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Exercises 21.7 Exercises
Verifying eigenvectors and eigenvalues. In each case, use matrix multiplication to verify that
\(\uvec{x}\) is an eigenvector for matrix
\(A\) corresponding to the eigenvalue
\(\lambda\text{.}\)
1.
\(\displaystyle
A = \begin{abmatrix}{rr}
-8 \amp 14 \\
-7 \amp 13
\end{abmatrix}
\text{,}
\quad
\uvec{x} = \begin{bmatrix} 2 \\ 2 \end{bmatrix}
\text{,}
\quad
\lambda = 6\)
Solution .
Calculate:
\begin{align*}
A \uvec{x} \amp =
\begin{abmatrix}{rr}
-8 \amp 14 \\
-7 \amp 13
\end{abmatrix}
\begin{bmatrix} 2 \\ 2 \end{bmatrix}
\amp
6 \uvec{x} \amp =
6 \begin{bmatrix} 2 \\ 2 \end{bmatrix}\\
\amp = \begin{bmatrix} 12 \\ 12 \end{bmatrix} \text{,}
\amp
\amp = \begin{bmatrix} 12 \\ 12 \end{bmatrix} \text{.}
\end{align*}
Since the two results match, \(A \uvec{x} = \lambda \uvec{x}\) is true in this case for \(\lambda = 6\text{.}\)
2.
\(\displaystyle
A = \begin{abmatrix}{rr}
-5 \amp 5 \\
0 \amp 0
\end{abmatrix}
\text{,}
\quad
\uvec{x} = \begin{abmatrix}{r} -3 \\ 0 \end{abmatrix}
\text{,}
\quad
\lambda = -5\)
Solution .
Calculate:
\begin{align*}
A \uvec{x} \amp =
\begin{abmatrix}{rr}
-5 \amp 5 \\
0 \amp 0
\end{abmatrix}
\begin{abmatrix}{r} -3 \\ 0 \end{abmatrix}
\amp
-5 \uvec{x} \amp =
-5 \begin{abmatrix}{r} -3 \\ 0 \end{abmatrix}\\
\amp = \begin{bmatrix} 15 \\ 0 \end{bmatrix} \text{,}
\amp
\amp = \begin{bmatrix} 15 \\ 0 \end{bmatrix} \text{.}
\end{align*}
Since the two results match, \(A \uvec{x} = \lambda \uvec{x}\) is true in this case for \(\lambda = -5\text{.}\)
3.
\(\displaystyle
A = \begin{abmatrix}{rrr}
-4 \amp 0 \amp -3 \\
0 \amp 2 \amp 0 \\
6 \amp 0 \amp 5
\end{abmatrix}
\text{,}
\quad
\uvec{x} = \begin{abmatrix}{r} -1 \\ 3 \\ 2 \end{abmatrix}
\text{,}
\quad
\lambda = 2\)
Solution .
Calculate:
\begin{align*}
A \uvec{x} \amp =
\begin{abmatrix}{rrr}
-4 \amp 0 \amp -3 \\
0 \amp 2 \amp 0 \\
6 \amp 0 \amp 5
\end{abmatrix}
\begin{abmatrix}{r} -1 \\ 3 \\ 2 \end{abmatrix}
\amp
2 \uvec{x} \amp =
\begin{abmatrix}{r} -1 \\ 3 \\ 2 \end{abmatrix}\\
\amp = \begin{abmatrix}{r} -2 \\ 6 \\ 4 \end{abmatrix} \text{,}
\amp
\amp = \begin{abmatrix}{r} -2 \\ 6 \\ 4 \end{abmatrix} \text{.}
\end{align*}
Since the two results match, \(A \uvec{x} = \lambda \uvec{x}\) is true in this case for \(\lambda = 2\text{.}\)
4.
\(\displaystyle
A = \begin{abmatrix}{rrrr}
1 \amp 1 \amp -1 \amp -1 \\
-1 \amp 0 \amp -1 \amp 0 \\
0 \amp -1 \amp 2 \amp 1 \\
-1 \amp 1 \amp -3 \amp -1
\end{abmatrix}
\text{,}
\quad
\uvec{x} = \begin{abmatrix}{r} -3 \\ 4 \\ 3 \\ -2 \end{abmatrix}
\text{,}
\quad
\lambda = 0\)
Solution .
