Exercises on eigenvalues and eigenvectors Problem 21.1: (6.1 #19. Introduction to Linear Algebra: Strang) A three by three matrix B is known to have eigenvalues 0, 1 and 2. This information is enough to ﬁnd three of these (give the answers where possible): a) The rank of B b) The determinant of B T B c) The eigenvalues of B T B d) The eigenvalues of ( B2 + I )−1 Solution: a) B has 0 as an eigenvalue and is therefore singular (not invertible). Since B is a three by three matrix, this means that its rank can be at most 2. Since B has two distinct nonzero eigenvalues, its rank is exactly 2. b) Since B is singular, det( B) = 0 . Thus det( B T B) = det( B T ) det( B) = 0. c) There is not enough information to ﬁnd the eigenvalues of B T B. For example: ⎡ If B = ⎣ ⎤ 0 ⎡ ⎦ then B T B = ⎣ 1 ⎤ 0 1 2 ⎡ 4 ⎤ 0 1 ⎣ 1 If B = ⎡ ⎦ then B T B = ⎣ 2 ⎦. ⎤ 0 2 ⎦. 4 d) If p(t) is a polynomial and if x is an eigenvector of A with eigenvalue λ, then p( A)x = p(λ)x. We also know that if λ is an eigenvalue of A then 1/λ is an eigenvalue of A−1 . Hence the eigenvalues of ( B2 + I )−1 are 021+1 , 121+1 and 221+1 , or 1, 1/2 and 1/5. Problem 21.2: (6.1 #29.) Find the eigenvalues of A, B, and C when ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 2 3 0 0 1 2 2 2 A = ⎣ 0 4 5 ⎦ , B = ⎣ 0 2 0 ⎦ and C = ⎣ 2 2 2 ⎦ . 0 0 6 3 0 0 2 2 2 Solution: Since the eigenvalues of a triangular matrix are its diagonal entries, the eigenvalues of A are 1,4, and 6. For B we have: det( B − λI ) = (−λ)(2 − λ)(−λ) − 3(2 − λ) = (λ2 − 3)(2 − λ). √ Hence the eigenvalues of B are ± 3 and 2. Finally, for C we have: det(C − λI ) = (2 − λ)[(2 − λ)2 − 4] − 2[2(2 − λ) − 4] + 2[4 − 2(2 − λ)] = λ3 − 6λ2 = λ2 (λ − 6). The eigenvalues of C are 6, 0, and 0. We can quickly check our answers by computing the determinants of A and B and by noting that C is singular.