Linear-Algebra-100-Problems

Linear-Algebra-100-Problems - 100 ∗ 2011.1.24 2009 , , ,...

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Unformatted text preview: 100 ∗ 2011.1.24 2009 , , , , , , 3 2011 ‡ 1 , † 100 , , , , , , , , , , , , , ∗ † ‡ http://ccjou.twbbs.org/blog/ 1 Problem 1 In each part solve the matrix equation for X . (a) (b) AXB = C given that 1 4 A = −2 3 , 1 −2 Solution (a) Ax = b 3×3 X X −1 0 120 11 0= −3 1 5 3 1 −1 8 6 −6 0 1 B= 20 0 0 1 −1 , C= 6 −1 −4 1 0 , : −1 1 01 T 1 X = 2 0 1 0 −1 , 3×5 3 1 −3 1 5 , (reduced row echelon form), : −1 1 01 3 1 −3 12 1 5 −1 1 3 1 −3 0112 012 1 −1 0 1 0 0 2 1 1 2 0 0 1 −1 −3 −1 3 1 1 1 2 2 1 12 −1 6 3 0 −1 1 1 0 −1 0 1 −1 −3 −1 3 0 1 1 2 1 0 0 1 −1 1 0 −4 6 0 3 0 1 −1 1 100 0 1 0 001 X= (b) X 2×2 9 , AXB = C , XT , −1 3 −1 60 1 4 x, y, z, w, X , , , Y = XB , AY = C Y, (a) Y, , 1 XB = Y X , [ A C ], 1 4 8 6 −6 −2 3 6 −1 1 −2 −4 0 4 8 6 −6 1 0 , 0 1 2 1 −1 0 1 2 1 −1 0000 0 0000 0 ID 2 , E 00 EAC = E A EC = I2 0 I2 D 0 0 Y 0 1 4 8 6 −6 0 11 22 11 −11 0 −6 −12 −6 6 1 0 0 2 −2 EA = I2 0 D 0 , , E C = EAY = D 3 Y = , EC = Y, Y = 0 2 −2 2 1 −1 XB = Y [ BT 2 0 0 Y T ], 2 1 : 0 0 1 BT X T = Y T , 0 1 2 1 0 −1 −2 −1 , 0 1 2 1 0 −1 −2 −1 XT = 02 11 01 21 0 1 2 1 0000 , 1001 X= Problem 2 Let Determine the LU decomposition P A = LU , where P is the associated permutation matrix. Solution , lij , i > j , , , P , (1, 2, 3, 4), , L = [lij ] U = [uij ], i ≤ j A= 1 2 −3 4 2 3 2 1 −3 −1 1 −4 4 8 12 −8 4 ( 4 1 2 8 12 −8 4 1 2 1 1 4 2 2 −3 8 3 2 4 ), 2 : 2 −3 3 −3 −1 1 4 1 −4 4 2 3 4 2 12 −8 1 1 −4 2 −3 8 0 4 −7 8 4 −7 5 −3 −1 1 2 −3 0 8 24 −24 −7 8 4 −7 2 −1 −3 1 5 −8 2 −3 8 0 1 2 −1 4 −3 −5 24 −24 32 −27 1 3 4 2 3 1 4 1 3 4 2 2 −1 4 −3 1 24 −24 5 −8 2 −3 8 2 −1 4 0 24 −24 −3 −5 4/3 3 1 4 2 3 1 4 100 L= 1 2 −3 4 0 −1 8 −7 2 1 0 0 , U = 0 0 24 −24 4 0 1 0 0 0 0 5 −3 −5 4/3 1 0 00 0 0 , P = 1 0 0 1 0 0 0 0 001 Problem 3 An n × n matrix A is called idempotent if A2 = A. (a) Show that if A is idempotent and invertible, then A = I . (b) Describe all 2 × 2 real idempotent matrices. (c) If A is idempotent, find the inverse of I − cA (if possible) for some scalar c. 5 Solution (a) A − A2 = 0 A=I (b) 2 × 2 rankA = 2, A rankA = 1, A v vT u = 1, rankA = 0, (c) A2 = A, 2, 1, 0, , (a) A=I u, A−1 A−1 (A − A2 ) = A−1 A − A−1 AA = I − A = 0, A=0 A A = T , A2= uvT T , uv uv 4 −6 2 2 , A = , u = , v = 2 −3 −3 1 , (I − cA)(I − dA) = I − cA − dA + cdA2 = I + (cd − c − d)A cd − c − d = 0, d= c c −1 , (I − cA)−1 = I − c=1 Problem 4 Answer the following questions. c A c−1 (a) Show that if A and A + B are invertible, then I + BA−1 is also invertible. (b) If A, B , and A + B are invertible, express (A−1 + B −1 )−1 in terms of A, B , and A + B . (c) Suppose that a square matrix A satisfies A2 − 3A + I = 0. Find the inverse of A. (d) Suppose that I + A is invertible and B = (I + A)−1 (I − A). Show that I + B is also invertible. 6 Solution (a) (A + B )−1 , I + BA−1 (I + BA−1 )−1 = A(A + B )−1 (b) Z −1 Y −1 X −1 A−1 (A + B )B −1 = (I + A−1 B )B −1 = B −1 + A−1 A−1 +B −1 , (XY Z )−1 = , A (I + BA−1 )A(A + B )−1 = (A + B )(A + B )−1 = I , (A−1 + B −1 )−1 = B (A + B )−1 A (c) A, I = −A2 + 3A = A(−A + 3I ) A−1 = 3I − A (d) I + A, B = (I + A)−1 (I − A) (I + A)B = I − A, I + B, (I + B )−1 = 1 (I + A) 2 (I + A)−1 , AB + A + B = I , I A2 − 3A + I = 0 , (I + A)(I + B ) = 2I , Problem 5 Let x be a vector in R3 satisfying xT x = 1. (a) Find a vector x such that (I3 − 2xxT ) 7 1 3 2 3 2 3 =0 0 1 (b) If possible, find a vector x such that (I3 − 2xxT ) Otherwise, give a reason why there is no such x. Solution , (x 2 3=0 0 −2 −1 1 : = xT x = 1) (I − 2xxT )b = c (I − 2xxT )b = b − 2(xT b)x = c x, b, c ∈ R3 , x b, c x b−c , x = k(b − c), , 2(xT b)x = b − c, 2k2 (b − c)T b(b − c) = b − c x , b = c, 2k2 (b − c)T b = 1, k=± 1 2(b − c)T b 1 2(b − c)T b (b − c) (b − c)T b > 0, x=± , xT x = 1, b=c (a) b = c = 1, b, c 2(b − c)T b = (b − c)T (b − c), √ 3 2, (b) b= √ 1 x = ±√ 3 14, c = 1, 8 −1 1 1 k=± , Problem 6 Let B be a skew-symmetric matrix, B T = −B . If A = (I + B )(I − B )−1 prove that A is an orthogonal matrix. Solution B T = −B , AT = (I + B )(I − B )−1 = (I + B )−1 (I − B ) (I − B )(I + B ) = (I + B )(I − B ), AT A = (I + B )−1 (I − B )(I + B )(I − B )−1 = (I + B )−1 (I + B )(I − B )(I − B )−1 = I AT = A−1 Problem 7 Suppose that A is an n × n matrix such that for some k > 0, Ak = 0. Such matrices are called nilpotent. (a) Show that A is not invertible. (b) Show that I − A is invertible. (c) If the matrices A and B commute, show that I + AB is invertible. Note that B may not be nilpotent. T = (I − B T )−1 (I + B T ) 9 Solution (a) A , AA−1 = I , Ak (A−1 )k = (A · · · A) (A−1 · · · A−1 ) k k −1 = A · · · A(AA k −1 ) A−1 · · · A−1 k −1 = Ak−1 (A−1 )k−1 Ak (A−1 )k = AA−1 = I , , Ak det(Ak ) = det(A) · · · det(A) = det(A)k k Ak = 0, (A−1 )k det(Ak ) = det(0) = 0, (b) Ak = 0, I−A det(A) = 0, A k > 0, Ak I −A : 1 − xk = (1 − x)(1 + x + x2 + · · · + xk−1 ) I − Ak = (I − A)(I + A + A2 + · · · + Ak−1 ) Ak = 0 , (I − A)−1 = I + A + A2 + · · · + Ak−1 (c) (b) k , : (I − A)(I + A + A2 + · · · + Ak−1 ) = I , I −A 1 + xk = (1 + x)(1 − x + x2 − · · · + xk−1 ) 10 , I + (AB )k = (I + AB )(I − AB + (AB )2 − · · · + (AB )k−1 ) (AB )k B A = AB , (AB )k = AB (AB )k−1 = A(BA)B (AB )k−2 = A(AB )B (AB )k−2 = A2 B 2 (AB )k−2 (AB )k = Ak B k , I + AB (AB )m = (AB )k (AB ) = 0, , Problem 8 Let A be an n × n matrix. If A2 = 0, show that I − A is nonsingular. Solution I −A , (I − A)(I + A) = I − A2 = I , (I − A)−1 = I + A , Ax = A2 x = 0, , I −A 0, 1, Problem 9 Let A be an n by n real symmetric matrix, and s(A) be the sum of all entries of A. Prove that s(A2 ) ≥ (s(A))2 s(In ) x=0 (I − A)x = 0, , A2 x = λ2 x = 0, Ax = λx, , x = Ax, : I −A x = 0, I −A λ = 0, A n I−A A, Ak = 0 k Ak = 0, k m, , (AB )k = 0, m = k + 1, I + (AB )m (I − A)x = x − λx = x, 11 Solution A2 , n n n s(A2 ) = i=1 j =1 k =1 aik akj A , s(A2 ) = i, k, aik = aki , n n k =1 n n i=1 j =1 n aki akj n j =1 akj = k =1 n i=1 n aki 2 = k =1 n i=1 aki ρ2 k k =1 = ρk = n i=1 aki n k =1 Cauchy-Schwarz ρ2 ≥ k , 1 n n 2 , ρk k =1 = (s(A))2 s(In ) 1 = (1, 1, . . . , 1), 2 s(A2 ) = 1T A2 1 = 1T AT A1 = (A1)T (A1) = A1 Cauchy-Schwarz , 2 (s(A))2 = (1T A1)2 ≤ 1 · A1 2 = n A1 2 Problem 10 Let A and B be n × n matrices. Prove that if AB = 0, then for every positive integer k, trace(A + B )k = traceAk + traceB k 12 Solution (A + B )k = Ak + kAk−1 B + · · · + kAB k−1 + B k , (A + B )k AB = 0, trace(Y X ), , Ak , B k , Am B k−m, Am B k−m = Am−1 (AB )B k−m−1 = 0 XY YX , B m Ak−m , m ≥ 1 trace(XY ) = trace(B m Ak−m ) = trace(Ak−m B m ) = trace0 = 0 trace(A + B )k = trace(Ak + B k ) = traceAk + traceB k , trace(X + Y ) = traceX + traceY , X Y Problem 11 For n × n matrices A and B , the commutator of A and B is defined by [A, B ] ≡ AB − BA. Show that (a) [A, B ]T = [B T , AT ] (b) [A, [B, C ]] + [B, [C, A]] + [C, [A, B ]] = 0 (c) trace[A, B ] = 0 (d) [A, B ] is never similar to the identity matrix. (e) If the diagonal entries of A are all equal to zero, then there exists matrices U and V such that A = [U, V ]. Solution (a) [A, B ]T = (AB − BA)T = B T AT − AT B T = [B T , AT ] 13 (b) [A, [B, C ]] + [B, [C, A]] + [C, [A, B ]] = A(BC − CB ) − (BC − CB )A + B (CA − AC ) − (CA − AC )B + C (AB − BA) − (AB − BA)C =0 (c) trace[A, B ] = trace(AB − BA) = trace(AB ) − trace(BA) = trace(BA) − trace(BA) = 0 (d) (c) , (e) U , trace[A, B ] = 0, [A, B ] I A = [U, V ] = U V − V U , , 1 2 .. . n traceI = n = 0, [A, B ] I U = A = [U, V ] = U V − V U , aij = ivij − vij j = (i − j )vij i = j, Problem 12 Suppose A is m × m, U is m × n, V is n × m, and D is n × n. The following AU . questions relate to the inverse of the block matrix VD 14 vij = aij /(i − j ), i = j, vij = 0, (a) Suppose A is invertible. Find W , X , Y , and Z satisfying Im Y W0 AU = X In VD 0Z (b) Suppose Z is invertible. Find the inverse of Im Y 0 Z . (c) Use the results of (a) and (b) to compute the inverse of . To have VD this result, what assumptions of the matrix invertibility should be made? AU (d) An alternative way to derive the inverse of a block matrix is to start with AU Z′ 0 Im X ′ = VD Y ′ In 0 W′ Follow the steps described in (a), (b), (c) and then derive a new formula AU . By equating the (1,1) entry for the inverse of the block matrix VD of the obtained inverse and that obtained in (c), show that (A − U D−1 V )−1 = A−1 + A−1 U (D − V A−1 U )−1 V A−1 (e) By the formula in (d), derive a formula for (A + uvT )−1 . (f) By the formula in (d), show that (A + U BV )−1 = A−1 − A−1 U (I + BV A−1 U )−1 BV A−1 15 Solution (a) , : W A = Im XA + In V = 0 WU = Y XU + In D = Z A , W = A−1 X = −V A−1 Y = A−1 U Z = D − V A−1 U (b) , , Im Y 0 Z Im P 0 : Q = Im 0 0 In , , Im P + Y Q = 0 ZQ = In Z , Q = Z −1 , P = − Y Q = − Y Z −1 , −1 Im Y Im −Y Z −1 = −1 0Z 0 Z : A−1 0 AU V D 16 = Im 0 A−1 U D − V A−1 U (c) (a) −V A−1 In (b) , AU V D −1 = Im 0 Im 0 A−1 U −1 , D − V A−1 U = = A−1 + A−1 U (D − V A−1 U )−1 V A−1 −A−1 U (D − V A−1 U )−1 −(D − V A A−1 U )−1 V A−1 (D − V A−1 U )−1 D − V A−1 U , : −(A − U D−1 V )−1 U D−1 (A − U D−1 V )−1 U D−1 + −V A−1 In A−1 0 −A−1 U (D − V A−1 U )−1 −1 I −1 U )−1 −V A (D − V A n A−1 0 (d) (e) D AU V D −1 (c) , = (A − U D−1 V )−1 (A , (1,1) (d) − D −1 V − U D−1 V )−1 D −1 V D −1 D = −1, U = u, V = vT , , (A + uvT )−1 = A−1 + A−1 u(−1 − vT A−1 u)−1 vT A−1 = A−1 − 1 + vT A−1 u (f) B = − D −1 , D = − B −1 , (d) , A−1 uvT A−1 1 + vT A−1 u (A + U BV )−1 = A−1 − A−1 U (B −1 + V A−1 U )−1 V A−1 B (B −1 + V A−1 U ) = I + BV A−1 U , BV A−1 U )−1 B , (A + U BV )−1 = A−1 − A−1 U (I + BV A−1 U )−1 BV A−1 , A I + BV A−1 U 17 (B −1 + V A−1 U )−1 = (I + Problem 13 Suppose A and B are n × n matrices, and B is not nilpotent. If AB + BA = 0, nilpotent if Ak = 0 for some positive integer k. Solution k , 2k − 1 AB = −BA, then AX + XA = B has no solution. Note that an n × n matrix A is called AB 2k−1 = (AB )B 2k−2 = −BAB 2k−2 = B 2 AB 2k−3 = · · · = −B 2k−1 A , B = AX + XA, B 2k = (AX + XA)B 2k−1 = AXB 2k−1 + XAB 2k−1 = AXB 2k−1 − XB 2k−1 A traceB 2k = trace(AXB 2k−1 − XB 2k−1 A) = trace(AXB 2k−1 ) − trace(XB 2k−1 A) = trace(AXB 2k−1 ) − trace(AXB 2k−1 ) = 0 B λi , i = 1, 2, . . . , n, n (trace) λ2k = 0 i i=1 , traceB 2k = k , λi = 0, i = 1, 2, . . . , n, ,B , 18 Problem 14 Let 1 0 S= 1 −2 0 −1 2 0 Find an extension vector set T from R4 so that S ∪ T form a basis for R4 . Solution , , S 1 0 A, A , : e1 , . . . , e4 T , A 0 0 1 0 0 1 −1 0 0 1 0 −2 20001 1, 2, 4, 5 01000 , 0 1 1 0 0 1 2 0 0 0 1 0 0 000011 2 101000 0 0 1 0 T = , 0 1 0 0 19 Problem 15 Suppose A is a 3 × 4 matrix. The block matrix [ A I3 ] is reduced by row operations to 1210 2 −1 0 0 0 0 1 −1 1 0 0000 1 −2 1 (a) Find a basis for the row space of A. (b) Find a basis for the nullspace of A. (c) Find a basis for the left nullspace of A. (d) Find a basis for the column space of A. Solution , , E A Im m×n rankA = r , R = EA , R= r × (n − r ) E2 (m − r ) × m F , RE = Ir F 0 0 = A R Ir F 0 0 E1 E2 : E A EIm = RE A r , A m×n , m×m E E= E1 E2 E2 = , E1 r×m , , rankA = 2, 21 2 −1 0 , E1 = , F = 00 −1 10 20 1 −2 1 , (a) , F E2 A , A C (AT ) = C (RT ), , R [ Ir F ] [1, 2, 1, 0]T , [0, 0, 0, 1]T (b) , N (A) = N (R) matrix) Ax = 0 R −F In−r ,A N , Rx = 0 , (nullspace RN = 0 R N N = A (c) A [−2, 1, 0, 0]T , [−1, 0, 1, 0]T AT y = 0 yT A = 0T Ir F E1 A = =R EA = 00 E2 y E2 , E2 AT , AT E , N = −2 −1 1 0 0 0 1 0 E2 A = 0, [1, −2, 1]T (d) , B N (AT ), C (A) = N (B ), (c) AT , C (A) = C (R), N (AT ) , (c) R C (A) B = E2 , A C (B T ) = , C (A) = N (AT )⊥ A 21 B C (B T )⊥ = N (B ), , B = [ 1 −2 1 ], B , A A, E −1 , [2, 1, 0]T [−1, 0, 1]T E A = R, A = E −1 R , : 1211 110 1210 A = 1 2 0 0 0 0 1 = 1 2 1 2 1213 131 0000 R 1, 4 , A 1, 4 : [1, 1, 1]T [1, 2, 3]T Problem 16 Consider the following matrices: 1 4 −4 151 A = 2 6 0 , B = 4 6 −8 . 0 5 −4 172 nullspace. Solution A B, A B , , 0 −4 −2 0 2 1 1 4 −4 1 5 1 0 1 00 14 15 1 1 2 Determine if A and B have the same column space, row space, nullspace, or left (reduced row echelon form), AT BT , AT BT A 0 B 0 −10 8 0 5 −4 4 0 1 −5 00 0 22 −4 A, B 3 1 0 −2 1 0 1 2 00 0 0 1 −4 5 00 0 4 1 0 −5 A B , C (AT ) = C (B T ), N (A) = N (B ) AT , B T : 0 −4 2 0 −2 1 1 4 0 0 −10 5 0 8 −4 , 1 2 1 1 0 1 −2 00 0 14 0 0 1 −1 2 0 8 −4 12 1 1 0 1 −2 00 0 10 2 0 1 −1 2 00 0 10 2 A T BT AT BT C (A) = C (B ), N (AT ) = N (B T ) Problem 17 If possible construct a 3 × 3 real matrix A with the following properties: (a) A is a symmetric matrix. Its row space is spanned by the vector [1, 1, 2]T and its column space is spanned by the vector [1, 2, 3]T . (b) All three of these equations have no solution but A = 0: Ax = e1 , Ax = e2 , Ax = e3 where e1 = [1, 0, 0]T , e2 = [0, 1, 0]T , e3 = [0, 0, 1]T . (c) The vector [1, 1, 1]T is in the row space of A but the vector [1, −1, 0]T is not in the nullspace of A. Solution (a) A A = uvT , 112 1 A = k 2 [ 1 1 2 ] = k 2 2 4 336 3 23 , rankA = 1 1 k = 0, (b) e1 , e2 , e3 A 111 A=1 1 1 111 (c) [1, 1, 1]T , , [1, −1, 0]T [1, 1, 1]T , , , [1, −1, 0]T 111 A=0 1 1 000 , [1, −1, 0]T , , , [0, −1, 1]T Problem 18 [1, −1, 0]T Suppose that U and W are two subspaces in R4 . The subspace U is spanned by u1 = [1, 1, 0, 0]T and u2 = [1, 0, 1, 0]T , and the subspace W is spanned by w1 = [0, 1, 0, 1]T and w2 = [0, 0, 1, 1]T . (a) Find a basis for the sum U + W . (b) Find a basis for the intersection U ∩ W . (c) Explain why the following relation is correct. dimU + dimW = dim(U + W ) + dim(U ∩ W ) Solution (a) A A u1 , u2 , w1 , w2 , A U + W = span(U ∪ W ) , 24 : 1 0 1 0 A= 0 1 0 1 0011 U +W 1100 0 −1 1 0 0 1 0 1 0 011 A 1 1 , 0 0 1 100 0 −1 1 0 0 0 1 1 0 011 1 100 0 1 −1 0 0 0 1 1 00 00 11 00 (b) x ∈ N (A) U ∩W Ax = 0, , 1, 2, 1 0 , 1 0 3: 0 1 0 1 (a) A N (A) x1 u1 + x2 u2 + x3 w1 + x4 w2 = 0 ui wi : x1 u1 + x2 u2 = −x3 w1 − x4 w2 U, (a) , A 1 −1 0 0 1 1 0 00 A 1 00 1 0 0 0 0 , 10 W, : U ∩W 1 0 0 0 0 0 −1 10 01 00 , 1 0 0 0 1 −1 0 0 1 1 0 00 [1, −1, −1, 1]T 1 1 1 1 0 U ∩W 1 0 1 − = 1 · u1 + (−1) · u2 = 0 1 −1 0 0 0 25 0 (c) (a) (b) U +W U ∩W , rankA = dim(U + W ), dimU + dimW , – dimN (A) = dim(U ∩ W ) A dimN (A) + rank(A) = dimU + dimW , dimU + dimW = dim(U + W ) + dim(U ∩ W ) Problem 19 Let W1 and W2 be two subspaces in vector space V . Show that the following two statements are equivalent. (a) W1 ∩ W2 = {0}. (b) dim(W1 + W2 ) = dimW1 + dimW2 Solution (a) ⇒ (b) (a) , {x1 , . . . , xm } W1 , {y1 , . . . , yn } W2 , , W1 ∩ W2 = {0}, (b) ⇒ (a) , , (b) {z1 , . . . , zk } k > 0, dim(W1 + W2 ) < dimW1 + dimW2 (b) Problem 20 Prove that the intersection of three 6-dimensional subspaces in R8 is not the single point {0}. 26 , k = 0, W1 ∩ W2 = {0} W1 ∩ W2 = {0} W1 W2 , {z1 , . . . , zk } W1 ∩ W2 dim(W1 + W2 ) = {z1 , . . . , zk , x1 , . . . , xm } {x1 , . . . , xm , y1 , . . . , yn } W1 + W2 dim(W1 + W2 ) = m + n = dimW1 + dimW2 {z1 , . . . , zk , y1 , . . . , yn }, k+n+m dimW1 = k + m, dimW2 = k + n, Solution , V, W R8 , dimU = dimV = dimW = 6 U +V U, R8 , dim(U + V ) ≤ 8, dim(U + V ) = dimU + dimV − dim(U ∩ V ) 6 + 6 − dim(U ∩ V ) ≤ 8, dim(U ∩ V ) ≥ 4, dim(V ∩ W ) ≥ 4 dim((U ∩ V ) + (U ∩ W )) = dim(U ∩ V ) + dim(U ∩ W ) − dim(U ∩ V ∩ W ) (U ∩ V ) + (U ∩ W ) ⊆ U , dim((U ∩ V ) + (U ∩ W )) ≤ 6, dim(U ∩ V ) + dim(U ∩ W ) − dim(U ∩ V ∩ W ) ≤ 6 dim(U ∩ V ) ≥ 4, dim(V ∩ W ) ≥ 4, dim(U ∩ V ∩ W ) ≥ dim(U ∩ V ) + dim(U ∩ W ) − 6 ≥ 4 + 4 − 6 = 2 U, V , W Problem 21 Let (a) Find the intersection of the nullspace of A and the nullspace of B . (b) Find the intersection of the column space of A and the column space of B . A=2 2 0 , 1 −5 3 2 −4 3 B=5 4 6 0 3 −3 12 0 27 Solution (a) x ∈ N (A) ∩ N (B ), 0 0 , 2 −4 3 Ax = 0 B x = 0, 0 1 A B 0 100 A B x = N (A) ∩ N (B ) , 1 : 2 2 −2 −7 −8 −6 3 B 2 2 0 1 −5 3 1 2 0 5 4 6 0 3 −3 2 2 0 1 −5 3 2 −4 3 5 4 6 0 3 −3 , A 0 0 0 0 0 0 3 3 6 −3 0 0 0 0 0 {0} 1 0 0 1 0 0 0 0 00 (b) , A, B D, E , (a) N (B T ) BT : 2 2 1 0 6 −3 3 0 −3 2 1 5 0 N (AT ) C (A) = N (AT )⊥ , , E , D , , C (A) = C (D T ) = N (AT ), C (D T )⊥ = N (D ) N (E ) AT 22 1 AT = −4 2 −5 30 3 B =2 4 3 0 6 −3 T C (B ) = 15 0 0 −6 3 0 6 −3 0 1 00 15 11 1 2 −1 2 0 0 1 0 1 −2 00 0 0 1 −1 2 00 0 10 5 2 −1 2 10 1 0 1 00 0 28 D = [ −2 1 2 ], E = [ −5 1 2 ] , D E −2 1 2 −5 1 2 , −2 1 2 −3 0 0 100 −2 1 2 , A B 100 012 D E [0, −2, 1]T Problem 22 E1 Suppose A is m by n, and rankA = r . Let E = be a nonsingular matrix E2 such that B EA = 0 where B is r by n, and B is in row echelon form. Prove that C (A) = N (E2 ), where C (A) is the column space of A, and N (E2 ) is the nullspace of E2 . Solution ) C (A) ⊂ N (E2 ) C (A = N (E2 ), E B 1 A = , E2 A = 0, A E2 E2 0 b ∈ N (E2 ), E1 b E1 c = Eb = b = E2 E2 b 0 c = E1 b rA b, , b ∈ C (A), N (E2 ) ⊂ C (A) , C (A) ⊂ N (E2 ) E A b = EA Eb = rank A b = rankA = r , N (E2 ) ⊂ C (A) 29 Bc 0 0 Ax = b Problem 23 Let A be an n by n nilpotent matrix of index k, i.e., k is the smallest positive integer such that Ak = 0. If x is a vector such that Ak−1 x = 0, show that the set {x, Ax, A2 x, . . . , Ak−1 x} is linearly independent. Solution c1 x + c2 Ax + · · · + ck Ak−1 x = 0 Ak−1 , Aj = 0, j ≥ k, c1 Ak−1 x + c2 Ak x + · · · + ck A2k−2 x = c1 Ak−1 x = 0 Ak−1 x = 0 c1 = 0, c2 Ax + c3 A2 x + · · · + ck Ak−1 x = 0 Ak−2 , c2 = 0, c3 A2 x + · · · + ck Ak−1 x = 0 ci = 0, i = 1, . . . , k Problem 24 Let {v1 , . . . , vn } be a basis for a vector space V and n ≥ 2. Is the set {v1 + v2 , v2 + v3 , . . . , vn−1 + vn , vn + v1 } also a basis for V ? Solution c1 (v1 + v2 ) + c2 (v2 + v3 ) + · · · + cn (vn + v1 ) = 0 30 (c1 + cn )v1 + (c1 + c2 )v2 + · · · + (cn−1 + cn )vn = 0 : c1 + cn = 0, c1 + c2 = 0, . . ., cn−1 + cn = 0, ci 1 0 0 ··· 0 1 1 1 0 ··· 0 0 0 1 1 ··· 0 0 n+1 =0 . . . . . . . = 1 + (−1) ... ... ... .. 