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{\bf Question}

Find the eigenvectors for the following matrices
\begin{description}
\item[(a)] $\left(\begin{array}{cc} 3 &amp; 4 \\ 2 &amp; 1 \end{array} \right)$
\item[(b)]$\left( \begin{array}{ccc} 1&amp; 0 &amp; 1 \\ -1 &amp; 1 &amp; 0 \\ 1 &amp;
0 &amp; 1 \end{array}\right)$

\end{description}


\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
$\left(\begin{array}{cc} 3 &amp; 4 \\ 2 &amp; 1 \end{array} \right)\left(
\begin{array}{c}x\\y\end{array}\right) = \lambda
\left(\begin{array}{c}x\\y\end{array}\right)$

To find eigenvalues

 \begin{eqnarray*} \left| \begin{array}{cc} 3 -
\lambda &amp; 4 \\ 2 &amp; 1 - \lambda \end{array} \right| &amp; = &amp; (\lambda
- 3)(\lambda - 1) - 8 \\ &amp; = &amp; \lambda^2 - 4\lambda  + 3 - 8 \\ &amp;
= &amp;  \lambda^2 - 4\lambda  + 5  \\ &amp; = &amp; (\lambda - 5)(\lambda +
1) = 0 \end{eqnarray*}

$\Rightarrow {\rm eigenvalues\ are\ } \lambda_1 = 5, \lambda_2 =
-1$

eigenvector let ${\bf e_1} = \left( \begin{array}{c} e_{11} \\
e_{12} \end{array} \right)$

and $A {\bf e_1}= \lambda_1{\bf e_1}$ implies $A\left(
\begin{array}{c} e_{11} \\ e_{12} \end{array} \right) =
\lambda_1\left( \begin{array}{c} e_{11} \\ e_{12}
\end{array} \right)$

From the first equation we get

$3e_{11} + 4e_{12} = 5e_{11} \Rightarrow 4e_{12}= 2e_{11}
\Rightarrow 2e_{12}= e_{11}$

The second equation will be equivalent.

Let $e_{12} = c$

The first eigenvector is ${\bf e_1} = \left( \begin{array}{c} 2c
\\ c
\end{array} \right)$

For the second eigenvector let ${\bf e_2} = \left(
\begin{array}{c} e_{21}
\\ e_{22} \end{array} \right)$

and $A {\bf e_2}= \lambda_2{\bf e_2} \Rightarrow
\left(\begin{array}{cc} 3&amp; 4 \\ 2 &amp; 1 \end{array}\right) \left(
\begin{array}{c} e_{21} \\ e_{22} \end{array} \right) =
-\left( \begin{array}{c} e_{21} \\ e_{22}
\end{array} \right)$

From the first equation we get

$3e_{21} + 4e_{22} = -e_{21} \Rightarrow 4e_{22}= -4e_{21}
\rightarrow -e_{22}= e_{21} = -c$

Hence ${\bf e_2} = \left( \begin{array}{c} -c \\ c
\end{array} \right)$ is the eigenvector corresponding to
$\lambda_2 = -1$

\item[(b)]
$B{\bf x} = \lambda{\bf x}$

$B = \left( \begin{array}{ccc} 1&amp; 0 &amp; 1 \\ -1 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 1
\end{array}\right)$

To find eigenvectors $ |B - \lambda I|  =  0$

 \begin{eqnarray*}
 \left| \begin{array}{ccc} 1 - \lambda &amp; 0 &amp; 1 \\ -1 &amp; 1- \lambda
&amp; 0 \\ 1 &amp; 0 &amp; 1- \lambda \end{array}\right| &amp; = &amp; (1 - \lambda)^3
- (1 - \lambda) \\ &amp; = &amp; (1 - \lambda) [ (1- \lambda)^2 - 1] \\ &amp;
= &amp; (1- \lambda) \lambda (2 - \lambda) = 0
\end{eqnarray*}

Hence the eigenvalues are $\lambda_1 = 0, \, \lambda_2 = 1, \,
\lambda_3 = 2$

To find the eigenvectors:

 ${\bf e_1} = \left( \begin{array}{c}
e_{11}
\\ e_{12}
\\ e_{13}
\end{array} \right) \hspace{.2in} B{\bf e_1} =  \lambda_1{\bf e_1}
= 0$

$ \Rightarrow  \left( \begin{array}{ccc} 1&amp; 0 &amp; 1 \\ -1 &amp; 1 &amp; 0
\\ 1 &amp; 0 &amp; 1 \end{array}\right) \left( \begin{array}{c} e_{11} \\ e_{12} \\ e_{13}
\end{array} \right) =  0$

$\Rightarrow e_{11}=-e_{13}= e_{12} = C \Rightarrow {\bf e_1} = C
\left(
\begin{array}{c} 1 \\ 1 \\ -1 \end{array} \right) = \left(
\begin{array}{c} C \\ C \\ -C \end{array} \right)$


${\bf e_2} = \left( \begin{array}{c} e_{21} \\ e_{22} \\ e_{23}
\end{array} \right) \hspace{.2in} B{\bf e_2} =  \lambda_2{\bf e_2}
= {\bf e_2} $

$\Rightarrow  \left( \begin{array}{ccc} 1&amp; 0 &amp; 1 \\ -1 &amp; 1 &amp; 0 \\
1 &amp; 0 &amp; 1 \end{array}\right) \left( \begin{array}{c} e_{21} \\
e_{22} \\ e_{23} \end{array} \right) =  \left(
\begin{array}{c} e_{21} \\ e_{22} \\ e_{23} \end{array} \right)$

$\Rightarrow e_{21} + e_{23}= e_{21}, e_{21} + e_{23} = e_{23},
-e_{21} + e_{22} = e_{22} \Rightarrow e_{21} = e_{23} = 0$

$\Rightarrow {\bf e_2} = \left( \begin{array}{c} 0 \\ D \\ 0
\end{array} \right)$

${\bf e_3} = \left( \begin{array}{c} e_{31} \\ e_{32} \\ e_{33}
\end{array} \right) \hspace{.2in} B{\bf e_3} =  \lambda_3{\bf e_3}
= 2{\bf e_3} $

$\Rightarrow  \left( \begin{array}{ccc} 1&amp; 0 &amp; 1 \\ -1 &amp; 1 &amp; 0 \\
1 &amp; 0 &amp; 1 \end{array}\right) \left( \begin{array}{c} e_{31} \\
e_{32} \\ e_{33} \end{array} \right) =  2\left(
\begin{array}{c} e_{31} \\ e_{32} \\ e_{33} \end{array} \right)$

$\Rightarrow e_{31}=-e_{32}= e_{33} = E \Rightarrow {\bf e_3} = E
\left(
\begin{array}{c} 1 \\ -1 \\ 1 \end{array} \right) = \left(
\begin{array}{c} E \\ -E \\ E \end{array} \right)$
\end{description}


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