Exam-Style Problem

9231 P13 - Jun 2013 - Q9 - 11 marks

6405

The square matrix \(\mathbf{A}\) has an eigenvalue \(\lambda\) with corresponding eigenvector \(\mathbf{e}\). The non-singular matrix \(\mathbf{M}\) is of the same order as \(\mathbf{A}\). Show that \(\mathbf{M e}\) is an eigenvector of the matrix \(\mathbf{B}\), where \(\mathbf{B}=\mathbf{M} \mathbf{A} \mathbf{M}^{-1}\), and that \(\lambda\) is the corresponding eigenvalue.

Let
\(\mathbf{A}=\left(\begin{array}{rrr} -1 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & 2 \end{array}\right)\)

Write down the eigenvalues of \(\mathbf{A}\) and obtain corresponding eigenvectors.

Given that
\(\mathbf{M}=\left(\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)\)
find the eigenvalues and corresponding eigenvectors of \(\mathbf{B}\).

Solution

Answer: Let \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\). Then

\(\mathbf{B}(\mathbf{M}\mathbf{e})=\mathbf{MAM}^{-1}(\mathbf{M}\mathbf{e})=\mathbf{MAe}=\mathbf{M}(\lambda\mathbf{e})=\lambda(\mathbf{M}\mathbf{e}).\)

Since \(\mathbf{M}\) is non-singular, \(\mathbf{M}\mathbf{e}\neq \mathbf{0}\). Therefore \(\mathbf{M}\mathbf{e}\) is an eigenvector of \(\mathbf{B}\) with corresponding eigenvalue \(\lambda\).

The eigenvalues of \(\mathbf{A}\) are \(-1,1,2\).

Corresponding eigenvectors for \(\mathbf{A}\) may be taken as

for \(\lambda=-1\): \(\left(\begin{array}{c}1\\0\\0\end{array}\right)\)
for \(\lambda=1\): \(\left(\begin{array}{c}1\\1\\0\end{array}\right)\)
for \(\lambda=2\): \(\left(\begin{array}{c}9\\4\\1\end{array}\right)\)

The eigenvalues of \(\mathbf{B}\) are also \(-1,1,2\).

Using \(\mathbf{M}\mathbf{e}\):

for \(\lambda=-1\): \(\mathbf{M}\left(\begin{array}{c}1\\0\\0\end{array}\right)=\left(\begin{array}{c}1\\0\\0\end{array}\right)\)
for \(\lambda=1\): \(\mathbf{M}\left(\begin{array}{c}1\\1\\0\end{array}\right)=\left(\begin{array}{c}1\\1\\0\end{array}\right)\)
for \(\lambda=2\): \(\mathbf{M}\left(\begin{array}{c}9\\4\\1\end{array}\right)=\left(\begin{array}{c}10\\4\\1\end{array}\right)\)

So corresponding eigenvectors of \(\mathbf{B}\) may be taken as \(\left(\begin{array}{c}1\\0\\0\end{array}\right)\), \(\left(\begin{array}{c}1\\1\\0\end{array}\right)\), \(\left(\begin{array}{c}10\\4\\1\end{array}\right)\).

Let \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\), where \(\mathbf{e}\neq \mathbf{0}\). Since

\(\mathbf{B}=\mathbf{MAM}^{-1},\)

we have

\(\mathbf{B}(\mathbf{M}\mathbf{e})=\mathbf{MAM}^{-1}(\mathbf{M}\mathbf{e})=\mathbf{MAe}=\mathbf{M}(\lambda\mathbf{e})=\lambda(\mathbf{M}\mathbf{e}).\)

Also, \(\mathbf{M}\) is non-singular, so \(\mathbf{M}\mathbf{e}\neq \mathbf{0}\). Hence \(\mathbf{M}\mathbf{e}\) is an eigenvector of \(\mathbf{B}\) with eigenvalue \(\lambda\).

Now

\(\mathbf{A}=\begin{pmatrix}-1&2&1\\0&1&4\\0&0&2\end{pmatrix}.\)

Since \(\mathbf{A}\) is upper triangular, its eigenvalues are the diagonal entries:

\(\lambda=-1,\ 1,\ 2.\)

Eigenvectors of \(\mathbf{A}\)

For \(\lambda=-1\):

\(\mathbf{A}+\mathbf{I}=\begin{pmatrix}0&2&1\\0&2&4\\0&0&3\end{pmatrix}.\)

So \(z=0\), then \(y=0\), and \(x\) is free. Take \(x=1\):

\(\mathbf{e}_{-1}=\left(\begin{array}{c}1\\0\\0\end{array}\right).\)

For \(\lambda=1\):

\(\mathbf{A}-\mathbf{I}=\begin{pmatrix}-2&2&1\\0&0&4\\0&0&1\end{pmatrix}.\)

So \(z=0\), and \(-2x+2y=0\Rightarrow x=y\). Take \(x=y=1\):

\(\mathbf{e}_{1}=\left(\begin{array}{c}1\\1\\0\end{array}\right).\)

For \(\lambda=2\):

\(\mathbf{A}-2\mathbf{I}=\begin{pmatrix}-3&2&1\\0&-1&4\\0&0&0\end{pmatrix}.\)

Let \(z=t\). Then from the second row, \(-y+4t=0\Rightarrow y=4t\). From the first row, \(-3x+2y+z=0\), so

\(-3x+8t+t=0\Rightarrow 3x=9t\Rightarrow x=3t.\)

Taking \(t=1\), one eigenvector is

\(\mathbf{e}_{2}=\left(\begin{array}{c}3\\4\\1\end{array}\right).\)

So a suitable set of eigenvectors for \(\mathbf{A}\) is

\(\left(\begin{array}{c}1\\0\\0\end{array}\right),\quad \left(\begin{array}{c}1\\1\\0\end{array}\right),\quad \left(\begin{array}{c}3\\4\\1\end{array}\right).\)

Now

\(\mathbf{M}=\begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}.\)

The eigenvalues of \(\mathbf{B}=\mathbf{MAM}^{-1}\) are the same as those of \(\mathbf{A}\), namely

\(-1,\ 1,\ 2.\)

Corresponding eigenvectors are \(\mathbf{M}\mathbf{e}\):

For \(\lambda=-1\):

\(\mathbf{M}\left(\begin{array}{c}1\\0\\0\end{array}\right)=\left(\begin{array}{c}1\\0\\0\end{array}\right).\)

For \(\lambda=1\):

\(\mathbf{M}\left(\begin{array}{c}1\\1\\0\end{array}\right)=\left(\begin{array}{c}1\\1\\0\end{array}\right).\)

For \(\lambda=2\):

\(\mathbf{M}\left(\begin{array}{c}3\\4\\1\end{array}\right)=\left(\begin{array}{c}4\\4\\1\end{array}\right).\)

Therefore corresponding eigenvectors of \(\mathbf{B}\) may be taken as

\(\left(\begin{array}{c}1\\0\\0\end{array}\right),\quad \left(\begin{array}{c}1\\1\\0\end{array}\right),\quad \left(\begin{array}{c}4\\4\\1\end{array}\right).\)