0 0 0 ··· 1 0 0 0 0 ··· 1 1 n , ci V , V ; n , ci = 0, i = 1, . . . , n, Problem 25 Vandermonde matrices are descirbed by n 1 x1 x2 · · · x1 −1 1 n−1 2 1 x2 x2 · · · x2 V = . . . . .. . . . . . . . . . n−1 2 1 xm xm · · · xm in which xi = xj , for all i = j . Show that if n ≤ m, the columns of V are linearly independent. Solution 0 n x1 −1 c1 0 n x2 −1 c2 = 0 . . . . . . . . . n− xm 1 0 cn−1 c0 31 1 x1 x2 · · · 1 2 1 x2 x2 · · · . . .. . . . . . . . . 1 xm x2 · · · m i = 1, 2, . . . , m, n c0 + c1 xi + c2 x2 + · · · + cn−1 xi −1 = 0 i p(x) = c0 + c1 x + c2 x2 + · · · + cn−1 xn−1 m (n − 1) cn−1 = 0, Problem 26 If A is an m × n matrix and B is an n × p matrix, prove that rankA + rankB − n ≤ rankAB This is called Sylvester’s inequality. Solution – , dim N (A) = n − rankA rankB − rankAB ≤ dim N (A) rankB T : C (B ) → Cm , T (x) = Ax, T (y) = AB y, T y ∈ Cp rankAB , C (B ), – x ∈ C (B ), : V , xi , i = 1, 2, . . . , m , n ≤ m, , (n − 1) p(x) c0 = c1 = · · · = dim ima(T ) + dim ker(T ) = dim C (B ) dim ima(T ) = dim C (AB ) = rankAB , dim C (B ) = rankB , dim ker(T ) = rankB − rankAB C (B ) ⊆ Cn , dim ker(T ) ≤ dimN (A), 32 T A N (A) ( ), Problem 27 Suppose A is an m × n matrix. (a) Prove that there exist an m × m invertible matrix P and an n × n invertible matrix Q such that P AQ = Ir 0 0 0 where r is the rank of A. (b) Suppose that A with rank r can be factorized as Ir 0 V A=U 00 where U is an m × m invertible matrix and V is an n × n invertible matrix. Decribe the column space of A and the row space of A in terms of U and V. Solution (a) A , rankA = r , P AS = , , n×n P Ir F 0 0 S =R TT , m×m P , S RT n×n n×m , rankRT = r , JT Q = ST , n×m T T RT = T T , Ir FT 0 0 J = RT , = Ir 0 0 0 R = P AS = JT P AST = J , P AQ = J , 33 (b) U V U= U1 U r A= A A= : U1 U2 , , V1 U1 U2 A V V = A , V1 V2 A : Ir 0 0 0 r V1 V2 V : , C (V ) = Cn , U1 U2 Ir 0 0 0 V = U1 0 V U1 V1 0 V1 U Problem 28 , C (U ) = Cm , A=U Ir 0 0 0 A V1 V2 =U Let A be an n × n matrix. Prove that there is an invertible matrix B such that BA is a projection. Solution A A C (A) , r, Cn yi {y1 , . . . , yn } yi ∈ N (A), i = r + 1, . . . , n, n r r N (A) n−r Ayi = xi , {x1 , . . . , xr } {yr+1 , . . . , yn } Cn : {x1 , . . . , xr , xr+1 , . . . , xn } i = 1, . . . , r , xi ∈ C (A), N (A) , c1 y1 + · · · + cn yn = 0, A i=1 ci yi = i=1 ci Ayi = i=1 ci xi = 0 34 x1 , . . . , xr C (A) , Cn , yr+1 , . . . , yn : {x1 , . . . , xn } c1 = · · · = cr = 0, N (A) , {y1 , . . . , yn } cr+1 yr+1 + · · · + cn yn = 0, cr+1 = · · · = cn = 0 B i = r + 1, . . . , n, BAyi = B 0 = 0, Problem 29 B xi = yi , i = 1, . . . , n BA i = 1, . . . , r , BAyi = B xi = yi , Let A and B be n by n matrices. If AB = 0, show that the dimension of the nullspace of BA is at least n/2. Solution B = b1 · · · bn , bj ∈ Cn , AB = A b1 · · · bn = Ab1 · · · Abn = 0 bj ∈ N (A), j = 1, . . . , n, , rankA + dimN (A) = n, min{rankA, rankB }, rank(BA) ≤ min{rankA, n − rankA} rank(BA) ≤ n/2, – , C (B ) ⊆ N (A), rankB ≤ dimN (A) – rankB ≤ n − rankA, rank(BA) ≤ dimN (BA) = n − rank(BA) ≥ n/2 Problem 30 Let A be an n by n matrix of the form D0 P −1 A=P 00 Prove the following statements. 35 where P and D are n × n and r × r , r ≤ n, nonsingular matrices, respectively. (a) There exists a nonsingular matrix B such that A2 = BA. (b) rankA = rankA2 (c) The column space of A and the nullspace of A are disjoint, i.e., C (A) ∩ N (A) = {0}. Solution (a) D2 0 D0 D0 P −1 = P P −1 P P −1 = BA A2 = P 00 0D 00 D0 P −1 B=P 0D (b) (c) (a) ⇒ (b) (b) ⇒ (c) y , rankA2 = rank(BA) = rankA dimN (A) = dimN (A2 ) x ∈ C (A)∩N (A), Ax = 0 rankA = rankA2 , N (A) ⊆ N (A2 ), x = Ay, N (A) = N (A2 ) Ax = A(Ay) = A2 y = 0 y ∈ N (A2 ) = N (A), Ay = 0, Problem 31 Let A and B be nonsingular n × n matrices. Prove that rank(A − B ) = rank(A−1 − B −1 ) Solution A−B A−1 − B −1 , : x=0 B −1 − A−1 = B −1 (A − B )A−1 36 A−1 B −1 , , rank(A−1 − B −1 ) = rank(B −1 − A−1 ), A m×m : , N (AB ) = N (B ) A−1 , rank(B −1 − A−1 ) = rank(A − B ), ,B B x = 0, m × n, AB x = 0 – rank(AB ) = rankB rankAT = rankA, : AB , B AB x = 0, A−1 AB x = B x = 0 rank(AB ) = n − dimN (AB ), rankB = n − dimN (B ), , BA , A , rank(BA) = rank(BA)T = rank(AT B T ) = rankB T = rankB Problem 32 Suppose A and B are n × n matrices and A is nonsingular. Let B = A − U QV , I − V A−1 U Q is nonsingular, and vice versa. Solution B V x = y, , x=0 , y=0 A, (I − V A−1 U Q)y = 0 Ax = U Qy, x = A−1 U Qy, where U is n × p, Q is p × m, and V is m × n. Show that if B is nonsingular then B x = Ax − U QV x = Ax − U Qy = 0 B V A−1 U Q I − V A−1 U Q Ax = U QV x , A−1 , , I − V A−1 U Q , x=0 , I− B x = 0, x = A−1 U QV x = A−1 U Qy (I − V A−1 U Q)y = y − V x = 0 I − V A−1 U Q U QV x = 0, Problem 33 Suppose A is an m × n real matrix. 37 A , , y = 0, B V x = 0, ,B Ax = B x + (a) If AT Ax = 0, show that Ax = 0. (b) Show that A and AT A have the same nullspace. (c) Show that A and AT A have the same row space, and thus rankA = rankAT A. Solution (a) 0 Ax Ax = 0 (b) x ∈ N (A), N (AT A) (c) (b) Ax = 0, (a) AT Ax = AT 0 = 0, N (AT A) ⊆ N (A), x ∈ N (AT A), N (A) = C (A), xT AT Ax = 0, xT AT Ax = (Ax)T Ax = Ax , AT (Ax) = 0 2 = 0, Ax = N (AT ) ∩ C (A) = {0} Ax ∈ N (AT ), N (A) ⊆ N (AT A) , N (A) = N (AT A), C (AT ) = N (A)⊥ , C (AT A) = N (AT A)⊥ C (AT ) = C (AT A), rankA = dimC (AT ) = dimC (AT A) = rankAT A Problem 34 (a) Suppose A is an n by n matrix of rank r . Show that A may be written as a sum of rank-one matrices: A = x1 yT + · · · + xr yT 1 r where xi and yi , i = 1, . . . , r , are n-dimensional vectors. (b) Suppose B is an n by n matrix of the form B = x1 yT + · · · + xr yT 1 r If n-dimensional vectors x1 , . . . , xr are independent, what conditions of yi ’s must be satisfied so that the rank of B is r ? 38 (c) Suppose X is a 3 × 2 matrix, and Y is a 2 × 5 matrix. How many possible dimensions of the nullspace of XY are there? Solution (a) A A aj yij ’s aj = y1j x1 + · · · + yrj xr , : y11 · · · y1n . . .. . . . . . yr1 · · · yrn , r, A {x1 , . . . , xr }, j = 1, . . . , n, A= a1 · · · an = x1 · · · xr , A= x1 · · · xr yT 1 . . = x1 yT + · · · + xr yT . 1 r T yr : {x1 , . . . , xr , xr+1 , . . . , xn }, yT .1 . . T yr T 0 . . . 0T (b) B {x1 , . . . , xr } n- B= x1 · · · xr xr+1 . . . xn n×n ( X , ) n , rankB = rankY , 39 y1 , . . . , yr = XY Y Y X r rankB = r , (c) XY Y, X 0 ≤ rankXY ≤ 2 , 3×2 XY 3×5 , dimN (XY ) X - 2, dimN (XY ) = 5 − rankXY , Problem 35 3, 4, 5 Suppose A is an m × n real matrix with full row rank, i.e., rankA = m. Which of the following equations always have a solution (possibly infinitely many) for any legal b? (a) Ax = b (b) AT x = b (c) AT Ax = b (d) AAT x = b (e) AT Ax = AT b (f) AAT x = Ab Solution (a) Ax = b (b) AT x = b n>m (c) AT Ax = b dimC (AT ) = m, , , , AT , : dimC (A) = rankA = m, dimC (AT ) = rankA = m, ( m) N (A) = N (AT A), AT A Rn C (A) = Rm b Rn , (b), dimC (AT A) = m Rn N (A)⊥ = N (AT A)⊥ , C (AT ) = C (AT A) n×n (d) AAT x = b , AAT dimC (AAT ) = dimC (A) = m, AAT m×m , , (c) C (AAT ) = C (A), 40 (e) AT Ax = AT b AT b (f) AAT x = Ab , , (a) , Ax = b AT Ax = (d), AAT 41 Problem 36 Let V be the vector space of polynomials of degree at most 3 with real coefficients. Let T be the map defined by T (f (x)) = for all f (x) ∈ V . (a) Show that T is a linear transformation. (b) Find the matrix [T ]B and [T ]C representing T with respect to the bases B = {1, x, x2 , x3 } and C = {1, 1+ x, 1+ x + x2 , 1+ x + x2 + x3 }, respectively. (c) Find the matrix M representing the change of bases from B to C. Solution (a) f (x), g(x) ∈ V , T (f (x) + g(x)) = d(f + g) d2 (f + g) = +2 2 dx dx d2 f df +2 2 dx dx + dg d2 g +2 2 dx dx d2 f df +2 2 dx dx = T (f (x)) + T (g(x)) T (cf (x)) = T (b) B T (1) = 0 T (x) = 2 T (x2 ) = 2 + 4x T (x3 ) = 6x + 6x2 d2 cf dcf +2 =c dx2 dx d2 f df +2 dx2 dx = cT (f (x)) 42 T (a + bx + cx2 + dx3 ) 0 0 [T (x)]B = [T ]B [x]B = 0 0 , : C a 0 4 6 b 0 0 6 c d 000 220 T (1) = 0 T (1 + x) = 2 T (1 + x + x2 ) = 4 + 4x = 4(1 + x) T (1 + x + x2 + x3 ) = 4 + 10x + 6x2 = −6 + 4(1 + x) + 6(1 + x + x2 ) T (s + t(1 + x) + u(1 + x + x2 ) + v (1 + x + x2 + x3 )) s 0 2 0 −6 0 0 4 4 t [T (x)]C = [T ]C [x]C = 0 0 0 6 u v 000 0 (c) M B B , 0 1 1 1 = 0 0 1 1 0001 1 −1 0 0 43 0 1 −1 0 1 0 0 1111 C , M −1 C B M −1 0 M = 0 0 1 −1 M [x]B = [x]C M [T (x)]B = [T (x)]C , (b) : M [T ]B [x]B = : 2 0 −6 04 00 00 4 6 0 [T ]C M [x]B , M [T ]B = [T ]C M , 022 1 −1 0 0 0 1 −1 0 0 0 4 0 0 1 −1 0 0 0 000 0 0 0 1 Problem 37 [T ]C = M [T ]B M −1 0 1111 0 6 0 1 1 1 0 = 6 0 0 1 1 0 0001 0 0 Let A be an m × n matrix. Prove or disprove the following statements. (a) If the vectors v1 , v2 , . . . , vk in Cn are linearly independent, so