Calculate:
\begin{equation*}
A \uvec{x} =
\begin{abmatrix}{rrrr}
1 \amp 1 \amp -1 \amp -1 \\
-1 \amp 0 \amp -1 \amp 0 \\
0 \amp -1 \amp 2 \amp 1 \\
-1 \amp 1 \amp -3 \amp -1
\end{abmatrix}
\begin{abmatrix}{r} -3 \\ 4 \\ 3 \\ -2 \end{abmatrix}
=
\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}
\end{equation*}
Clearly we also have \(0 \uvec{x} = \zerovec\text{,}\) and so \(A \uvec{x} = \lambda \uvec{x}\) is true in this case for \(\lambda = 0\text{.}\)
Calculating eigenvalues and eigenspaces.
For each matrix:
Compute the characteristic polynomial in factored form. (For larger matrices, you may wish to use
Proposition 9.4.2 and/or
Proposition 9.4.4 to simplify the calculation.)
Determine the eigenvalues.
Based on the eigenvalues, state whether the matrix is invertible or singular.
For each eigenvalue, determine a basis for the corresponding eigenspace.
5.
\(\displaystyle
\begin{abmatrix}{rr}
-1 \amp -4 \\
1 \amp -5
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = {(\lambda + 3)}^2 \)
\(\displaystyle \lambda = 3 \)
invertible
\(\displaystyle \displaystyle E_{-3}(A) = \Span \left\{ \begin{bmatrix} 2 \\ 1 \end{bmatrix} \right\} \)
6.
\(\displaystyle
\begin{abmatrix}{rr}
-11 \amp 18 \\
-9 \amp 16
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = (\lambda - 7) (\lambda + 2) \)
\(\displaystyle \lambda = 7, -2 \)
invertible
\begin{align*}
\amp E_7 (A) = \Span \left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\} \\
\amp E_{-2}(A) = \Span \left\{ \begin{bmatrix} 2 \\ 1 \end{bmatrix} \right\}
\end{align*}
7.
\(\displaystyle
\begin{abmatrix}{rrr}
8 \amp - 4 \amp 0 \\
2 \amp 2 \amp 0 \\
10 \amp -10 \amp 1
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = (\lambda - 1) (\lambda - 4) (\lambda - 6) \)
\(\displaystyle \lambda = 1, 4, 6 \)
invertible
\begin{align*}
\amp E_1(A) = \Span \left\{ \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\} \\
\amp E_4(A) = \Span \left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \right\} \\
\amp E_6(A) = \Span \left\{ \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix} \right\}
\end{align*}
8.
\(\displaystyle
\begin{abmatrix}{rrr}
-21 \amp 0 \amp 7 \\
28 \amp -7 \amp -14 \\
-42 \amp 0 \amp 14
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = \lambda {(\lambda + 7)}^2 \)
\(\displaystyle \lambda = 0, -7 \)
singular
\begin{align*}
\amp E_0(A) = \Span \left\{ \begin{abmatrix}{r} 1 \\ -2 \\ 3 \end{abmatrix} \right\} \\
\amp
E_{-7}(A) = \Span \left\{
\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix},
\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}
\right\}
\end{align*}
9.
\(\displaystyle
\begin{abmatrix}{rrr}
0 \amp -14 \amp -15 \\
-1 \amp 7 \amp 9 \\
1 \amp -10 \amp -12
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = {(\lambda + 1)}^2 (\lambda + 3) \)
\(\displaystyle \lambda = -1, -3 \)
invertible
\begin{align*}
\amp E_{-1}(A) = \Span \left\{ \begin{abmatrix}{r} 1 \\ -1 \\ 1 \end{abmatrix} \right\} \\
\amp E_{-3}(A) = \Span \left\{ \begin{abmatrix}{r} 6 \\ -3 \\ 4 \end{abmatrix} \right\}
\end{align*}
10.
