ T2 ) = 1 6= 2 = rank(T1 ) rank(T2 ). (d) For any two linear transformations T1 : V → W and T2 : U → V , K(T1 ◦ T2 ) ⊂ K(T2 ). Proof. False. Using the same example in (a), we have K(T1 ◦ T2 ) = Span{(1, 0)} 6⊂ {(0, 0)} = K(T2 ). (4) Let T : M2×2 (R) → M2×2 (R) be the map given by T (A) = A + AT . Do the following: (a) Show that T is a linear transformation. (b) Find K(T ) and R(T ). (c) Find the matrix representation [T ]B←B of T under the standard basis 1 0 0 1 0 0 0 0 B= , , , . 0 0 0 0 1 0 0 1 (d) Find the matrix representation [T ]C←C 1 0 0 0 0 1 0 C= , , , 0 0 0 1 1 0 −1

of T under the basis 1 . 0

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(e) Verify that [T ]B←B and [T ]C←C are similar. Solution. Since T (A + cB) = (A + cB) + (A + cB)T = (A + cB) + (AT + cB T ) = (A + AT ) + c(B + B T ) = T (A) + cT (B) for all A, B ∈ M2×2 (R) and c ∈ R, T is a linear transformation. The kernel of T is K(T ) = {A ∈ M2×2 (R) : T (A) = 0} = {A ∈ M2×2 (R) : A + AT = 0} = {A ∈ M2×2 (R) : AT = −A} = {2 × 2 skew symmetric matrices} For every A ∈ M2×2 (R), T (A)T = (A + AT )T = AT + (AT )T = A + AT = T (A). So T (A) is symmetric for all A and hence R(T ) ⊂ {B ∈ M2×2 (R) : B = B T }. On the other hand, for every symmetric 2 × 2 matrix B, 1 1 1 T ( B) = T (B) = (B + B T ) = B ⇒ B ∈ R(T ) 2 2 2 and hence {B ∈ M2×2 (R) : B = B T } ⊂ R(T ). Therefore, R(T ) = {B ∈ M2×2 (R) : B = B T } = {2 × 2 symmetric matrices}. Since 1 0 2 = 0 0 0 0 1 0 = 0 0 1 0 0 0 = 1 0 1 0 0 0 = 0 1 0

T T T T

0 1 =2 0 0 1 1 =0 0 0 1 1 =0 0 0 0 1 =0 2 0

0 0

0 0 1 0 0 + + 0 0 0 1 0 0 0 1 0 0 + + 0 0 0 1 0 0 0 1 0 0 0 0 +0 +0 +2 , 0 0 0 1 0 0 1

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we obtain [T ]B←B Since 1 0 2 T = 0 0 0 0 0 0 T = 0 1 0 0 1 0 T = 1 0 2 0 1 0 T = −1 0 0

0 1 =2 0 0 0 1 =0 2 0 2 1 =0 0 0 0 1 =0 0 0

2 0 = 0 0 0 0

0 1 1 0

0 1 1 0

0 0 . 0 2

0 0 +2 0 0 0 0 +0 0 0 0 0 +0 0 0

0 1

0 0 +2 1 1 0 0 +0 1 1

1 0

1 0 1 +0 , 0 −1 0

we obtain [T ]C←C

2 2 = 2

. 0

To see that [T ]B←B and [T ]C←C are similar, it is enough to verify −1 [T ]B←B = PB←C [T ]C←C PB←C where PB←C

1 0 = 0 0

−1 = PC←B PB←C

0 0 1 1 and 1 −1 0 0 1 0 0 0 0 0 0 1 = 0 1 1 0 . 2 2 0 12 − 12 0 0 0 0 1

(5) Let T1 : V → W and T2 : V → W be two linear transformations. Show that if K(T1 ) ∩ K(T2 ) = {0}, then K(T1 + T2 ) ∩ K(T1 − T2 ) = {0}.

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Proof. It suffices to prove that K(T1 ) ∩ K(T2 ) = K(T1 + T2 ) ∩ K(T1 − T2 ). For v ∈ K(T1 ) ∩ K(T2 ), T1 (v) = T2 (v) = 0. So (T1 + T2 )(v) = T1 (v) + T2 (v) = 0 ⇒ v ∈ K(T1 + T2 ) (T1 − T2 )(v) = T1 (v) − T2 (v) = 0 ⇒ v ∈ K(T1 − T2 ) and hence v ∈ K(T1 + T2 ) ∩ K(T1 − T2 ) for all v ∈ K(T1 ) ∩ K(T2 ), i.e., K(T1 ) ∩ K(T2 ) ⊂ K(T1 + T2 ) ∩ K(T1 − T2 ). For v ∈ K(T1 + T2 ) ∩ K(T1 − T2 ), (T1 + T2 )(v) = (T1 − T2 )(v) = 0 ⇒ T1 (v) + T2 (v) = T1 (v) − T2 (v) = 0. So

T1 (v) + T2 (v) = 0 T1 (v) − T2 (v) = 0

⇒

T1 (v) = 0 T2 (v) = 0

and hence v ∈ K(T1 ) ∩ K(T2 ) for all v ∈ K(T1 + T2 ) ∩ K(T1 − T2 ), i.e., K(T1 + T2 ) ∩ K(T1 − T2 ) ⊂ K(T1 ) ∩ K(T2 ). In conclusion, K(T1 ) ∩ K(T2 ) = K(T1 + T2 ) ∩ K(T1 − T2 ). If K(T1 )∩K(T2 ) = {0}, then K(T1 +T2 )∩K(T1 −T2 ) = {0}. (6) Let T : V → W be a linear transformation. Show that {T (v1 ), T (v2 ), ..., T (vn )} is linearly dependent for every linearly dependent set {v1 , v2 , ..., vn } in V . Proof. Since v1 , v2 , ..., vn are linearly dependent, there exist c1 , c2 , ..., cn ∈ R, not all zero, such that c1 v1 + c2 v2 + ... + cn vn = 0.

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Since T is a linear transformation, c1 v1 + c2 v2 + ... + cn vn = 0 ⇒ T (c1 v1 + c2 v2 + ... + cn vn ) = 0 ⇒ c1 T (v1 ) + c2 T (v2 ) + ... + cn T (vn ) = 0. This implies that T (v1 ), T (v2 ), ..., T (vn ) are linearly dependent since c1 , c2 , ..., cn are not all zero. (7) Let T : V → W and S : U → V be two linear transformations between vector spaces U, V and W . Show that T ◦ S = 0 if and only if R(S) ⊂ K(T ). Proof. This follows from T ◦ S = 0 ⇔ T ◦ S(u) = 0 for all u ∈ U ⇔ T (S(u)) = 0 for all u ∈ U ⇔ S(u) ∈ K(T ) for all u ∈ U ⇔ R(S) ⊂ K(T ). (8) Let T : V → V be a linear transformation from a finitedimensional vector space V to itself. Suppose that T 2 = 0. Show that 1 rank(T ) ≤ dim V. 2 Proof. Since T 2 = T ◦ T = 0, R(T ) ⊂ K(T ) by the previous problem. Therefore, dim R(T ) ≤ dim K(T ) ⇒ rank(T ) ≤ dim K(T ). By Rank Theorem, dim K(T ) = dim V − rank(T ). So 1 rank(T ) ≤ dim V − rank(T ) ⇒ rank(T ) ≤ dim V. 2