are Av1 , Av2 , . . . , Avk . (b) If the vectors Av1 , Av2 , . . . , Avk in Cm are linearly independent, so are v1 , v2 , . . . , vk . Solution (a) A = 0, Av1 , Av2 , . . . , Avk ? , , A c1 Av1 + c2 Av2 + · · · + ck Avk = A(c1 v1 + c2 v2 + · · · + ck vk ) = 0 c1 = c2 = · · · = ck = 0, , (b) c2 , . . . , ck c1 v1 + c2 v2 + · · · + ck vk = 0 A, c1 Av1 + c2 Av2 + · · · + ck Avk = 0 Av1 , Av2 , . . . , Avk 44 , , , v1 , · · · , vk v1 , v2 , . . . , vk A A , c1 , v1 , · · · , vk Problem 38 Suppose [T ]E = 11 01 is the transformation matrix for a linear transformation −9 −25 is the transfor4 11 mation matrix with respect to B basis. Find a set of basis vectors of B. Solution B B [T (x)]B = [T ]B [x]B [T ]B B −1 = B −1 [T ]E , L 2{b1 , b2 }, [T ]E B = [ b1 b2 ], B B [x]B , T with respect to the standard basis, and [T ]B = T (x) = [T ]E x, [x]B = B −1 x [T (x)]B = B −1 T (x) = B −1 [T ]E x, ) L(B = B [T ]B − [T ]E B , ab , B= L(B ) , cd −25c + 10d B [T ]B = [T ]E B L(B ) = 0 (kernel), L(B ) = −10a + 4b − c −25a + 10b − d −10c + 4d a −10 4 −1 0 a b −25 10 0 −1 b L = c 0 0 −10 4 c d 0 0 −25 10 d : −10 4 −1 0 1 −2 5 0 1 25 2 −5 , : −25 10 0 −1 0 0 −10 4 0 0 −25 10 0 0 0 01 00 00 0 0 45 [ 2 , 1, 0, 0]T 5 1 [− 25 , 0, 2 , 1] 5 , − 1 25 β 2×2 α β α β 12 β =0 25 , L(B ) = 0 B B , det(B ) = β=0 Problem 39 : 2 5α 2 5α 2 5β − 1 25 β 2 5β , α β 2 5α − 1 25 β 2 5β =− Consider the following linear transformation T (X ) = AX − XA where X is a 2 × 2 matrix. If A is diagonalizable, prove that T is also diagonalizable. Solution Axi = λi xi , i = 1, 2, Bij j xi , A , , {x1 , x2 } B = {B11 , B12 , B21 , B22 } Bij : T (B11 ) = AB11 − B11 A 2×2 2×2 T (Bij ) = ABij − Bij A, = A x1 0 − x1 0 a11 a12 a21 a21 = λ1 x1 0 − a11 x1 a12 x1 = λ1 B11 − a11 B11 − a12 B12 46 T (B12 ) = AB12 − B12 A = λ1 B12 − a21 B11 − a22 B12 T (B21 ) = AB21 − B21 A = λ2 B21 − a11 B21 − a12 B22 T (B22 ) = AB22 − B22 A = λ2 B22 − a21 B21 − a22 B22 T B [T ]B = [T (B11 )]B [T (B12 )]B [T (B21 )]B [T (B22 )]B λ1 − a11 −a21 0 0 −a12 λ I − AT λ1 − a22 0 0 = 1 = 0 0 0 λ2 − a11 −a21 0 0 −a12 λ2 − a22 A A = S ΛS −1 , cI − AT = (S T )−1 (cI − Λ)S T 0 λ2 I − AT , Problem 40 If {u1 , . . . , un } and {v1 , . . . , vn } are bases of Rn , and if n Auj = i=1 n cij ui cij vi i=1 B vj = what is the relation between the matrices A and B ? Solution C, U , V 1, 2, . . . , n n×n , C = [cij ], U : c · · · c1n 11 . . .. . . . . . cn1 · · · cnn uj , V vj , j = A u1 · · · un = u1 · · · un 47 AU = U C , ,U V , A C A = U V −1 BV U −1 P = U V −1 , , P vj = uj , j = 1, . . . , n, n n BV = V C A = P BP −1 , A U, V C = V −1 BV , B , P A = U CU −1 Auj = AP vj n Auj = i=1 cij ui = i=1 cij P vi = P i=1 cij vi = P B vj AP − P B A, B j , AP vj = P B vj , n, (AP − P B )vj , j = 1, 2, . . . , n, AP = P B P rank(AP − P B ) = 0, , 48 Problem 41 Suppose W is a subspace of an inner product space V and W ⊥ is the orthogonal projection of x onto W , i.e., x = y + y′ , for some y′ ∈ W ⊥ , if and only if for every z ∈ W , x − y ≤ x − z . Solution y ∈ W, y − z ∈ W, x = y + y′ , y′ ∈ W ⊥ , x − y ∈ W⊥ z ∈ W, complement of W in V . For any x ∈ V , show that y ∈ W is the orthogonal x − y, y − z = 0, x−z 2 = (x − y) + (y − z) 2 = x−y 2 + y−z 2 x−y ≤ x−z , w′ ∈ W ⊥ , w x W w=y y, w ∈ W , x−y 2 x = w + w′ , = (x − w) − (w − y) = x−w ≥ x−w 2 2 2 2 + w−y , Problem 42 Let {v1 , . . . , vn } be an orthonormal basis for an inner product space V over C. If x ∈ V , show that (a) x = n i=1 x, vi vi k i=1 | (b) x, x ≥ x, vi |2 , 1 ≤ k ≤ n. Note that x, y denotes the inner product of x and y. 49 Solution (a) x x = c1 v1 + · · · + cn vn vi , vi , vj = 1 i = j , vi , vj = 0 i = j, ci = x, vi , i = 1, . . . , n (b) (a) n n x, x = = = ci vi , i=1 n n j =1 cj vj ≥ ci vi , cj vj i=1 j =1 n |ci |2 i=1 k |ci |2 (k = 1, . . . , n) i=1 Problem 43 Suppose x and y are vectors in an inner product space V . Show that | x − y | ≤ x−y This is called the backward triangle inequality. Solution x+y ≤ x + y x, : x = x−y+y ≤ x−y + y 50 x − y ≤ x−y y = x−y+x ≤ x−y + x −( x − y ) ≤ x − y Problem 44 Suppose x1 , x2 , x3 are three vectors in Rn , and xT xj < 0, for i, j = 1, 2, 3, i = j . i Prove that x1 , x2 , x3 are linearly independent. Solution , x, x2 , x3 , xi , i = 1, 2, 3, , x3 = c1 x1 + c2 x2 , xT x3 = xT (c1 x1 + c2 x2 ) = c1 + c2 xT x2 < 0 1 1 1 xT x3 = xT (c1 x1 + c2 x2 ) = c1 xT x1 + c2 < 0 2 2 2 xT x2 < 0, 1 −xT x2 1 −c1 (xT x2 ) < c2 (xT x2 )2 , 1 1 c2 < −c1 xT x1 < c2 (xT x2 )2 2 1 (xT x2 )2 > 1, 1 x1 , x2 , x1 , x2 , x3 Problem 45 Let x1 , . . . , xn be vectors in Rm , and let A = [aij ], where aij = xT xj i i, j = 1, . . . , n. Show that x1 , . . . , xn are linearly independent if and only if A is nonsingular. 51 Solution c1 x1 + · · · + cn xn = 0 xT , i = 1, 2, . . . , n, i , xT (c1 x1 + · · · + cn xn ) = c1 xT x1 + · · · + cn xT xn = c1 ai1 + · · · + cn ain = 0 i i i n , : x1 , . . . , xn , rankA = n, A QR Q T Q = In , 0 c a11 · · · a1n 1 . . . . .. . . = . Ac = . . . . . . 0 cn an1 · · · ann , c1 = · · · = cn = 0, QR Ac = 0 X = x1 · · · xn , X = QR, A = X T X = (QR)T (QR) = RT QT QR = RT R X A Problem 46 Let A be an n by n real matrix satisfying (Ax)T (Ay) = xT y for every x, y in Rn . Show that A is an orthogonal matrix. Solution (Ax)T (Ax) = xT AT Ax = xT x xT (AT A − I )x = 0 52 , R rankA = rank(RT R) = rankR, xT B x = 0, B = AT A − I , B B=0 , : , B, x (x + y)T B (x + y) (x + y)T B (x + y) − xT B x − yT B y = xT B y + yT B x , , AT A = I , Problem 47 Suppose Q is a 4 × 3 real matrix with orthonormal columns q1 , q2 , q3 . (a) Suppose v is not in the column space of Q. Use Gram-Schmidt process to obtain the fourth orthonormal vector q4 . (b) Describe the nullspaces of Q, QT , QT Q and QQT . (c) Suppose b = q1 + 2q2 + 3q3 + 4q4 . Find the least-squares solution to Qx = b. What is the projection p of this b onto the column space of Q? Solution (a) v qi , i = 1, 2, 3, : By 2 , = 0, AT = A−1 , A 2xT B y, B y = 0, y xT B y = 0 , x = By B = 0, p4 = v − (vT q1 )q1 − (vT q2 )q2 − (vT q3 )q3 : q4 = p4 / p4 (b) Q = [ q1 q2 q3 ] C (Q) q1 , q2 , q3 q4 ) , , Q T Q = I3 , 3 N (QT ) ,Q (a) q4 4×4 QT ( q4 Q QT Q QT q 4 = Q · 0 = 0 {0}, 53 (c) QT Q = I , QT b , qT 1 ˆ (normal equation) QT Qx = QT b 1 ˆ x= p Problem 48 ˆ x = qT 2 qT 3 ˆ p = Qx = q1 + 2q2 + 3q3 (q1 + 2q2 + 3q3 + 4q4 ) = 2 3 Suppose C is an n × n real symmetric positive-definite matrix. (a) Suppose C = B T B for some m × n matrix B . Which of the following (if any) must be properties of B : full column rank, full row rank? (b) Suppose A has linear independent columns, and C = B T B as in (a). In terms of B , A, and b, write down an explicit formula for the x that minimize (Ax − b)T C (Ax − b). (c) Suppose that C was only positive semi-definite. Is there still a minimum value of (Ax − b)T C (Ax − b)? Is there still a unique solution x? Solution (a) C , x = 0, xT C x = xT B T B x = B x 2 0, > 10 B , , B = 0 1 , 00 N (B ) = {0}, B 10 01 (b) C = BT B = : (Ax − b)T C (Ax − b) = (Ax − b)T B T B (Ax − b) = (BAx − B b)T (BAx − B b) = B Ax − B b 54 2 (BA)T (BA)ˆ = (BA)T B b x B B Ax = 0, (BA)T (BA) , BA Ax = 0, , ˆ x = (AT B T BA)−1 AT B T B b = (AT CA)−1 AT C b (c) B C , (Ax − b)T C (Ax − b) Problem 49 Suppose A is an n × n real matrix. , C = BT B, (Ax − b)T C (Ax − b) , C = 0, x = 0, , N (A) = {0} N (BA) = {0} B A (a), A N (B ) = {0}, , , x (a) Show that AT A may be written as AT A = LLT , where L is a real lower triangular matrix with nonnegative diagonal entries. (b) Show that the factorization AT A = LLT , L with positive diagonal entries, is unique if A is nonsingular. This is called the Cholesky factorization of AT A; every positive definite matrix may be factored in this way. Solution (a) A ,R QR n×n A = QR, Q = q1 · · · qn n×n R (R)ii ≥ 0 qT 1 . R = Q− 1 A = QT A = . a 1 · · · a n . qT n (R)ii = qT ai , i QR L = RT , Q− 1 = QT , qi AT A = (QR)T (QR) = RT QT QR = RT R = LLT L 55 (b) A Ax 2 ,A > 0, AT A , x = 0, xT AT Ax = (Ax)T (Ax) = AT A B = AT A, B = LLT , (AT A)ii = aT ai = eT AT Aei > 0 i i bT 21 0T B= b11 b21 B22 b11 > 0, = 1 l11 l11 0 lT 21 LT 22 l21 L22 l11 = √ b11 , l21 = = 2 l11 l11 lT 21 l11 l21 l21 lT + L22 LT 21 22 (2,2) √1 b21 , b11 L22 LT = B22 − l21 lT = B22 − 22 21 B , x yT b11 bT 21 x, y x y 1 b21 bT 21 b11 b21 B22 = b11 x2 + 2x(bT y) + yT B22 y 21 = b11 x − = b11 x − B22 − b1 b21 bT 21 11 B22 − Problem 50 Suppse u and v are linear independent vectors in Rn . (a) Find a nonzero vector w in Rn as a linear combination of u and v so that w is perpendicular to u. (b) Suppose u and v are two columns of A, i.e., A = [ u v ], find Q and R so that A = QR, where Q has orthonormal columns and R is a 2 by 2 upper triangular matrix. (c) Let P be the orthogonal projection matrix onto the subspace spanned by u and v. Express P in terms of Q only. 56 T 1 b22 b21 b21 1T by b11 21 1T by b11 21 2 + yT B22 y − 2 1 (bT y)2 b11 21 1 b21 bT y > 0 21 b11 , L22 + yT B22 − = L22 LT 22 Solution (a) Gram-Schmidt w u , v u w = v − cu, wT u = vT u − cuT u = 0, w=v− vT u u uT u , c = (vT u)/(uT u), (b) (a) Q , w u , q1 = u/ u , q2 = w/ w , A q1 A = QR: u u u u w w w w r1 r2 q2 , Q = [ q1 q2 ] R r1 r2 , u = Qr1 = [ q1 q2 ]r1 = v = Qr2 = [ q1 q2 ]r2 = r1 = u 0 , r2 = A cu w , , c = (vT u)/(uT u) (b) QR QT Q = I , (c) P P = A(AT A)−1 AT = (QR)(RT QT QR)−1 (RT QT ) = (QR)(RT R)−1 (RT QT ) = Q(RR−1 (RT )−1 RT )QT = QQT Problem 51 Suppoes A, B and Q are n × n real matrices, and B = QT AQ, where Q is an orthogonal matrix. Denote the (i, j ) entry of A to be aij . Show that n n n n a2 = ij i=1 j =1 i=1 j =1 b2 ij 57 Solution n n a2 = tr(AT A) ij i=1 j =1 tr(AT A) Q− 1 , tr(AT A), , tr(CD) = tr(DC ) Problem 52 = tr(B T B ) B = QT AQ, QT = tr(B T B ) = tr(QT AT QQT AQ) = tr(QT AT AQ) = tr(AT AQQT ) = : CD DC , Suppose u1 , u2 , u3 form an orthonormal basis for R3 and v1 , v2 form an orthonormal basis for R2 . Let A = u1 vT + u2 vT . 