\(\displaystyle
\begin{abmatrix}{rrr}
0 \amp 3 \amp - 6 \\
-13 \amp 16 \amp -26 \\
- 5 \amp 5 \amp - 7
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = {(\lambda - 3)}^3 \)
\(\displaystyle \lambda = 3 \)
invertible
\(\displaystyle \displaystyle
E_3(A) = \Span \left\{
\begin{abmatrix}{r} -2 \\ 0 \\ 1 \end{abmatrix},
\begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}
\right\}\)
11.
\(\displaystyle
\begin{abmatrix}{rrr}
-1 \amp 0 \amp 1 \\
-3 \amp 1 \amp 1 \\
-2 \amp 1 \amp 0
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = {\lambda}^3 \)
\(\displaystyle \lambda = 0 \)
singular
\(\displaystyle \displaystyle
E_0(A) = \Span \left\{ \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} \right\}\)
12.
\(\displaystyle
\begin{abmatrix}{rrrr}
-19 \amp 0 \amp -30 \amp 0 \\
0 \amp -4 \amp 0 \amp 0 \\
9 \amp 0 \amp 14 \amp 0 \\
- 3 \amp 0 \amp - 6 \amp -4
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = (\lambda + 1) {(\lambda + 4)}^3 \)
\(\displaystyle \lambda = -1, -4 \)
invertible
\begin{align*}
\amp E_{-1}(A) = \Span \left\{ \begin{abmatrix}{r} 5 \\ 0 \\ -3 \\ 1 \end{abmatrix} \right\} \\
\amp
E_{-4}(A) = \Span \left\{
\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix},
\begin{abmatrix}{r} -2 \\ 0 \\ 1 \\ 0 \end{abmatrix},
\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}
\right\}
\end{align*}
13.
\(\displaystyle
\begin{abmatrix}{rrrr}
-19 \amp 5 \amp 1 \amp -18 \\
-46 \amp 1 \amp -31 \amp -49 \\
18 \amp 1 \amp 17 \amp 19 \\
1 \amp -6 \amp -18 \amp - 1
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = \lambda (\lambda - 4) {(\lambda + 3)}^2 \)
\(\displaystyle \lambda = 0, 4, -3 \)
singular
\begin{align*}
\amp E_0 (A) = \Span \left\{ \begin{abmatrix}{r} -8 \\ -17 \\ 5 \\ 4 \end{abmatrix} \right\} \\
\amp E_4 (A) = \Span \left\{ \begin{abmatrix}{r} -1 \\ - 1 \\ 0 \\ 1 \end{abmatrix} \right\} \\
\amp E_{-3}(A) = \Span \left\{ \begin{abmatrix}{r} -2 \\ - 3 \\ 1 \\ 1 \end{abmatrix} \right\}
\end{align*}
14.
\(\displaystyle
\begin{abmatrix}{rrrr}
-4 \amp 34 \amp 4 \amp 36 \\
0 \amp 10 \amp 0 \amp 12 \\
-1 \amp 23 \amp 0 \amp 24 \\
0 \amp -8 \amp 0 \amp -10
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = (\lambda - 2) {(\lambda + 2)}^3 \)
\(\displaystyle \lambda = 2, -2 \)
invertible
\begin{align*}
\amp E_2(A) = \Span \left\{ \begin{abmatrix}{r} -9 \\ -3 \\ -6 \\ 2 \end{abmatrix} \right\} \\
\amp
E_{-2}(A) = \Span \left\{
\begin{bmatrix} 2 \\ 0 \\ 1 \\ 0 \end{bmatrix},
\begin{abmatrix}{r} 1 \\ -1 \\ 0 \\ 1 \end{abmatrix}
\right\}
\end{align*}
15.
