MATH 217 Linear Algebra
Homework 8
December 19, 2021
1. For the vector ~v and matrix A given below, determine whether ~v is an eigenvector of A or not. If it is, find the
corresponding eigenvalue.
1
1 −1
(a) ~v =
, A=
.
3
6 −4
−1
5 2
(b) ~v =
, A=
.
1
3 6
 


1
3 6 7
(c) ~v = −2, A = 3 2 7.
2
5 6 4
2. For λ and A given below, determine whether λ is an eigenvalue of A or not. If it is, find the corresponding eigenvector.
3 2
(a) λ = 2, A =
.
3 8
−1 4
(b) λ = −3, A =
.
6 9


3 0 −1
(c) λ = 4, A =  2 3 1 .
−3 4 5


4 −2 3
(d) λ = 1, A =  0 −1 3 .
−1 2 −2
3. Compute the characteristic polynomials of the following matrices.


a 0 0
a c
(a)
.
(b)  b c 0 .
b d
d e f


0 0 −2
1 2
(c)
.
(d) 0 2 3 .
3 4
6 1 −1
4. Determine the eigenvalues and corresponding eigenvectors of the following



1 0
0
0
3 −1
2  (c) 2
(a)
(b) 0 3
−5 −1
2 −2 −1
2
1
5. Let A =
2
matrices.

2 2
0 2 .
2 0
−1
.
4
(a) Show that the characteristic polynomial of A is p(λ) = λ2 − 5λ + 6.
(b) Show that A satisfies its characteristic polynomial, that is, A2 − 5A + 6I2 = 02 . (Note that this is known as
Cayley-Hamilton Theorem which is true for any n × n matrix.)
(c) Use the result in part (b) to determine A−1 . (Hint: Multiply the equation in part (b) by A−1 .)
1
6. If ~v1 = (1, −1) and ~v2 = (2, 1) are eigenvectors of the matrix A corresponding to the eigenvalues λ1 = 2 and λ2 = −3
respectively, find A(3~v1 − ~v2 ).
7. Let A be an n × n invertible matrix. Prove that if λ is a nonzero eigenvalue of A, then 1/λ is an eigenvalue of A−1 .
8. Let A be an n × n matrix. Prove that λ = 0 is an eigenvalue of A if and only if det(A) = 0.
9. Let A be a 4 × 4 matrix with characteristic polynomial p(λ) = λ4 + 2λ3 − 5λ2 + λ. Is A invertible? Explain.
Answers
1. (a) Yes, λ = −2. (b) Yes, λ = 3. (c) No.
2. (a) Yes, ~v = t(−2, 1), t ∈ R \ {0}. (b) Yes, ~v = t(−2, 1), t ∈ R \ {0}. (c) Yes, ~v = t(−1, −1, 1), t ∈ R \ {0}. (d) Yes,
~v = t(0, 1, 32 ), t ∈ R \ {0}.
3. (a) λ2 + (−a − d)λ + (ad − bc) = 0. (b) (a − λ)(c − λ)(f − λ) = 0. (c) λ2 − 5λ − 2 = 0. (d) λ3 − λ2 + 10λ − 24 = 0.
4. (a) λ1 = −2 of algebraic multiplicity one, ~v = t(1, 5), t ∈ R \ {0} and λ2 = 4 of algebraic multiplicity one,
~v = t(1, −1), t ∈ R \ {0}. (b)λ = 1 of algebraic multiplicity three, ~v = t(0, −1, 1), t ∈ R \ {0}. (c) λ1 = 4 of algebraic
multiplicity one, ~v = r(1, 1, 1), r ∈ R\{0} and λ2 = −2 of algebraic multiplicity two, ~v = s(−1, 1, 0)+t(−1, 0, 1), s, t ∈
R such that s, t are not both zero.
1 4 1
5. (c) A−1 =
.
6 −2 1
6. (12, −3).
7.
8.
9. No, because, one of the eigenvalues is zero. Now use the previous problem.
2