1 2 (a) Show that AAT is an orthogonal projection matrix. (b) Find the eigenvalues and corresponding eigenvectors of AAT . (c) Show that AT A is the identity matrix. Solution (a) vT v2 = 0 1 vT vi = 1, i AAT = (u1 vT + u2 vT )(v1 uT + v2 uT ) = u1 uT + u2 uT 1 2 1 2 1 2 AAT , (AAT )2 = AAT , (AAT )2 = (u1 uT + u2 uT )(u1 uT + u2 uT ) = u1 uT + u2 uT 1 2 1 2 1 2 uT u2 = 0 1 uT ui = 1 i PT = P P2 = P (b) AAT R3 {u1 , u2 }, AAT AAT , u1 (a) u2 , AAT : AAT u1 = (u1 uT + u2 uT )u1 = u1 1 2 AAT u2 = (u1 uT + u2 uT )u2 = u2 1 2 58 AAT 1, 1, , , u3 , AAT u1 , u2 u3 , N ((AAT )T ), 1, 1, 0, , AAT u3 AAT , (AAT )T u3 = AAT u3 = 0 u1 , u2 , u3 (c) AT A = (v1 uT + v2 uT )(u1 vT + u2 vT ) = v1 vT + v2 vT 1 2 1 2 1 2 AT Av1 = (v1 vT + v2 vT )v1 = v1 2 1 v1 , v2 R2 , AT Av2 = (v1 vT + v2 vT )v2 = v2 , 2 1 x = c1 v1 + c2 v2 , x ∈ R2 AT Ax = AT A(c1 v1 + c2 v2 ) = c1 AT Av1 + c2 AT Av2 = c1 v1 + c2 v2 = x x ∈ R2 , Problem 53 Suppose we are given the following measurements: y = 2 at t = −1, y = 0 at t = 0, y = −3 at t = 1, and y = −5 at t = 2. (a) Find the best straight line fit to the above measurements. (b) Suppose that instead of a straight line, we fit the data by a parabola: y = c0 + c1 t + c2 t2 . Without actually solving the best parabola fit, will it be possible to determine whether the best parabola fit has less fitting error than that generated by the best line fit? Explain the reason. Solution (a) y = c + dt, c−d =2 c=0 c + d = −3 c + 2d = −5 59 , AT Ax = x, AT A = I2 Ax = b, , , x = [c, d]T , 2 1 −1 0 1 0 , b= A= −3 1 1 −5 1 2 A , : 42 26 , AT b = −6 −15 −6 −15 ˆ AT Ax = AT b AT A = ˆ x = [−0.3, −2.4]T , (b) (a) 42 26 x = ˆ y = −0.3 − 2.4t 1 −1 2 −0.1 0 1 0.3 0 −0.3 − = e = b − Ax = −3 1 −0.3 1 −2.4 1 2 −5 0.1 , 1 −1 1 0 1 1 1 2 e A′ x′ = b: 2 1 0 0 x = −3 1 −5 4 A′ : 1 0 −0.1 0.3 −0.3 0.1 = 0 1 4 60 e A′ , Problem 54 The trace of an n by n matrix A is defined to be the sum of its diagonal entries, denoted by tr(A). (a) Suppose A is an m × n matrix and B is an n × m matrix. Show by direct calculation that tr(AB ) = tr(BA). (b) If A and B are 2 × 2 matrix such that AB − BA = ab cd show that a + d = 0. (c) Suppose m × n matrix A has linear independent columns, P is the projection matrix onto the column space of A, and Q is the projection matrix onto the left nullspace of A. Show that tr(P ) = n and tr(Q) = m − n. Solution (a) C, (C )ij tr(AB ) = i=1 n (i, j ) m m (trace) n (AB )ii = i=1 j =1 nm (A)ij (B )ji (B )ji (A)ij j =1 i=1 tr(BA) = j =1 (BA)jj = tr(AB ) = tr(BA) (b) : tr(AB − BA) = tr(AB ) − tr(BA) = 0 = a + d 61 (c) C (A) P = A(AT A)−1 AT , (a) , tr(P ) = tr(A(AT A)−1 AT ) = tr((AT A)−1 AT A) = tr(In ) = n Q = Im − P , Problem 55 Suppose A is an m × n real matrix. Given some b ∈ Rm , it is known that the equation Ax = b is solvable. (a) Show that there is a unique vector y in the row space of A such that Ay = b. (b) Show that among all possible solutions of Ax = b, the one in the row space has the minimum length. (c) Suppose A has independent rows and y is in the row space of A so that Ay = b. Express y in terms of A and b. Solution (a) x ∈ Rn ,x Ax = b, x = y + z, C (AT ) N (A) tr(Q) = tr(Im − P ) = tr(Im ) − tr(P ) = m − n y ∈ C (AT ), z ∈ N (A) y Ay = b, Ay′ = b, y y′ y′ = y, , , Ax = A(y + z) = Ay + Az = Ay = b, y, y′ ∈ C (AT ) A(y − y′ ) = 0, y − y′ ∈ C (AT ), y − y′ ∈ N (A), y − y′ ∈ N (A) ∩ C (AT ) = {0}, (b) y (a), z z=0 , x = y + z, x 2 = (y + z)T (y + z) = y y ∈ C (AT ), z ∈ N (A), y 2+ z 2 ≥ y 2, (c) A , : y = AT c, c y A Ay = AAT c = b 62 , rankAAT = rankAT = m, AAT y = AT (AAT )−1 b Ay = b , c = (AAT )−1 b, Problem 56 Let W = span{1, x} ⊂ P2 , where P2 is a vector space of polynomials of degree at most 2 has the standard inner product for p, q ∈ P2 as follows: 1 p, q = 0 p(x)q (x)dx Compute the orthogonal complement W ⊥ . Solution p(x) = a0 + a1 x + a2 x2 ∈ W ⊥ , 1 p(x) ⊥ 1 1 0 p(x) ⊥ x, 1 1 0 = p, 1 = 0 (a0 + a1 x + a2 x2 )1dx = a0 (a0 + a1 x + a2 x2 )xdx = a0 1dx + a1 0 1 xdx + a2 0 1 x2 dx 1 1 0 = p, x = 0 xdx + a1 0 0 x2 dx + a2 0 x3 dx 1 a0 + a1 + 2 1 1 a0 + a1 + 2 3 1 a0 = 6 α, a1 = −α, a2 = α, 1 a2 = 0 3 1 a2 = 0 4 , α p(x) = α , W ⊥ = span 1 6 1 − x + x2 6 − x + x2 63 Problem 57 Explain why the following determinant is equal to zero: 0 0 0 0 0 0 0 0 0 a14 a15 a24 a25 a34 a35 a41 a42 a43 a44 a45 a51 a52 a53 a54 a55 For an n by n matrix A = [aij ], what is the smallest m × m zero principal submatrix, aij = 0, i, j = 1, . . . , m, making det A = 0? Solution , m×m , , ⌈ n+1 ⌉ 2 m m > n − m, m , , m ; n, , n−m n − m ≥ m; m , Problem 58 Let An be the tridiagonal n × n matrix with 2’s on the main diagonal, 1’s immediately above the main diagonal, 3’s immediately below the main diagonal, and 0’s everywhere else: 2 1 0 0 ... 0 2 1 0 ... 3 2 1 ... 0 3 2 ... . . . .. ... . ... 0 0 0 ... 3 0 An = 0 . . . 0 0 0 0 . . . 2 64 (a) Explicitly calculate detAn , for n = 1, 2, 3, 4, and express detAn in terms of detAn−1 and detAn−2 . (b) Derive the formula for detAn . Solution (a) detA1 = det[2] = 2 detA2 = 21 32 210 detA3 = 321 032 =2 21 32 10 32 =1 −3 = 2detA2 − 3detA1 = 2 · 1 − 3 · 2 = −4 2100 detA4 = 3210 0321 0032 210 =2 3 2 1 032 100 −3 3 2 1 032 = 2detA3 − 3detA2 = 2 · (−4) − 3 · (1) = −11 detAn detAn = 2detAn−1 − 3detAn−2 (b) (a) detAk+1 detAk+2 detAk+1 , A A 65 : = 2 −3 1 0 , A = S ΛS −1 , S Λ detAk+1 detAk uk+1 = Auk , uk = detAk u1 = [1, 2]T : A= 2 −3 1 0 = λ1 λ2 1 1 λ1 0 1 √ λ2 2 2 i − 1 0 1 −λ2 λ1 λ1 = −1 + √ √ 2i, λ2 = −1 − 2i n λ1 −1 un = An−1 u1 , An−1 1 1 − λ2 √ = n 11 0 λ2 −1 2 2i −1 λ1 λn − λn −λn λ2 + λ1 λn 1 1 2 1 2 =√ 2 2i λn−1 − λn−1 −λn−1 λ2 + λ1 λn−1 1 2 1 2 λ1 λ2 0 An−1 un = An−1 u1 u1 1 n n n n detAn = √ (λ1 −1 − λ2 −1 − 2λ1 −1 λ2 + 2λ1 λ2 −1 ) 2 2i Problem 59 Calculate the determinant of the following 111 1 0 1 0 1 1 A= 1 1 1 1 1 0 111 6 × 6 matrix: 101 1 1 1 1 1 1 1 1 0 1 1 1 011 66 Solution , , B , : 011111 A P B 000010 P B 1 0 1 A = PB = 0 0 0 521634, 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 1 0 0 0 1 00100 1 3 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 11110 , 6×6 uuT B = uuT − I , , u = [1, 1, 1, 1, 1, 1]T , uT v = 0, detP = −1 uuT (uuT )u = (uT u)u = 6u, dimSpan{u}⊥ = 5, 5, −1, −1, −1, −1, −1, 5 Problem 60 Let v ∈ Span{u}⊥ , (uuT )v = (uT v)u = 0 B uuT 6, 0, 0, 0, 0, 0, detB = −5, detA = (detP )(detB ) = (−1)(−5) = where A, B, C, D are n × n matrices. If D is invertible and CD = DC , show that detP = det(AD − BC ) Solution : , 1 , : P ( ), P = AB CD 67 , AB CD In 0 = , : B D −D−1 C In , In 0 A − BD−1 C 0 (2,1) , AB CD · − D −1 C −D −1 C In = detP · detIn · detIn = detP A − BD−1 C 0 B D = det(A − BD−1 C ) · detD = det(AD − BD−1 CD) = det(AD − BD−1 DC ) = det(AD − BC ) C D = DC Problem 61 , Let and be m × m and n × n matrices, respectively. Derive a formula for A B 0A . det BC Solution , 0A 0 In A0 = BC Im 0 CB , 0 A 0 In 0 68 A 0 = B C Im CB m+n 0 Im , In 0 = (−1)m+n Im 0 0 In , = (−1)m+n , A 0 = (detA)(detB ) CB 0 A BC = (−1)m+n (detA)(detB ) Problem 62 Let A and B be n by n matrices over C. Prove that AB BA Solution (A + B )x = λx, (A − B )y = µy, AB x x AB y y = λ , = µ BA x x BA −y −y AB BA B A − tIn A+B λ A−B µ, = det(A + B ) det(A − B ) : A − tIn B t=0 = det ((A + B ) − tIn ) det ((A − B ) − tIn ) 69 Problem 63 Suppose A and B are n × n real matrices. Let AB C= −B A Show that detC ≥ 0. Solution i= √ I iI AB I −iI A − iB 0 = 0I −B A 0I −B A + iB −1 I iI 0 I , = 1, I −iI 0 I =1 A B = A − iB −B 0 A + iB −B A = det(A + iB )det(A + iB ) = det(A + iB )det(A + iB ) ≥ 0 Problem 64 Suppose A, B , C and D are n × n matrices. If rank |A| |B | |C | |D | 70 =0 AB CD = n, show that Solution A I −CA−1 , rankA = n, 0 AB A B = I CD 0 