Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} 1-\lambda & 1 & 1 \\ 3 & -1-\lambda & 2\\ 3 & 1 & 4-\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$
$\begin{bmatrix} 1-\lambda & 1 & 1 \\ 3 & -1-\lambda & 2\\ 3 & 1 & 4-\lambda \end{bmatrix}=0$
$\lambda_1=3-\sqrt 6,\lambda_2=3+\sqrt 6, \lambda_3=-2$
2. Find eigenvectors:
For $\lambda=3-\sqrt 6$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} -2+\sqrt 6 & 1 & 1 \\ 3 & -4+\sqrt 6 & 1\\ 1 & 1 & 1+\sqrt 6 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix} $
Let $r$ be free variables.
$\vec{V}=r(\sqrt 6,\sqrt 6-1,\sqrt 6-5)\\
E_1=\{(\sqrt 6,\sqrt 6-1,\sqrt 6-5)\} \\
\rightarrow dim(E_2)=1$
For $\lambda=3+\sqrt 6$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} -2-\sqrt 6 & 1 & 1 \\ 3 & -4-\sqrt 6 & 1\\ 1 & 1 & 1-\sqrt 6 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix} $
Let $s$ be free variables.
$\vec{V}=s(\sqrt 6,\sqrt 6+1,\sqrt 6+5)\\
E_2=\{(\sqrt 6,\sqrt 6+1,\sqrt 6+5)\} \\
\rightarrow dim(E_2)=1$
For $\lambda=-2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 3 & 1 & 1 \\ 3 & 1 & 1\\ 3 & 1 & 6 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix} $
Let $t$ be free variables.
$\vec{V}=t(-1,3,0)\\
E_3=\{(\sqrt 6,\sqrt 6-1,\sqrt 6-5)\} \\
\rightarrow dim(E_2)=1$
