Solved Problems for Chapter 8

If \(A\) is an invertible \(n\times n\) matrix, compare the eigenvalues of \(A\) and \(A^{-1}\). More generally, for \(m\) an arbitrary integer, compare the eigenvalues of \(A\) and \(A^{m}\).

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\(A^{m}X=\lambda ^{m}X\) for any integer. In the case of \(-1,A^{-1}\lambda X=AA^{-1}X=X\) so \(A^{-1}X =\lambda ^{-1}X\). Thus the eigenvalues of \(A^{-1}\) are just \(\lambda ^{-1}\) where \(\lambda \) is an eigenvalue of \(A\).
If \(A\) is an \(n\times n\) matrix and \(c\) is a nonzero constant, compare the eigenvalues of \(A\) and \(cA\).

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Say \(AX=\lambda X.\) Then \( cAX=c\lambda X\) and so the eigenvalues of \(cA\) are just \( c\lambda \) where \(\lambda \) is an eigenvalue of \(A\).
Let \(A,B\) be invertible \(n\times n\) matrices which commute. That is, \(AB=BA\). Suppose \(X\) is an eigenvector of \(B\). Show that then \(AX\) must also be an eigenvector for \(B\).

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\(BAX=ABX =A\lambda X=\lambda AX\). Here it is assumed that \(BX=\lambda X\).
Suppose \(A\) is an \(n\times n\) matrix and it satisfies \(A^{m}=A\) for some \(m\) a positive integer larger than 1. Show that if \(\lambda \) is an eigenvalue of \(A\) then \(\left \vert \lambda \right \vert \) equals either 0 or \( 1\).

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Let \(X\) be the eigenvector. Then \(A^{m}X=\lambda ^{m} X,A^{m}X=AX=\lambda X\) and so
\[ \lambda ^{m}=\lambda \]
Hence if \(\lambda \neq 0,\) then
\[ \lambda ^{m-1}=1 \]
and so \(\left \vert \lambda \right \vert =1.\)
Show that if \(AX=\lambda X\) and \(AY=\lambda Y\), then whenever \(k,p\) are scalars,
\begin{equation*} A\left ( kX+pY\right ) =\lambda \left ( kX+pY\right ) \end{equation*}
Does this imply that \(kX+pY\) is an eigenvector? Explain.

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The formula follows from properties of matrix multiplications. However, this vector might not be an eigenvector because it might equal \(0\) and eigenvectors cannot equal \(0\).
Find the eigenvalues and eigenvectors of the matrix
\begin{equation*} \left [ \begin{array}{rrr} 6 & 76 & 16 \\ -2 & -21 & -4 \\ 2 & 64 & 17 \end{array} \right ] \end{equation*}
One eigenvalue is \(-2.\)

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The characteristic polynomial of this matrix is \(-\lambda ^3+2\lambda ^2+5\lambda -6\). Knowing one of the eigenvalues gives us one factor, \((\lambda +2)\). Use long division of polynomials to finish factoring \(-\lambda ^3+2\lambda ^2+5\lambda -6=-(\lambda +2)(\lambda -1)(\lambda -3)\). Now we have the following eigenvalues: \(\lambda _1=-2\), \(\lambda _2=1\), and \(\lambda _3=3\).

The corresponding eigenvectors are: \(\vec {v}_1=\begin{bmatrix}7\\-2\\6\end{bmatrix}\), \(\vec {v}_2=\begin{bmatrix}8\\-2\\7\end{bmatrix}\), and \(\vec {v}_3=\begin{bmatrix}4\\-1\\4\end{bmatrix}\).

Find the eigenvalues and eigenvectors of the matrix
\begin{equation*} \left [ \begin{array}{rrr} 3 & 5 & 2 \\ -8 & -11 & -4 \\ 10 & 11 & 3 \end{array} \right ] \end{equation*}
One eigenvalue is \(-3\).

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Characteristic polynomial: \(-\lambda ^3-5\lambda ^2-7\lambda -3=-(\lambda +3)(\lambda +1)^2\).
\[\lambda _1=-3,\quad \vec {v}_1=\begin{bmatrix}1\\-2\\2\end{bmatrix}\]
\[\lambda _2=-1,\quad \begin{bmatrix}1\\-2\\3\end{bmatrix}\]
Is it possible for a nonzero matrix to have only \(0\) as an eigenvalue?

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Yes. \(\left [ \begin{array}{cc} 0 & 1 \\ 0 & 0\end{array} \right ] \) works.
Let \(T\,\) be the linear transformation which reflects vectors about the \(x\) axis. Find a matrix for \(T\) and then find its eigenvalues and eigenvectors.