D − CA−1 B , A , detB = 0 D −CA−1 B = 0, B D = CA−1 B detD = (detC )(detA−1 )(detB ), detC = 0 |A| · |D | = |B | · |C | C = DB −1 A, Problem 65 , rankB = n, I 0 AB A B = −DB −1 I CD C − DB −1 A 0 detC = (detD )(detB −1 )(detA) = 0 Suppose A and B are n by n orthogonal matrices such that detA + detB = 0 Prove that det(A + B ) = 0. Solution det(A + B ) = det(A + B )T = det(AT + B T ) detA detB , , (detA)det(A + B ) = (detA)det(AT + B T ) = det(AAT + AB T ) = det(I + AB T ) (detB )det(A + B ) = det(BAT + BB T ) = det(BAT + I )T = det(I + AB T ) , (detA − detB )det(A + B ) = 0 det(A + B ) = 0, A , detA = detB detA + detB = 0 detA = 0, det(A + B ) = 0 71 Problem 66 Suppose A and B are n × n matrices satisfying AB = −BA. (a) Find the flaw in the statement: Taking determinants gives (detA)(detB ) = −(detB )(detA), so either A or B must have zero determinant. Thus AB = −BA is only possible if A or B is singular. ab st , B = . Express the relation (b) In the case of n = 2, let A = cd uv AB + BA = 0 in the form of matrix equation C x = 0, where C is a 4 by 4 matrix, and x = [s, t, u, v ]T . Find the determinant of C , and identify the condition for detC = 0. If B = 0, find all possible 2 × 2 A’s and B ’s satisfying AB + BA = 0. Solution (a) , A n×n , det(kA) = kn detA, n , det(−BA) = det(−B )(detA) = (−1)n (detB )(detA) (b) A, B x = [s, t, u, v ]T , 2a c b 0 s = AB + BA = 0, ab st st ab + cd uv uv cd 2as + bu + ct au + cs + cv + du at + bv + bs + dt ct + bu + 2dv , =0 0 b Cx = c 0 a+d 0 b t 0 = 0 a + d c u 0 0 v c b 2d 72 , 2a detC = b c 0 c a+d 0 c a+d = 2a 0 c , 0 a+d b b 0 a+d b b c 2d 0 b c 2d = C, 2a b c 0 0 a+d 0 c 0 0 a+d b 0 a+d b , −2d b c 2d : b a+d + 2d c 0 , 0 c : detC = 2a [(a + d)(2d(a + d) − bc) − bc(a + d))] − 2d [2bc(a + d)] = 4a(a + d)(ad + d2 − bc) − 4bcd(a + d) = 4(a + d)(a2 d + ad2 − abc − bcd) = 4(a + d)2 (ad − bc) a+d = 0 C B Problem 67 Suppose A = [aij ] is an n × n matrix. Let adjA denote the adjoint of A; that is, adjA is the n × n matrix whose (i, j )-entry is the cofactor (−1)i+j detAji of aji , where Aji is the submatrix obtained from A by deleting the j -th row and the i-th column. Prove the following statements. (a) det(adjA) = (detA)n−1 (b) adj(adjA) = (detA)n−2 A (c) adj(AB ) = (adjB )(adjA) (d) adjAT = (adjA)T 73 ad − bc = 0 ,C , detC = 0 x = 0, A x , (e) adj(kA) = kn−1 adjA (f) adj(SAS −1 ) = S (adjA)S −1 for nonsingular S . (g) If A is symmetric, so is adjA. (h) If A is upper triangular, so is adjA. Solution adjA : A(adjA) = (adjA)A = (detA)I A , adjA = (detA)A−1 A , adjA , A = 0, rankA = n − 1, rank(adjA) = 1, (a) A rankA < n − 1, rank(adjA) = 0, , det(adjA) = det (detA)A−1 = (detA)n det(A−1 ) = (detA)n (detA)−1 = (detA)n−1 A (b) A , detA = 0, det(adjA) = 0, , , A adjA, adj(adjA) = det(adjA)(adjA)−1 , (adjA)−1 = ((detA)A−1 )−1 = (detA)−1 A, (a) , adj(adjA) = (detA)n−1 (detA)−1 A = (detA)n−2 A A , rank(adjA) , rank(adj(adjA)) = 0, adj(adjA) = 0 (c) A, B , AB , adj(AB ) = (det(AB ))(AB )−1 = (detA)(detB )B −1 A−1 = (adjB )(adjA) 74 A B , A, B A + ǫI B + ǫI , adj((A + ǫI )(B + ǫI )) = (adj(B + ǫI ))(adj(A + ǫI )) , (d) (adjA)T (e) (f) ǫ → 0, adjAT (i, j ) (i, j ) (−1)i+j detAT ji (−1)i+j detAij , (−1)i+j det(kAji ) = (−1)i+j kn−1 detAji (c), adj(SAS −1 ) = (adjS −1 )(adjA)(adjS ) = (detS −1 )S (adjA)(detS )S −1 = S (adjA)S −1 (g) (h) A (d) (adjA)T = adjAT = adjA, , aij = 0, i > j i > j, Aji AT = A adjA , (i, j ) (−1)i+j detAji , detAji = 0 Problem 68 Suppose A is an n × n matrix. What are the possible values of the rank of adjA? Solution rankA adjA = det(A)A−1 rankA = n, A , rank(adjA) = n , det(A) = 0, A , E , n×n R A adjEA = (adjA)(adjE ), , E , , , adjE rank(adjEA) = rank(adjA) rank(adjR) = rank(adjA) 75 R= Ir F 0 0 , rankR = r = n − 1 , adjR 0 0 ··· 0 adjR = . . . . . . .. . .. . 0 0 ··· 1 adjR , n − 1, 3 Problem 69 Consider the following linear equation: Au b x = vT d bn+1 xn+1 : 0, 1, n rank(adjR) = 1 adjR = 0, rankR = r < n − 1 ,R , rank(adjA) rank(adjR) = 0 0 0 ··· 0 where A is a nonsingular n × n matrix, and u, v, and b are vectors in Rn . (a) Find an n × n matrix Y and a vector z in Rn so that multiplying both sides Y0 yields an equivalent echelon form, i.e., of the system by zT 1 Y zT 0 1 A vT u d = In 0T ∗ ∗ where ∗ means arbitrary vector or scalar. Au has the form (b) Show that the determinant of vT d A vT u d = d (detA) − vT (adjA)u 76 (c) Show that the determinant of A + uvT has the formula det(A + uvT ) = detA + vT (adjA)u (d) Show that if d − vT A−1 u = 0, then the solution of the system is given by x = A−1 (b − xn+1 u) xn+1 = Solution (a) , : Y A = In zT A + vT = 0T Y = A−1 , (b) (a) A−1 0 A−1 A−1 0 zT = −vT A−1 , In 0T A−1 u d − vT A−1 u , : bn+1 − vT A−1 b d − vT A−1 u −vT A−1 1 A vT u d −vT A−1 1 = A−1 −vT A−1 In 0T d− u d 0 1 = detA−1 = (detA)−1 A−1 u vT A−1 u = d − vT A−1 u A vT = (detA)(d − vT A−1 u) = d (detA) − vT (adjA)u 77 A(adjA) = (adjA)A = (detA)I , (detA)A−1 (c) A −vT u 1 In vT 0 1 = A + uvT : A + uvT 0T , A , adjA = u 1 In vT 0 1 =1 u 1 A + uvT 0T (b) A −vT , u 1 = det(A + uvT ) = detA + vT (adjA)u det(A + uvT ) = detA + vT (adjA)u (d) : (a) A−1 0 A vT u A vT u b d bn+1 , A−1 u : A−1 b b −vT A−1 1 d bn+1 = x xn+1 In 0T d − vT A−1 u bn+1 − vT A−1 b In 0T A−1 u d − vT A−1 u = A−1 b bn+1 − vT A−1 b 78 : x + A−1 uxn+1 = A−1 b (d − vT A−1 u)xn+1 = bn+1 − vT A−1 b d − vT A−1 u = 0 , x = A−1 b − A−1 xn+1 u, xn+1 = (bn+1 − vT A−1 b)(d − vT A−1 u)−1 Problem 70 Suppose A and B are n by n matrices and suppose there exists a nonzero vector x such that Ax = 0 and a vector y such that Ay = B x. If Aj is the matrix obtained from A by replacing its j th column by the j th column of B , show that n detAj = 0. j =1 Solution Ax = 0 rankA < n−1, x C = [cij ] C T = adjA Aj j A (cofactor), , (adjoint) , Aj rankA = n − 1 x = 0, A , , , detA = 0, rankA < n , dimN (A) = 1, n j =1 detAj =0 N (A) detAj = b1j c1j + b2j c2j + · · · + bnj cnj n n detAj = j =1 j =1 (b1j c1j + b2j c2j + · · · + bnj cnj ) = tr(BC T ) = tr(C T B ) tr(C T B ) = 0 AC T = C T A = (detA)I CT B A , 79 AC T = C T A = 0, CT B AC T B = 0, x , C T B x = C T Ay = 0y = 0, CT B , λ, tr(C T B ) x (C T B )2 , CT B λ2 = 0, tr(C T B ) = 0 , (C T B )2 = 0 CT B 80 Problem 71 Let 10101 0 1 0 1 0 A = 1 0 1 0 1 0 1 0 1 0 10101 Compute the eigenvalues and eigenvectors of A. Solution : i+j , , 1 0 P = 0 0 0 , aij = 0, , , : (1, 2, 3, 4, 5) → (1, 3, 5, 2, 4), 0000 0 0 1 0 1 0 0 0 0 0 0 1 0100 81 10000 10 0 0 1 0 0 0 1 P T AP = 0 0 0 0 1 1 0 0 1 0 0 0 0 1 00010 10 11100 1 1 1 0 0 = 1 1 1 0 0 = B 0 0 0 1 1 00011 ,B ,B , P T P = I, B y = λy , P, A 101 01 10 01 10 10000 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 1 1 00100 111 1 1 1 111 B 11 11 , P B y = AP y = λP y A n x = Py e = (1, 1, . . . , 1), E n E = [eij ], eij = 1, E = eeT , E e = (eeT )e = (eT e)e = ne E n−1 n, e span{e} n − 1, {u1 , . . . , un−1 }, E ui = (eeT )ui = (eT ui )e = 0 E n−1 B , ui , i = 1, . . . , n − 1 B , A λ1 = 3, λ2 = 0, λ3 = 0, λ4 = 2, λ5 = 0, 82 : 0 0 1 −1 1 y1 = 1 , y2 = 0 , y3 = −2 , y4 = 0 , y5 = 0 1 1 0 0 0 −1 1 0 0 0 1 1 1 0 0 A xi = P yi , i = 1, . . . , 5, 1 1 1 0 0 0 x1 = 1 , x2 = −1 , x3 = 1 0 0 0 −2 0 1 1 1 , x4 = 0 , x5 = 0 −1 1 0 0 0 0 Problem 72 Let cab ca A = b c a , B = c a b , C = a b c . bca abc cab ab c b (a) For any scalars, a, b, and c, show that A, B , and C are similar. (b) If BC = CB , show that A has two zero eigenvalues. Solution (a) B C A, bca 001 010 A = b c a = 0 0 1 c a b 1 0 0 = P T BP 010 100 abc cab ab c 83 (b) B C = CB , 001 cab 010 A = b c a = 1 0 0 a b c 0 0 1 = P CP T 010 bca 100 cab ab c , a2 + b2 + c2 − ab − bc − ca = 0 (a + b + c), a3 + b3 + c3 − 3abc = 0 A t−a pA (t) = b c , b c−t a c a t−b = t3 − (a + b + c)t2 Problem 73 Let A = 0 2 0 003 100 . Which of the following matrices are similar to A? 300 , B2 = 0 2 6 003 145 , B3 = 7 2 0 891 300 B1 = 0 1 0 002 140 200 234 B4 = 5 2 0 , B5 = 0 1 4 , B6 = 0 2 1 063 563 012 84 Solution n×n , 1, 2, 3, B , −3, 2, 7, A Problem 74 Suppose A is an n × n real matrix. (a) Show that A and AT have the same set of eigenvalues. (b) If λ and µ are eigenvalues of A, with λ = µ, then any eigenvector of A corresponding to λ is orthogonal to any eigenvector of AT corresponding to µ. Solution (a) A AT , detB T = detB : B4 1, 2, 3, B1 , B2 , B3 B5 A A B6 B B4 A −3, 3, 6, B5 1, 2, 3, B6 {λ1 , . . . , λn }, , λi = λj , i = j , A det(AT − λI ) = det((AT − λI )T ) = det(A − λI T ) = det(A − λI ) , (b) µ yT Ax : yT Ax = yT (λx) = λ(yT x) , , yT Ax = (yT Ax)T = xT AT y, xT AT y = xT (µy) = µ(xT y) λ=µ , λ(xT y) = µ(yT x) xT y = 0, x y , , y, A A λ AT x, Ax = λx, AT y = µy AT 85 Problem 75 Let A be an n × n real matrix. If rankA = 1, show that either A2 = 0 or A is diagonalizable. Solution nV = span{v}, i = 1, . . . , n − 1, vT u, , n−1 