Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
2-\lambda & 1 & 1 \\
2 & 1-\lambda & -2\\
-1 & 0 & -2 - \lambda\end{bmatrix}
\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix}$
$\begin{bmatrix}
2-\lambda & 1 & 1 \\
2 & 1-\lambda & -2\\
-1 & 0 & -2 - \lambda\end{bmatrix}
=0$
$\lambda_1=-1, \lambda_2=3$
2. Find eigenvectors:
For $\lambda=-1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2-\lambda & 1 & 1 \\
2 & 1-\lambda & -2\\
-1 & 0 & -2 - \lambda\end{bmatrix}
=\begin{bmatrix}
3 & 1 & 1\\
2 & 2 & -2\\
-1 & 0 & -1
\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0\end{bmatrix} \\$
We obtain reduced row echelon form:
$\begin{bmatrix}
3 & 1 & 1\\
2 & 2 & -2\\
-1 & 0 & -1
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & -2\\
0 & 0 & 0
\end{bmatrix}$
It has one unpivot column. (1)
For $\lambda=3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2-\lambda & 1 & 1\\
2 & 1-\lambda & -2\\
-1 & 0 & -2-\lambda
\end{bmatrix} \approx \begin{bmatrix}
-1 & 1 & 1 \\
2 & -2 & -2\\
-1 & 0 & -5
\end{bmatrix} \begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0\end{bmatrix} \\$
We obtain reduced row echelon form:
$ \begin{bmatrix}
-1 & 1 & 1 \\
2 & -2 & -2\\
-1 & 0 & -5
\end{bmatrix} \approx \begin{bmatrix}
1 & -1 & -1 \\
0 & 1 & 6\\
0 & 0 & 0
\end{bmatrix} $
It has one unpivot column (2)
From (1) and (2), $A$, is diagonalizable.
The canonical Jordan form is
$J=\begin{bmatrix} 3 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} $
