MA2101 Linear Algebra II AY 2012/2013 Sem 2 NATIONAL UNIVERSITY OF SINGAPORE MATHEMATICS SOCIETY PAST YEAR PAPER SOLUTIONS Written by Chua Hongshen Audited by Le Hoang Van MA2101 Linear Algebra II AY 2012/2013 Sem 2 Question 1 (a) For any A, A0 ∈ W , k ∈ R, (A + kA0 )u = Au + kAu = 0 ∈ W . ∴ W is a subspace of V . (b) For any a b c d ∈ W , we have 0 1 a b ⇒ a = b, c = d = 0 −1 c d 0 0 1 1 is linearly independent. , Clearly 1 1 0 0 n 1 1 0 0 o and dim(W ) = 2. , ∴ A basis for W is 1 1 0 0 n 1 1 0 0 0 1 0 0 o n 1 1 0 0 o for V . , , , to a basis , (c) Extend 0 1 0 0 1 1 0 0 1 1 0 0 n 0 1 0 0 o , then V = W ⊕ W 0 . , Let W 0 = 0 1 0 0 Question 2 (a) For any a + bx + cx2 ∈ Ker(T ), we have          1 i 1 a 0 a −2 0 1 i   b  = 0 ⇒  b  = t  −i  for some t ∈ C 1 2i 0 c 0 c 1 ∴ {−2 − ix + x2 } is a basis for Ker(T ). ∴ nullity(T ) = dim(Ker(T )) = 1, rank(T ) = dim (P2 (C))− nullity(T ) = 2.  (b) From [T ]E,B  1 i 1 = 0 1 i , we have 1 2i 0 T (1) = (1, 0, 1), T (x) = (i, 1, 2i), T (x2 ) = (1, i, 0) NUS Math LaTeXify Proj Team Page: 1 of 6 NUS Mathematics Society MA2101 Linear Algebra II AY 2012/2013 Sem 2 ∴ [T ]E,C = [(1, 0, 1)]E [(i, 1, 2i)]E [(1, i, 0)]E   1 0 2 = 0 i i  1 −1 1  (c) Let P = [I]B,C = [1]B [1 + ix]B [1 + x2 ]B  1 1 1 = 0 i 0. 0 0 1 Then P is invertible and [T ]E,B P = [T ]E,C . Question 3 (a) Let E = {E11 , E12 , E21 , E22 } be the standard basis for V = M2×2 (R). Then  0 1 [T ]E =  0 0 1 0 0 0  0 1 0 0  3 −1 2 0 x −1 0 −1 −1 x 0 0 4 3 2 Thus, cT (x) = = x − 3x + x + 3x − 2. 0 0 x − 3 1 0 0 −2 x (b) By Cayley-Hamilton Theorem, T 4 − 3T 3 + T 2 + 3T − 2IV = 0V 1 4 T − 3T 3 + T 2 + 3T = IV 2 1 T ◦ ( (T 3 − 3T 2 + T + 3IV )) = IV 2 Let p(x) = 12 (x3 − 3x2 + x + 3), then T −1 = p(T ). (c) For any a b c d ∈ E1 (T ), we have          −1 1 0 1 a 0 a 1 b 1  1 −1 0 0   b  0     =   ⇒   = t   for some t ∈ R 0 c 0 0 2 −1  c  0 0 0 2 −1 d 0 d 0 ∴ dimE1 (T ) = 1. NUS Math LaTeXify Proj Team Page: 2 of 6 NUS Mathematics Society MA2101 Linear Algebra II AY 2012/2013 Sem 2 Since cT (x) = (x − 1)2 (x + 1)(x − 2) and dim(E1 (T )) = 1, so the Jordan canonical form of T is similar to   1 1 0 0 0 1 0 0    0 0 −1 0 0 0 0 2 Question 4 0 1 (a) An example is . −1 0 ∗ ∗ 0 1 0 1 1 0 0 1 0 1 By direct calculation, = = −1 0 −1 0 0 1 −1 0 −1 0 (b) Since A is normal, A is unitary diagonalizable. Hence, there exists a unitary matrix P such that   λ1 0   ∗ .. P∗ AP =   = Q ⇒ A = PQP . 0 λn where λ1 , ..., λn are the eigenvalues of A. ∴ AT = A∗ = (PQP∗ )∗ = P∗ Q∗ P = P∗ QP = A i.e. A is symmetric. (c) Let T be a normal operator on a finite dimensional real innner product space. If the characteristic polynomial of T can be factorized into linear factor over R, then T is self-adjoint. Question 5 (a) Let {v1 , v2 , ..., vn } be a basis for Ker(S) and{S(w1 ), S(w2 ), ..., S(wn )} be a basis for R(S). Define B = {w1 , w2 , ..., wm , v1 , v2 , ..., vn } which is a basis for V . Extend {S(w1 ),S(w2 ), ..., S(w n )} to a basis of C = {S(w1 ), S(w2 ), ..., S(wn ), u1 , u2 , ..., un } for V . Im 0m×n Then [S]C,B = . 0n×m 0n×n W 0m×n (b) [T ◦ S]B = . Y 0n×n W X [S ◦ T ]B = . 