Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} 5-\lambda & 0 & -1\\ 1 & 4-\lambda & -1\\
1 & 0 & 3-\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$
$\begin{bmatrix} 5-\lambda & 0 & -1\\ 1 & 4-\lambda & -1\\
1 & 0 & 3-\lambda \end{bmatrix}=0$
$(4- \lambda)(\lambda^2+8\lambda+16)=0$
$\lambda_1=\lambda_2=\lambda_3=4$
2. Find eigenvectors:
For $\lambda=4$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 5-\lambda & 0 & -1\\ 1 & 4-\lambda & -1\\
1 & 0 & 3-\lambda \end{bmatrix}=\begin{bmatrix} 1 & 0 & -1 \\ 1 & 0 & -1 \\ 1 & 0 & -1\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix} $
Let $r,s,t$ be free variables.
$\vec{V}=r(1,1,1)+s(1,0,0)+t(1,0,1)\\
E_1=\{(1,1,1);(1,0,0);(1,0,1)\} \\
\rightarrow dim(E_2)=3$
Hence, $S=\begin{bmatrix} 1 & 1 & 1\\ 1 & 0 & 0 \\
1 & 0 & 1 \end{bmatrix} \\
\rightarrow S^{-1}AS=D=\begin{bmatrix} 4 & 1 & 0\\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} $
