Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & -2 & 0\\
-2 & 1-\lambda & 0\\
0& 0 & 3-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & -2 & 0\\
-2 & 1-\lambda & 0\\
0& 0 & 3-\lambda
\end{bmatrix}=0$
$(\lambda-3)^2(\lambda+1)=0$
$\lambda_1=\lambda_2=3,\lambda_3=-1$
2. Find eigenvectors:
For $\lambda=-1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2 & -2 & 0\\
-2 & 2 & 0\\
0& 0 & 4
\end{bmatrix}=\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} $
Let $r$ be a free variable.
$\vec{V}=r(1,1,0) \\
E_1=\{(1,1,0)\}
\rightarrow dim(E_1)=1$
The eigenvectors span $\{(1,1,0)\}$ in $R$
For $\lambda=3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-2 & -2 & 0\\
-2 & -2 & 0\\
0& 0 & 0
\end{bmatrix}=\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} $
Let $r,t$ be a free variable.
$\vec{V}=r(-1,1,0)+t(0,0,1)\\
E_2=\{(-1,1,0);(0,0,1)\}
\rightarrow dim(E_2)=2$
The eigenvectors span $\{(-1,1,0);(0,0,1)\}$ in $R$
Since there are three independent eigenvectors exist, $A$ is diagonalizable.
Obtain $S=\begin{bmatrix}
1 & -1 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{bmatrix} $
