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QUESTION

Find the eigenvalues of the matrix

$$\left(\begin{array}{ccc}1&3&0\\3&1&0\\0&0&2\end{array}\right),$$

and determine the corresponding eigenvectors.

Verify that the eigenvectors are perpendicular to each other.


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ANSWER

$\mathbf{A}=\left(\begin{array}{ccc}1&3&0\\3&1&0\\0&0&2\end{array}\right)$
The eigenvalues satisfy $\det(A-\lambda I)=0$

\begin{eqnarray*}
\det(A-\lambda
I)&=&\det\left(\begin{array}{ccc}1-\lambda&3&0\\3&1-\lambda&0\\0&0&2-\lambda\end{array}\right)\\
&=&(2-\lambda)\{(1-\lambda)^2-3^2\}=(2-\lambda)\{(1-\lambda-3)(1-\lambda+3)\}\\
&=&(2-\lambda)(-\lambda-2)(4-\lambda)=(2-\lambda)(\lambda+2)(\lambda-4)\\
&=&0 {\rm\ if\ }\lambda=2,-2,4
\end{eqnarray*}



$\lambda=4:\\ (A-4I)\textbf{X}=0\\
\left(\begin{array}{ccc}-3&3&0\\3&-3&0\\0&0&-2\end{array}\right)
\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)\\
\left.\begin{array}{r}-3x_1+3x_2=0\\3x_1-3x_2=0\\-2x_3=0\end{array}\right\}\
x_1=x_2=C,\ x_3=0$

Therefore the eigenvector is
$C\left(\begin{array}{c}1\\1\\0\end{array}\right)$

$\lambda=-2:\\ (A-(-2)I)\textbf{X}=0\\
\left(\begin{array}{ccc}3&3&0\\3&3&0\\0&0&4\end{array}\right)
\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)\\
\left.\begin{array}{r}3x_1+3x_2=0\\3x_1+3x_2=0\\4x_3=0\end{array}\right\}\
x_2=-x_1,\ x_3=0$

Therefore the eigenvector is
$D\left(\begin{array}{c}1\\-1\\0\end{array}\right)$

$\lambda=2:\\ (A-2I)\textbf{X}=0\\
\left(\begin{array}{ccc}-1&3&0\\3&-1&0\\0&0&0\end{array}\right)
\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)\\
\left.\begin{array}{r}-x_1+3x_2=0\\3x_1-x_2=0\\0=0\end{array}\right\}\rightarrow
-x_1+3x_2=0,\ 8x_2=0,\rightarrow  x_1=x_2=0,\ x_3=$E

Therefore the eigenvector is
$E\left(\begin{array}{c}0\\0\\1\end{array}\right)$

$C(1,1,0)\cdot D(1,-1,0)=CD(1-1+0)=0\\ C(1,1,0)\cdot
E(0,0,1)=CE(0+0+0)=0\\ D(1,-1,0)\cdot E(0,0,1)=DE(0+0+0)=0$

So the eigenvectors are perpendicular to each other.



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