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The matrix of \(T\) is \(\left [ \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \right ]\).

The eigenvalues and eigenvectors are:

\[\lambda _1=-1,\quad \vec {v}_1=\begin{bmatrix}0\\1\end{bmatrix}\]
\[\lambda _2=1,\quad \vec {v}_2=\begin{bmatrix}1\\0\end{bmatrix}\]
Let \(T\) be the linear transformation which reflects all vectors in \( \mathbb {R}^{3}\) through the \(xy\) plane. Find a matrix for \(T\) and then obtain its eigenvalues and eigenvectors.

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The matrix of \(T\) is \(\left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right ]\) The eigenvalues are \(\lambda _1=-1\), \(\lambda _2=1\). Bases for the corresponding eigenspaces are:
\[ \mathcal {S}_{\lambda _1}=\left \{ \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right \}, \mathcal {S}_{\lambda _2}=\left \{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right \} \]
Find the eigenvalues and eigenvectors of the matrix
\begin{equation*} \left [ \begin{array}{rrr} 5 & -18 & -32 \\ 0 & 5 & 4 \\ 2 & -5 & -11 \end{array} \right ] \end{equation*}
One eigenvalue is \(1.\) Diagonalize if possible.

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The eigenvalues are \(-1, -1, 1\). The eigenvectors corresponding to the eigenvalues are:
\[ \left \{ \left [ \begin{array}{c} 10 \\ -2 \\ 3 \end{array} \right ] \right \} \leftrightarrow -1, \left \{ \left [ \begin{array}{c} 7 \\ -2 \\ 2 \end{array} \right ] \right \} \leftrightarrow 1 \]
Therefore this matrix is not diagonalizable.
Find the eigenvalues and eigenvectors of the matrix
\begin{equation*} \left [ \begin{array}{rrr} -13 & -28 & 28 \\ 4 & 9 & -8 \\ -4 & -8 & 9 \end{array} \right ] \end{equation*}
One eigenvalue is \(3.\) Diagonalize if possible.

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The eigenvectors and eigenvalues are:
\[ \left \{ \left [ \begin{array}{c} 2 \\ 0 \\ 1 \end{array} \right ] \right \} \leftrightarrow 1, \left \{ \left [ \begin{array}{c} -2 \\ 1 \\ 0 \end{array} \right ] \right \} \leftrightarrow 1, \left \{ \left [ \begin{array}{c} 7 \\ -2 \\ 2 \end{array} \right ] \right \} \leftrightarrow 3 \]
The matrix \(P\) needed to diagonalize the above matrix is
\[ \left [ \begin{array}{rrr} 2 & -2 & 7 \\ 0 & 1 & -2 \\ 1 & 0 & 2 \end{array} \right ] \]
and the diagonal matrix \(D\) is
\[ \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right ] \]
Find the eigenvalues and eigenvectors of the matrix
\begin{equation*} \left [ \begin{array}{rrr} 89 & 38 & 268 \\ 14 & 2 & 40 \\ -30 & -12 & -90 \end{array} \right ] \end{equation*}
One eigenvalue is \(-3.\) Diagonalize if possible.

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The eigenvectors and eigenvalues are:
\[ \left \{ \left [ \begin{array}{c} -6 \\ -1 \\ -2 \end{array} \right ] \right \} \leftrightarrow 6, \left \{ \left [ \begin{array}{c} -5 \\ -2 \\ 2 \end{array} \right ] \right \} \leftrightarrow -3, \left \{ \left [ \begin{array}{c} -8 \\ -2 \\ 3 \end{array} \right ] \right \} \leftrightarrow -2 \]
The matrix \(P\) needed to diagonalize the above matrix is
\[ \left [ \begin{array}{rrr} -6 & -5 & -8 \\ -1 & -2 & -2 \\ 2 & 2 & 3 \end{array} \right ] \]
and the diagonal matrix \(D\) is
\[ \left [ \begin{array}{rrr} 6 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -2 \end{array} \right ] \]
If \(A \sim B\) and \(A\) has any of the following properties, show that \(B\) has the same property.
1.
A is Idempotent, that is \(A^{2} = A\).
2.
A is Nilpotent, that is \(A^{k} = 0\) for some \(k \geq 1\).

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If \(B = P^{-1}AP\) and \(A^{k} = 0\), then \(B^{k} = (P^{-1}AP)^{k} = P^{-1}A^{k}P = P^{-1}0P = 0\).
3.
A is Invertible.

Bibliography

These problems come from Chapter 7 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)

Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 359–401.