Problem 76 Let A be an n × n real matrix, and T be the linear transformation defined by T (A) = AT . A, rankA = 1, n−1 A n−1 n n , u v A = uvT xi ∈ V ⊥ , , A dimV ⊥ = n − 1, Axi = uvT xi = u0 = 0, xi , i = 1, . . . , n − 1 u u u ∈ V ⊥, , A Au = uvT u = (vT u)u, V ⊥ , vT u = 0, A vT u = 0, A A2 = uvT uvT = u0vT = 0 (a) Show that ±1 are the only eigenvalues of T . (b) Describe the eigenvectors corresponding to each eigenvalue of T . (c) Find an ordered basis for 2 × 2 real matrices such that the matrix representation with respect to that basis is diagonal. Solution (a) T (A) = AT = λA, A = λAT = λ2 A λ = ±1 (b) −A, λ = 1, AT = A, (skew-symmetric) , λ = −1, AT = λ2 = 1, A , A = λAT , ,λ , 86 (c) 2×2 1 0 01 00 0 0 , , , S= 0 0 00 10 0 1 T : ab , A= cd S ab ac = T (A) = T cd bd 1 0 [T ]S = 0 0 000 0 1 0 1 0 0 001 (a), [T ]S λ = −1 λ = ±1 λ=1 0 0 1 0 1 0 x1 = , x2 = , x3 = 0 1 0 1 0 0 −1 x4 = 1 0 0 T 1 0 01 00 0 −1 , , , 0 0 10 01 1 0 87 : Problem 77 If AB = BA, show that A and B have a common eigenvector. Solution n×n x1 · · · xk , A N (A − λI ) λ, , AB = BA, dimN (A − λI ) = k, k ≥ 1, {x1 , . . . , xk } n×k X= , (A − λI )X = 0 (A − λI )BX = ABX − λBX = BAX − λBX = B (A − λI )X = 0 B X ∈ N (A − λI ), BX = XC , C k×k , BX C {x1 , . . . , xk } C z = β z, , , B X z = XC z = X (β z) = βX z Xz B A B Xz x1 , . . . , xk , Xz N (A − λI ), Problem 78 Suppose A is an n by n real matrix with eigenvalues λ1 , . . . , λn . Note that the √ eigenvalues are not necessarily real. Denote λj = aj + ibj , where i = −1. Prove that (a) (b) Solution (a) (trace) traceA = j =1 n j =1 bj =0 =0 n j =1 aj bj , n n n λj = j =1 aj + i j =1 n j =1 bj bj A , traceA ∈ R, 88 =0 (b) aj bj , n n n n traceA2 = j =1 λ2 = j j =1 a2 − j b2 + 2i j j =1 j =1 aj bj =0 , Problem 79 A , traceA2 , n j =1 aj bj Let A and B be n × n matrices satisfying AB = BA. Prove the following statements. (a) If λ is an eigenvalue of A, then the eigenspace corresponding to λ is invariant under B . (b) The column space of A and the nullspace of A are invariant under B . Note that s subspace X is invariant under linear transformation T if T (X ) ⊆ X . Solution (a) Ax = λx, , A(B x) = B (Ax) = B (λx) = λ(B x) Bx (b) x ∈ N (A), λ Ax = 0, A(B x) = B (Ax) = B 0 = 0 B x ∈ N (A), x ∈ C (A), y N (A) x = Ay, B x = BAy = A(B y) , B x ∈ C (A), C (A) B B , 89 Problem 80 Suppose A is an n by n matrix. The adjoint of A, denoted by adjA, is defined to be the transposed of cofactors of elements of A. Show that each eigenvector of A is also an eigenvector of adjA. Solution Ax = λx λ = 0, (adjA)x = (adjA) ,x λ = 0, adjA Ax = 0, x = 0, A , detA = 0 rankA < n 1 Ax λ = detA x λ x = 0, A(adjA) = (adjA)A = (detA)I : rankA < n − 1 n − 1, rankA = n − 1, (adjA)x = µx adjA Problem 81 adjA rankA = n − 1 adjA = 0, 1, (adjA)x λ=0 A rankA < n − 1, A (adjA)x = 0 · x N (A) = Span{x} N (A), λ = 0, A x µ A(adjA)x = (detA)I x = 0 · x = 0, , Assume that a = 0. Show that the eigenvalues and eigenvectors of the following n by n matrix b a a .. . c are given by c b .. A= . λj = b + 2a .. . b a cb jπ n+1 c cos a 90 and for j = 1, 2, . . . , n. Solution A (c/a)2/2 sin(2jπ/(n + 1)) xj = . . . (c/a)n/2 sin(njπ/(n + 1)) (c/a)1/2 sin(1jπ/(n + 1)) (A − λI )x = 0 cxi−1 + (b − λ)xi + axi+1 = 0 i = 1, 2, . . . , n, x0 = xn+1 = 0 xk+2 + b−λ a c xk+1 + xk = 0 a , x0 = xn+1 = 0 k = 0, 1, . . . , n − 1, xk = r k , ,r n r2 + r1 xk = c1 r k + c2 kr k , xk = 0, k k xk = c1 r1 + c2 r2 , b−λ a λ, r+ c =0 a , r1 = r2 = r , c1 = c2 = 0, , r1 = r2 k, r2 , x0 = xn+1 = 0 x , x0 = xn+1 = 0 0 = c1 + c2 n n 0 = c1 r1 +1 + c2 r2 +1 r1 r2 n+1 c2 = − c1 = 1, r1 = r2 ei2πj/(n+1) , i = √ −1, j = 1, 2, . . . , n : r1 r2 = c a b−λ a r1 + r2 = − 91 r1 = cos θ + i sin θ c iπj/(n+1) , r2 ae = c −iπj/(n+1) , ae eiθ = λ = b + a(r1 + r2 ) = b + 2a n λj , j = 1, 2, . . . , n c1 + c2 = 0, c a c cos a λj r1 k /2 jπ n+1 xj r2 j kπ n+1 , k k k xk = c1 r1 + c2 r2 , k k xk = c1 (r1 − r2 ) = 2ic1 sin c1 = 1/(2i), (c/a)2/2 sin(2jπ/(n + 1)) xj = . . . (c/a)n/2 sin(njπ/(n + 1)) Problem 82 Use the Cayley-Hamilton theorem to show that if A is an n by n nonsingular matrix, then there is a polynomial q (t), of degree at most n − 1, such that A−1 = q (A). Solution A p(t) = det(tI − A), p(t) n , (c/a)1/2 sin(1jπ/(n + 1)) p(t) = tn + an−1 tn−1 + · · · + a1 t + a0 p(0) = a0 = detA, p(A) = 0, An + an−1 An−1 + · · · + a1 A + a0 I = 0 A−1 , An−1 + an−1 An−2 + · · · + a1 I + a0 A−1 = 0 92 A , a0 = 0 Cayley-Hamilton A−1 = − q (t) = − 1 An−1 + an−1 An−2 + · · · + a1 I a0 1 n−1 t + an−1 tn−2 + · · · + a2 t + a1 a0 Problem 83 An n by n square matrix A = [aij ] has the properties that aii = 1 and aik akj = aij for all i, j, k. What are the eigenvalues of A? Solution i, j, k, n akk = 1, n aik akj = aij = akk aij , n aik akj = k =1 k =1 akk aij = k =1 akk aij = (trA)aij A λ, k, akk = 1, (n − 1) 0 A2 = (trA)A λ2 = (trA)λ, trA = n n Problem 84 Let A be the following n × n 0 0 . A= . . 0 −a0 matrix 1 0 . . . 0 0 1 . . . 0 ··· ··· .. . ··· 0 0 . . . 0 trA A λ=0 λ = trA , A 0 0 . . . 1 −a1 −a2 · · · −an−2 −an−1 Show that the characteristic polynomial of A is det(A − tI ) = (−1)n p(t), where p(t) = tn + an−1 tn−1 + · · · + a1 t + a0 The matrix A is known as the companion matrix of the polynomial p(t). 93 Solution n×n C = (A − tI )B , 1 0 ··· 0 t B = t2 t 1 ··· 0 . . . .. . . . . .. . . . . tn−1 t n−2 tn−3 · · · 1 0 0 . . . 1 0 ··· 0 1 0 0 ··· 0 ∗ , C= 0 1 ··· 0 . . .. . .. . .. .. 0 0 0 ··· 1 −p(t) ∗ ∗ · · · ∗ detC = (−1)n+1 (−p(t))detIn−1 = (−1)n p(t) B 1, detB = 1, detC = det((A − tI )B ) = det(A − tI )(detB ) = det(A − tI ) Problem 85 Let A and B be n × n matrices, either real or complex. Consider a linear transformation T (X ) = AXB , where X is also n × n. Suppose that A has distinct while B T has distinct eigenvalues {φj , j = 1, 2, . . . , n} and associated eigenvectors eigenvalues {λi , i = 1, 2, . . . , n} and associated eigenvectors {yi , i = 1, 2, . . . , n}, and the matrices {yi zT , i, j = 1, 2, . . . , n} are associated linearly independent j eigenvectors. 94 {zj , j = 1, 2, . . . , n}. Show that the eigenvalues of T are {λi φj , i, j = 1, 2, . . . , n} Solution X = yi zT , j T (X ) = Ayi zT B = (Ayi )(B T zj )T = λi φj yi zT = λi φj X j j λi φj T (X ) , yi zT j = y1 y2 · · · yn B , Y −1 , c11 c21 . . . c12 c22 . . . · · · c1n · · · c2n . .. . . . zT 1 zT 2 . . . zT n 0= n i=1 n T j =1 cij yi zj = Y CZ T Y yi , Z , C = 0, Problem 86 Y ZT zj A cn1 cn2 · · · cnn , (Z T )−1 , {yi zT , i, j = 1, 2, . . . , n} j Let u and v be vectors in Rn , and let A be the square matrix uvT . (a) What are the row space and nullspace of A in terms of u and v? (b) Show that u is an eigenvector of A, and find the corresponding eigenvalue. (c) Show that det(I + uvT ) = det(I + vuT ). (d) The inverse of I + uvT has the form I + cuvT . What condition must be satisfied so that I + uvT is invertible? And what is the value of c? (e) What condition must be satisfied by u and v for A to be skew-symmetric (A = −AT )? (f) What condition must be satisfied by u and v so that A2 = A? (g) What are the possible values of the rank of I + uvT ? 95 Solution (a) uvT span{v} Rn x u v , A v , C (AT ) = ,N (A) = C (AT )⊥ , A vT x = 0 vT u (b) Au = (uvT )u = (vT u)u, (c) det(I + uvT ) = det(I + uvT )T = det(I + vuT ) (d) , (I + uvT )(I + cuvT ) = I + uvT + cuvT + cu(vT u)vT = I + (1 + c + cvT u)uvT , vT u = −1, c = −1/(1 + vT u), I + uvT uvT 1 + uT v (skew-symmetric) (a)), u v = k u, v A2 = A, (I + uvT )−1 = I − (e) A kuuT = −kuuT , (f) uvT = −vuT , AT = (uvT )T = vuT , u k=0 v u = 0, ( A2 = (uvT )2 = u(vT u)vT = (vT u)uvT , vT u = 1 (g) vT u = −1, , I + uvT , rank(I + uvT ) = n x = −(vT x)u, I + uvT rank(I + uvT ) = n − 1 (I + uvT )x = 0, vT u = −1, u dimN (I + uvT ) = 1, Problem 87 Let A be an n by n real matrix and have eigenvalues λ1 , . . . , λn−1 , 0, so that rankA ≤ n − 1, and suppose that the last row of A is a linear combination of the others. 96 (a) If A is partitioned as C vT u d in which C is an (n − 1) × (n − 1) matrix, show that there is an (n − 1)dimensional vector b such that vT = bT C and d = bT u. (b) Show that C + ubT has eigenvalues λ1 , . . . , λn−1 . Solution (a) A 1 , (b) (a), b In−1 0 −bT 1 n−1 In−1 0 −b T , , : C vT u d = C −b C + , : In−1 0 −bT , 1 C vT u d = A u 0 C 0T u 0 T , A u vT d−b u T 1 b vT = bT C d = bT u : In−1 0 −b C 0T T C vT u d In−1 0 b T D D C + ubT = u 0 1 D 1 In−1 0 b T 1 A, A = C + ubT 0T , λ1 , . . . , λn−1 , 0 0, =D C + ubT λ1 , . . . , λn−1 97 Problem 88 Suppose A is a real symmetric n × n matrix. (a) Show that the rank of A is equal to the number of nonzero eigenvalues of A, but that this is not generally true for non-symmetric matrices. (b) If A = 0, show that rankA ≥ (trA)2 tr(A2 ) with equality if and only if there is an n × r matrix U with orthonormal columns and some a ∈ R such that A = aU U T . Solution (a) , r , λi = 0, i = 1, 2, . . . , r , ui , i = 1, 2, . . . , n n×n ,A : λ ··· 0 1 . . .. . .. . . 