\(\displaystyle
\begin{abmatrix}{rrrr}
-1 \amp 5 \amp -1 \amp -1 \\
5 \amp - 1 \amp -1 \amp -1 \\
28 \amp -27 \amp 3 \amp 0 \\
25 \amp -25 \amp 1 \amp 5
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = (\lambda + 6) {(\lambda - 4)}^3 \)
\(\displaystyle \lambda = -6, 4 \)
invertible
\begin{align*}
\amp E_{-6}(A) = \Span \left\{ \begin{bmatrix} 0 \\ 1 \\ 3 \\ 2 \end{bmatrix} \right\} \\
\amp E_4 (A) = \Span \left\{ \begin{abmatrix}{r} -1 \\ -1 \\ -1 \\ 1 \end{abmatrix} \right\}
\end{align*}
16.
\(\displaystyle
\begin{abmatrix}{rrrr}
0 \amp 0 \amp 0 \amp 1 \\
-2 \amp 2 \amp -3 \amp -1 \\
1 \amp -1 \amp 1 \amp 1 \\
-3 \amp 3 \amp -4 \amp -3
\end{abmatrix}\)
Answer .
\(\displaystyle c_A(\lambda) = \lambda^4 \)
\(\displaystyle \lambda = 0 \)
singular
\(\displaystyle \displaystyle
E_0(A) = \Span \left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} \right\}\)
Zero, identity, and scalar matrices.
For each matrix:
Determine the characteristic polynomial in factored form and the eigenvalues.
For each eigenvalue, determine a basis for the corresponding eigenspace.
26. The
\(n \times n\) zero matrix,
\(\zerovec_n\text{.}\)
Answer .
Characteristic polynomial:
\(c_{\zerovec_n}(\lambda) = \lambda^n\text{.}\) Eigenvalue:
\(\lambda = 0\text{.}\)
Since
\(\zerovec_n \uvec{x} = 0 \uvec{x}\) is true for all
\(\uvec{x}\) in
\(\R^n\text{,}\) the eigenspace
\(E_0(\zerovec_n)\) is all of
\(\R^n\text{.}\) Therefore, the standard basis for
\(\R^n\) is also a basis for
\(E_0(\zerovec_n)\text{.}\)
27. The
\(n \times n\) identity matrix,
\(I_n\text{.}\)
Answer .
Characteristic polynomial:
\(c_{I_n}(\lambda) = {(\lambda - 1)}^n\text{.}\) Eigenvalue:
\(\lambda = 1\text{.}\)
Since
\(I_n \uvec{x} = 1 \uvec{x}\) is true for all
\(\uvec{x}\) in
\(\R^n\text{,}\) the eigenspace
\(E_1(I_n)\) is all of
\(\R^n\text{.}\) Therefore, the standard basis for
\(\R^n\) is also a basis for
\(E_1(I_n)\text{.}\)
28. A scalar matrix
\(k I_n\text{,}\) for a fixed scalar
\(k\text{.}\)
Answer .
Characteristic polynomial:
\(c_{(k I_n)}(\lambda) = {(\lambda - k)}^n\text{.}\) Eigenvalue:
\(\lambda = k\text{.}\)
Since
\((k I_n) \uvec{x} = k \uvec{x}\) is true for all
\(\uvec{x}\) in
\(\R^n\text{,}\) the eigenspace
\(E_k(k I_n)\) is all of
\(\R^n\text{.}\) Therefore, the standard basis for
\(\R^n\) is also a basis for
\(E_k(k I_n)\text{.}\)
29. Diagonal matrices.
Verify that for an
\(n \times n\) diagonal matrix
\(D\text{,}\) each standard basis vector
\(\uvec{e}_1, \uvec{e}_2, \dotsc, \uvec{e}_n\) of
\(\R^n\) is an eigenvector.
30. Block matrices.
(a)
Compute the characteristic polynomial and eigenvalues of the \(2 \times 2\) matrix
\begin{equation*}
\begin{abmatrix}{rr}
-1 \amp -4 \\
1 \amp -5
\end{abmatrix}\text{.}
\end{equation*}
Answer .