0n×m 0n×n NUS Math LaTeXify Proj Team Page: 3 of 6 NUS Mathematics Society MA2101 Linear Algebra II AY 2012/2013 Sem 2 (c) xIm − W 0m×n = det(xIm − W)det(xIn ) = xn cW (x). cT ◦S (x) = −Y xIn xIm − W −X = det(xIm − W)det(xIn ) = xn cW (x). cS◦T (x) = 0n×m xIn ∴ cT ◦S (x) = cS◦T (x). (d) No. For example, let S and T be linear operator on R2 such that S((x, y)) = (x, 0) and T ((x, y)) = (y, 0) for (x, y) ∈ R2 Then, (T ◦ S)((x, y)) = T (S((x, y))) = T ((x, 0)) = (0, 0) for (x, y) ∈ R2 . So T ◦ S is the zero operator on R2 and hence mT ◦S = x. On the other hand, (S ◦ T )((x, y)) = S(T ((x, y))) = S((0, y)) = (y, 0) for (x, y) ∈ R2 . 2 Let E = {(1, 0), (0, a basis for R . 1)} be 0 1 and hence mS◦T = x2 . Then [S ◦ T ]B = 0 0 T ◦ S and S ◦ T do not have the same minimal polynomial. Question 6 (a) (i) u ∈ Ki ⇒ Qi (u) = 0 ⇒ Qi+1 (u) = Q(Qi (u)) = Q(0) = 0 ⇒ u ∈ Ki+1 . ∴ Ki ⊆ Ki+1 . (ii) We prove by induction that Km = Km+1 , ∀m ≥ k, which implies Kk = Km , ∀m ≥ k. Given that Kk = Kk+1 . Suppose that, Km = Km+1 where m ≥ k. For any u ∈ Km+2 , we have Qm+2 (u) = 0 ⇒ Qm+1 (Q(u)) = 0 ⇒ Q(u) ∈ Km+1 = Km Then Qm+1 (u) = Qm (Q(u)) = 0 and hence u ∈ Km+1 . So, we have shown that Km+2 ⊆ Km+1 and together with Km+1 ⊆ Km+2 from part (i), we have Km+1 = Km+2 . Thus, by Mathematical Induction, Km = Km+1 , ∀m ≥ k. NUS Math LaTeXify Proj Team Page: 4 of 6 NUS Mathematics Society MA2101 Linear Algebra II AY 2012/2013 Sem 2 (b) (i) Suppose u ∈ K ∩ R, i.e. Qs (u) = 0 and u = Qs (v) for some u ∈ V . Then Q2s (v) = Qs (u) = 0 and hence v) ∈ K2s = Ks = K. This means u = Qs (v) = 0. So, we have K ∩ R = {0} which implies that K + R is a direct sum. By the Dimension Theorem for linear transformation, dim(V ) = dim(K) + dim(R) = dim(K ⊕ R) As K ⊕ R ⊆ V , we have V = K ⊕ R. (ii) For all u ∈ K, (T |K − λIK )s (u) = (T − λIV )s (u) = Qs (u) = 0 i.e. (T |K − λIK )s = 0K and hence mT |K (x)|(x − λ)s ⇒ mT |K (x) = (x − λ)t for some t ≤ s. Assume that t < s. Since Kt $ Ks , there exists v ∈ Ks − Kt . Then (T |K − λIK )t (v) = (T − λIV )t (v) = Qt (v) 6= 0 which contradicts with the fact that (T − λIV )t (v) = 0K . ∴ mT |K (x) = (x − λ)s . Question 7 (a) Take any u ∈ Ker(T ). Then T (u) = 0 ⇒ (T ∗ ◦ T )(u) = T ∗ (T (u)) = T ∗ (0) = 0 ⇒ u ∈ Ker(T ∗ ◦ T ). ∴ Ker(T ) ⊆ Ker(T ∗ ◦ T ). Take any v ∈ Ker(T ∗ ◦ T ). Then hT (v), T (v)i = hv, T ∗ (T (v))i = hv, (T ∗ ◦ T )(v))i = hv, 0i = 0 ⇒T (v) = 0 ⇒v ∈ Ker(T ) ∴ Ker(T ∗ ◦ T ) ⊆ Ker(T ). Thus, we have Ker(T ∗ ◦ T ) = Ker(T ). (b) No, for example, let T be a linear operator on R2 such that x 0 1 x x T = for ∈ R2 y 0 0 y y Then Ker (T ) = span {(1, 0)T } while Ker (T ∗ ◦ T ) = span {(0, 1)T }. NUS Math LaTeXify Proj Team Page: 5 of 6 NUS Mathematics Society MA2101 Linear Algebra II AY 2012/2013 Sem 2 (c) T (u) is the orthogonal projection of b onto R(T ) ⇐⇒ b − T (u) is orthogonal to R(T ) ⇐⇒ hT (v), b − T (u)i = 0 for all v ∈ V ⇐⇒ hv, T ∗ (b) − (T ∗ ◦ T )(u)i = 0 for all v ∈ V ⇐⇒ T ∗ (b) − (T ∗ ◦ T )(u) = 0 ⇐⇒ T ∗ (b) = (T ∗ ◦ T )(u) ⇐⇒ x = u is a solution to T ∗ (b) = (T ∗ ◦ T )(u) (d) Take any w ∈ {u|T (u) = b}. Then, (T ∗ ◦ T )(w) = T ∗ (T (w)) = T ∗ (b) i.e. w ∈ {u|(T ∗ ◦ T )(u) = b}. ∴ {u|T (u) = b} ⊆ {u|(T ∗ ◦ T )(u) = b}. Take any w0 ∈ {u|(T ∗ ◦T )(u) = b}. Since (T ∗ ◦T )(w) = T ∗ (b), by part (c), T (w0 ) is the orthogonal projection of b onto R(T ). As b ∈ R(T ), the orthogonal projection is b itself, i.e. T (w0 ) = b and hence w0 ∈ {u|T (u) = b}. ∴ {u|(T ∗ ◦ T )(u) = b} ⊆ {u|T (u) = b}. Combining both, we have {u|T (u) = b} = {u|(T ∗ ◦ T )(u) = b}. NUS Math LaTeXify Proj Team Page: 6 of 6 NUS Mathematics Society