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