\(c_A(\lambda) = {(\lambda + 3)}^2\text{,}\) \(\lambda = -3\text{.}\)
(b)
Compute the characteristic polynomial and eigenvalues of the \(3 \times 3\) matrix
\begin{equation*}
\begin{abmatrix}{rrr}
3 \amp - 9 \amp - 6 \\
-5 \amp 16 \amp 15 \\
4 \amp -13 \amp -11
\end{abmatrix}\text{.}
\end{equation*}
Answer .
\(c_A(\lambda) = (\lambda - 6) {(\lambda - 1)}^2\text{,}\) \(\lambda = 6,1\text{.}\)
(c)
Compute the characteristic polynomial and eigenvalues of the \(5 \times 5\) matrix
\begin{equation*}
\begin{abmatrix}{rrrrr}
-1 \amp -4 \amp 99 \amp 99 \amp 99 \\
1 \amp -5 \amp 99 \amp 99 \amp 99 \\
0 \amp 0 \amp 3 \amp - 9 \amp - 6 \\
0 \amp 0 \amp -5 \amp 16 \amp 15 \\
0 \amp 0 \amp 4 \amp -13 \amp -11
\end{abmatrix}\text{.}
\end{equation*}
Answer .
\(c_A(\lambda) = {(\lambda + 3)}^2 (\lambda - 6) {(\lambda - 1)}^2\text{,}\) \(\lambda = -3,6,1\text{.}\)
(d) Do you notice any relationship between the results of the first two tasks of this exercise and the results of the third task? What do you think is the reason?
(e)
Suppose \(A\) is an \(\ell \times \ell\) matrix, \(B\) is an \(m \times m\) matrix, \(X\) is an \(\ell \times m\) matrix, and \(D\) is the \(n \times n\) matrix
\begin{equation*}
D = \begin{bmatrix} A \amp X \\ \zerovec \amp B \end{bmatrix} \text{,}
\end{equation*}
for \(n = \ell + m\) and where \(\zerovec\) represents the \(m \times \ell\) zero matrix.
Use the patterns from this exercise to make a conjecture about the relationship between the characteristic polynomial and eigenvalues of
\(D\) and the characteristic polynomial and eigenvalues of
\(A\) and
\(B\text{.}\)
31. Characteristic polynomial and eigenvalues of a transpose.
(a) Is there any relationship between the characteristic polynomial of a matrix and the characteristic polynomial of the transpose of that matrix?
(b) Is there any between the eigenvalues of a matrix and the eigenvalues of the transpose of that matrix?
32. Eigenspaces are subspaces.
In
Subsection 21.4.4 we argued that an eigenspace for eigenvalue
\(\lambda\) of an
\(n \times n\) matrix
\(A\) is a subspace of
\(\R^n\) because is the same space as the null space of the matrix
\(\lambda I - A\text{.}\)
Practise using the
Subspace Test by using it to verify that an eigenspace is a subspace. But do not use the matrix
\(\lambda I - A\) in your verifications; instead, directly use the definitions of
eigenvector and
eigenvalue . That is, assume
\(\lambda\) is a fixed eigenvalue of
\(n \times n\) matrix
\(A\text{.}\) Apply the
Subspace Test to the collection of all
\(\uvec{x}\) in
\(\R^n\) that satisfy
\(A \uvec{x} = \lambda \uvec{x}\text{.}\)
33. Eigenvalues of powers.
Suppose that
\(\lambda\) and
\(\uvec{x}\) are an eigenvalue-eigenvector pair of a square matrix
\(A\text{.}\)
(a) Use the definitions of
eigenvector and
eigenvalue to verify that
\(\lambda^2\) and
\(\uvec{x}\) are an eigenvalue-eigenvector pair of
\(A^2\text{.}\)
(b) Does the pattern of
Task a extend to higher powers of
\(A\) and
\(\lambda\text{?}\)
34. Eigenvalues of an inverse.
Suppose that
\(\lambda\) and
\(\uvec{x}\) are an eigenvalue-eigenvector pair of an
invertible square matrix
\(A\text{.}\)
Use the definitions of
eigenvector and
eigenvalue to verify that
\(\frac{1}{\lambda}\) and
\(\uvec{x}\) are an eigenvalue-eigenvector pair of
\(\inv{A}\text{.}\)
