Jianwen Zhao Department of Computer Science and Engineering The Chinese University of Hong Kong 1/16
Problem 1. Matrix Diagonalization Diagonalize the following matrix: A = [ ] 1 2 4 3 2/16
Solution The 2 2 matrix A has two distinct eigenvalues λ 1 = 1 and λ 2 = 5, which means it is diagonalizable. We then obtain an arbitrary eigenvector v 1 of λ 1 and also an arbitrary eigenvector v 2 of λ 2, say [ [ 1 1 v 1 =, v 1] 2 = 2] Next, apply the diagonalization method we discussed in class, form: [ ] 1 1 Q = 1 2 by using v 1 and v 2 as the first and second column respectively. 3/16
Solution cont. Q has the inverse Q 1 = [ 2/3 ] 1/3 1/3 1/3 We thus obtain the following diagonalization of A: A = Q diag[ 1, 5] Q 1 4/16
Problem 2. Matrix Power Consider again the matrix A in Problem 1, i.e,. [ ] 1 2 A = 4 3 Calculate A t for any integer t 1. 5/16
Solution We already know that A = Q diag[ 1, 5] Q 1 Hence, A t = Q diag[( 1) t, 5 t ] Q 1 [ ] [ ] [ ] 1 1 ( 1) t 0 2/3 1/3 = 1 2 0 5 t 1/3 1/3 [ ] (5 = t + 2 ( 1) t )/3 (5 t + ( 1) t+1 )/3 (2 5 t + 2 ( 1) t+1 )/3 (2 5 t + ( 1) t+2 )/3 6/16
Problem 3. Matrix Diagonalization Diagonalize the following matrix: 4 3 3 A = 3 2 3 1 1 2 7/16
Solution A has eigenvalues λ 1 = 1 and λ 2 = 2. EigenSpace(λ 1 ) includes all [ x1 x 2 x 3 ] T satisfying x1 = u + v, x 2 = u, x 3 = v for any u, v R. The vector space EigenSpace(λ 1 ) has dimension 2 with a basis {v 1, v 2 } where v 1 = [ 1 1 0 ] T (given by u = 1, v = 0) and v2 = [ 1 0 1 ] T (given by u = 0, v = 1). Similarly, EigenSpace(λ 2 ) includes all [ x 1 x 2 ] T x 3 satisfying x 1 = x 2 = 3u and x 3 = u for any u R. The vector space EigenSpace(λ 2 ) has dimension 1 with a basis {v 3 } where v 3 = [ 3 3 1 ] T (given by u = 1). 8/16
Solution-cont. So far, we have obtained three linearly independent eigenvectors v 1, v 2, v 3 of A. We then construct Q = 1 1 3 1 0 3 0 1 1 and Q has the inverse 3 4 3 Q 1 = 1 1 0 1 1 1 We thus obtain the following diagonalization of A: A = Q diag[1, 1, 2] Q 1 9/16
Problem 4. Matrix Similarity Suppose that matrices A and B are similar to each other, namely, there exists P such that A = P 1 BP. Prove: if x is an eigenvector of A under eigenvalue λ, then P x is an eigenvector of B under eigenvalue λ. 10/16
Problem 5. Matrix Trace Definition. The trace of an n n square matrix A, denoted by tr(a), is defined to be the sum of the elements on the main diagonal of A, i.e., tr(a) = n i=1 a ii. For example, if then tr(a) = 4 + ( 2) + 2 = 4. 4 3 3 A = 3 2 3 1 1 2 Prove: tr(ab) = tr(ba), where A is an m n matrix and B is an n m matrix. 11/16
Solution Proof. Denote by a ij the element of A at i-th row and j-th column, b ji the element of B at j-th row and i-th column, where i = 1, 2,, m and j = 1, 2,, n. Then n (AB) ii = a i1 b 1i + a i2 b 2i + + a in b ni = a ij b ji Similarly, Hence (BA) jj = b j1 a 1j + b j2 a 2j + + b jm a mj = tr(ab) = m i=1 j=1 n a ij b ji = n j=1 i=1 j=1 m b ji a ij i=1 m b ji a ij = tr(ba) 12/16
Problem 6. Traces & Eigenvalues & Determinants Suppose A is an n n diagonalizable matrix, namely, there exists Q such that A = QBQ 1, and B is a diagonal matrix. Denote by λ 1, λ 2,, λ n the n eigenvalues of A. Prove: (1) tr(a) = n i=1 λ i, (2) det(a) = Π n i=1 λ i. 13/16
Solution Proof. (1) tr(a) = tr(qbq 1 ) = tr(bq 1 Q) = tr(b) n = i=1 Where the second equality used the fact that tr(ab) = tr(ba) and the last equality used the facts (i) A and B have exactly the same eigenvalues due to their similarity, and (ii) the eigenvalues of a diagonal matrix are simply its diagonal elements. λ i 14/16
Solution-cont. (2) det(a) = det(qbq 1 ) = det(q) det(b) det(q 1 ) = det(b) det(q) det(q 1 ) = det(b) det(qq 1 ) = det(b) = Π n i=1λ i Where the last equality used the facts (i) A and B have exactly the same eigenvalues due to their similarity, and (ii) the eigenvalues of a diagonal matrix are simply its diagonal elements. 15/16
In fact, the conclusion of this problem is true in general, regardless of whether A is diagonalizable. For any n n square matrix A, if its n eigenvalues are λ 1, λ 2,, λ n, then tr(a) = n i=1 λ i and det(a) = Π n i=1 λ i. The proof is not difficult but a little tedious, students who are interested may refer to the proof at the following link: https://www.adelaide.edu.au/mathslearning/play/seminars/ evalue-magic-tricks-handout.pdf 16/16