0 · · · λn uT 1 . . . uT n , n×n λi = 0, i = r + 1, r + 2, . . . , n, U n×n ,Λ A A = U ΛU T = u1 · · · un = λ1 u1 uT + · · · + λr ur uT 1 r λi = 0 u1 , . . . , ur A rankA = rankB = rankB T , rankA = rankΛ = r rankA = 1, A , , , rankA = r , , A A U, A r B = ΛU T , U, 0, 0, B T = U Λ 01 ,A= 00 98 (b) (a) , trA = λ1 + · · · + λr , (xT y)2 ≤ (xT x)(yT y) – (Cauchy-Schwarz) r- x = [λ1 , . . . , λr ]T , y = [1, . . . , 1]T , r 2 r (trA) = i=1 2 λi A ≤r λ2 = (rankA)tr(A2 ) i i=1 A2 , , A = 0, rankA > 0, tr(A2 ) > 0, , (a) , λ1 = · · · = λr = a, A= a(u1 uT 1 + ··· + ur uT ) r = a u1 · · · ur U Problem 89 n×r , r uT 1 . . . uT r = aU U T Suppose that A is an m × n matrix, B is an n × m matrix, with m ≤ n. (a) Suppose that C is m by m, and E is by n. Show that the eigenvalues of n CD are those of C together with those the triangular block matrix 0E of E . AB 0 0 0 and are (b) Show that (m + n) × (m + n) matrices B0 B BA similar. (c) Show that BA has the same eigenvalues as AB , counting multiplicity, together with additional n − m eigenvalues equal to 0. 99 Solution (a) C λi , i = 1, 2, . . . , m, E σj , j = 1, 2, . . . , n, C xi = λi xi , E T yj = σj yj , xi , yj C ET CD C xi xi x = λi i = 0 0E 0 0 T 0 C 0 0 0 = = σj T ET Ty yj D Ej yj CD 0 , σj , j = 1, 2, . . . , n (b) S E T = DT E T CD 0E CT 0 λi , i = 1, 2, . . . , m, D S = SE , D E AB 0 AB ABA IA = B0 B BA 0I IA AB ABA 0 0 = 0I B BA B BA AB 0 B , 0 m m ≤ n, m , 100 0 : (c) (a) IA 0 I 1, , BA , AB (b) n×n n−m 0 n 0, 0 0 B BA , BA m×m AB Problem 90 Let (a) If Ak = ck A + dk I , for k ≥ 0, find ck and dk as functions of k. (b) Express A−1 as a linear combination of I and A. Solution (a) Cayley-Hamilton I , A, A2 , . . . , An−1 Cayley-Hamilton A2 = 3A − 2I A3 = A(A2 ) = 3A2 − 2A = 3(3A − 2I ) − 2A = 7A − 6I A4 = A(A3 ) = 7A2 − 6A = 7(3A − 2I ) − 6A = 15A − 14I Ak = ck A + dk I , (3ck + dk )A − 2ck I , ck+1 = 3ck + dk dk+1 = −2ck c0 = 0, d0 = 1 : c 31 c c k+1 = k = A k dk+1 −2 0 dk dk 31 −2 0 1 −1 −1 2 20 01 21 11 Ak+1 = ck A2 + dk A = ck (3A − 2I ) + dk A = A , A2 − 3A + 2I = 0, n×n Ak , k ≥ n, A= 31 −2 0 p(t) = t2 − 3t + 2, A A= = S ΛS −1 = 101 ck dk = Ak = = c0 d0 1 −1 2 = S Λk S −1 2k 0 −1 −2k 2k − 1 +2 d0 0 21 0 1 11 1 c0 (b) A2 − 3A + 2I = 0, I =A 3 A−1 = − 1 A + 2 I 2 2I = −A2 + 3A = A(−A + 3I ), 1 (−A + 3I ) 2 (a) , 3 1 −2A + 2I , k = −1, A−1 = c−1 A + d−1 I = (2−1 − 1)A + (−2−1 + 2)I = k≥0 , k = −1 Problem 91 Suppose a 2 by 2 real matrix has the form 1−a b A= a 1−b (a) By the eigenvalue method, show that Ak → P as k → ∞, where 1 b b P= a+b a a (b) Let B = A − I . By the binomial theorem, show that 1 Ak = I + [(1 + r )n − 1]B r where r = −(a + b). Then, show that Ak → P , where P is given in (a). 102 where 0 < a ≤ 1, 0 < b ≤ 1, a + b < 2. Solution (a) 1, A , 1−a b a 1−b λ1 = 1, A 1 1 1 1 x1 = [b, a]T AT = trA = 2 − a − b = λ1 + λ2 x2 = [1, −1]T A , 1 0 : λ 2 = 1 − a − b, a + b < 2, A= b 1 a −1 |1 − a − b| < 1 k → ∞, (1 − a − b)k → 0, b 1 10 1 1 1 1 b b = Ak → =P a+b a a a −1 0 0 a + b a −b (b) −a b r = −(a + b) k Bi i ki r −1 B i 1 − a − b > −1 1 0 1 − a − b a + b a −b a>0 1 1 b > 0, 1 − a − b < 1, B 2 = rB , Ak : B =A−I = a −b B k = r k −1 B , k Ak = (I + B )k = i=0 k Bi = I + i k i=1 k i=0 =I+ 1 r k i=1 1 ki r B=I+ i r 1 = I + [(1 + r )k − 1]B r (a) |1 + r | < 1, k → ∞, (1 + r )k → 0, Ak → I − 1 B , r P = I − 1B r 103 Problem 92 Let A = [aij ] be an n × n complex matrix. The numerical range of A is defined by W (A) = {x∗ Ax|x ∈ Cn , x = 1} (a) Show that W (A + cI ) = W (A) + c for every c ∈ C. (b) Show that W (cA) = cW (A) for every c ∈ C. (c) Explain why aii ∈ W (A). (d) Let λ be an eigenvalue of A. Explain why λ ∈ W (A). (e) Describe W (A) if A is Hermitian. Solution (a) (b) (c) (d) x∗ (A + cI )x = x∗ Ax + cx∗ x = x∗ Ax + c (a), x∗ (cA)x = c(x∗ Ax) ei Ax = λx, A , i 1, , e∗ Aei = aii ∈ W (A) i x = 1, x∗ Ax = x∗ (λx) = λ ∈ W (A) A = U ΛU ∗ , Λ = diag(λ1 , . . . , λn ), U ∗ = n 2 λi zi i=1 (e) Hermitian U −1 , x∗ Ax = x∗ U ΛU ∗ x = z∗ Λz = z= U ∗x ,z 2 = U ∗x 2 = x∗ U U ∗ x =x 2 = 1, z∗ Λz ≤ λmax , W (Λ), λmin ≤ W (A) = λmin = mini=1,...,n λi , λmax = maxi=1,...,n λ λmin ≤ W (A) ≤ λmax 104 Problem 93 Let where I and B are n by n matrices. If B is nonsingular, determine the number of positive, negative, and zero eigenvalues of A. Solution A , (congruence) , of inertia) I 0 I I BT B , S , A S −1 , Sylvester , , , S T AS (law , A= I BT B 0 −B T −B T B S T AS B I −B I 0 = 0 0I 0 −B T B ,n , I 1, n x = 0, B x = 0, 2 xT (−B T B )x = −(B x)T (B x) = − B x ,A Problem 94 n ,n <0 Suppose A is symmetric positive semi-definite. Show that there exists a unique symmetric positive semi-definite matrix B such that B 2 = A. Solution A n A U ∗U = I , B 2 = U Λ 2 U ∗ U Λ 2 U ∗ = U ΛU ∗ = A 105 1 1 λi ≥ 0, i = 1, . . . , n, B A = U ΛU ∗ , Λ = diag(λ1 , . . . , λn ), √ √ 1 1 Λ 2 = diag( λ1 , . . . , λn ) B = U Λ 2 U ∗, C C , C , 1 2 C 2 = B2 = A A C 2 = B2 C = V Λ V ∗, V ∗V = I V ΛV ∗ = U ΛU ∗ W = U ∗V , U ∗, V, W Λ = ΛW W = I, Problem 95 U =V, B Let A be an n by n real normal matrix. A matrix is called normal if AT A = AAT . Show that A and AT have the same nullspace and column space, i.e., N (AT ) = N (A) and C (AT ) = C (A). Solution N (AT ) ⊆ N (A) x ∈ N (AT ), 2 AT x = 0, x ∈ N (AT A), Ax = 0 AAT x = A0 = 0, AT Ax = 0 xT AT Ax = (Ax)T (Ax) = Ax N (A) ⊆ N (AT ), , AQQT AT = AAT C (AT ), N (AT ) = N (A) C (AT ) ⊆ C (A) Q x ∈ N (AAT ) AT A = AAT , = 0, x ∈ C (AT ), y x = AT y A AT = AQ, AT A = (AQ)(QT AT ) = C (A) ⊆ C (AT ) = C (A) x = AT y = AQy, x ∈ C (A) Problem 96 Suppose A is a real symmetric positive definite. For every vector x, show that xT A−1 x = maxy 2xT y − yT Ay 106 Solution A n×n , A z = QT x, A = QΛQT , QT = Q− 1 , Λ = diag(λ1 , . . . , λn ), λi > 0 xT A−1 x = xT QΛ−1 QT x = zT Λ−1 z w = QT y , maxw 2zT w − wT Λw , vT Λv ≥ 0 2zT w − wT Λw = − zT Λ−1 z − 2zT w + wT Λw + zT Λ−1 z = −(Λ−1 z − w)T (z − Λw) + zT Λ−1 z = −(Λ−1 z − w)T Λ(Λ−1 z − w) + zT Λ−1 z ≤ zT Λ−1 z Problem 97 Suppose A, B , and C are n by n real matrices. For each of the following statements, give an example. (a) A and B have only positive eigenvalues, but AB has only negative eigenvalues. (b) A, B , and C are positive definite, but ABC has only negative eigenvalues. Solution (a) , , A= 12 01 , 107 B= 10 −2 1 A B 1, 1 AB AB = , −3 2 −2 1 −1, −1 (b) , A= 31 11 , B= 1 0 0 30 , C= 5 −2 −2 1 −5, −12 Problem 98 ABC = −45 24 −55 28 If A is a 3 × 3 symmetric matrix with eigenvalues λ1 < λ2 < λ3 . Prove that if p, q are two eigenvalues of a 2 × 2 principal submatrix of A and p ≤ q , then λ1 ≤ p ≤ λ2 ≤ q ≤ λ3 . Solution A A = QΛQT , Λ = diag(λ1 , λ2 , λ3 ), Q = [qij ] t = λi , i = 1, 2, 3, tI − A (orthogonal matrix), QT = Q−1 , tI − A = Q(tI )QT − QΛQt = Q(tI − Λ)QT adj(tI − A) = det(tI − A)(tI − A)−1 , q, p≤q 2 q31 B 2 q32 A 2 q33 (tI − A)−1 = Q(tI − Λ)−1 QT t − λ1 + t − λ2 108 + t − λ3 2×2 (3, 3) , : p, (tI − A)−1 det(tI2 − B )/ det(tI − A), (3, 3) , (adj(tI − A))33 / det(tI − A), det(tI2 − B ) q2 q2 q2 = 31 + 32 + 33 det(tI − A) t − λ1 t − λ2 t − λ3 p, q det(tI2 − B ) , , 2 q31 q2 q2 (t − p)(t − q ) = + 32 + 33 (t − λ1 )(t − λ2 )(t − λ3 ) t − λ1 t − λ2 t − λ3 ǫ , t = λ3 + ǫ, ,p q p, q , λ1 − ǫ λ2 , p, q ǫ → 0, λ3 + ǫ ; t = λ1 − ǫ , λ1 ≤ p ≤ ǫ → 0, q ≤ λ3 , , Problem 99 t = λ2 + ǫ , , p ≤ λ2 ≤ q matrix, show that B T AB is positive semidefinite. Also show that rank(B T AB ) = rankB , so that B T AB is positive definite if and only if B has rank m. Solution , B T AB A , B T AB B T AB B x = 0, Bx = 0 mx, Let A be an n × n real symmetric positive definite matrix. If B is an n × m real xT B T AB x = (B x)T A(B x) ≥ 0, B x = 0, ,B , B T AB x = 0; B T AB Problem 100 , B x = 0, n×m – xT B T AB x > 0 m×m B , , B T AB , rank(B T AB ) = rankB xT B T AB x = 0, B T AB , , B T AB x = 0, rankB = m , rankB = rank(B T AB ) = m, Let A be an n × n real symmetric positive semidefinite matrix and x be an n- dimensional real vector. Show that xT Ax = 0 if and only if Ax = 0. Conclude 109 that an n × n positive semidefinite matrix has rank n if and only if it is positive definite. Solution p(t) = (x + ty)T A(x + ty) t , AT = A, A , t, p(t) ≥ 0 , A p(t) = xT Ax + 2tyT Ax + t2 yT Ay dp = 2yT Ax + 2tyT Ay dt xT Ax = 0, yT Ax = 0, Ax = 0, , xT Ax = 0 x = 0, Ax = 0, p(0) = 0 y , A , Ax = 0 , rankA = n, xT Ax > 0, N (A) t = 0, dp dt =0 , 110 : http://www.lib.ntu.edu.tw/exam/default.htm : http://web.mit.edu/18.06/www/ : H. Anton, C. Rorres, Elementary Linear Algebra, 8th ed., John Wiley & Sons, New York, 2000. S. H. Friedberg, A. J. Insel, and L. E. 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