Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} 2-\lambda & -2 & 14\\ 0 & 3-\lambda & -7\\
0 & 0 & 2-\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix}$
$\begin{bmatrix} 2-\lambda & -2 & 14\\ 0 & 3-\lambda & -7\\
0 & 0 & 2-\lambda \end{bmatrix}=0$
$(3- \lambda)(\lambda-2)^2=0$
$\lambda_1=\lambda_2=2,\lambda_3=3$
2. Find eigenvectors:
For $\lambda=2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 2-\lambda & -2 & 14\\ 0 & 3-\lambda & -7\\
0 & 0 & 2-\lambda \end{bmatrix}=\begin{bmatrix} 0 & -2 & 14\\ 0 & 1 & -7\\ 0 & 0 & 0 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix} $
Let $r,s$ be free variables.
$\vec{V}=r(1,0,0)+s(0,1,-7)\\
E_1=\{(1,0,0);(0,1,-7)\} \\
\rightarrow dim(E_2)=2$
For $\lambda=3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 2-\lambda & -2 & 14\\ 0 & 3-\lambda & -7\\
0 & 0 & 2-\lambda \end{bmatrix}=\begin{bmatrix} -1 & -2 & 14\\ 0 & 0 & -7\\ 0 & 0 & -1 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix} $
Let $r$ be a free variable.
$\vec{V}=r(-2,1,0)\\
E_1=\{(-2,1,0)\} \\
\rightarrow dim(E_2)=1$
Hence, $S=\begin{bmatrix} 1 & 0 & -2\\ 0 & 1 & 1 \\
0 & -7 & 0 \end{bmatrix}$
