Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
5-\lambda & 8 & 16 \\
4 & 1-\lambda & 8\\
-4 & -4 & -11 - \lambda\end{bmatrix}
\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix}$
$\begin{bmatrix}
5-\lambda & 8 & 16 \\
4 & 1-\lambda & 8\\
-4 & -4 & -11 - \lambda\end{bmatrix}
=0$
$\lambda_1=1, \lambda_2=-3$
2. Find eigenvectors:
For $\lambda=1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
5-\lambda & 8 & 16 \\
4 & 1-\lambda & 8\\
-4 & -4 & -11 - \lambda\end{bmatrix}
=\begin{bmatrix}
4 & 8 & 16\\
4 & 0 & 8\\
-4 & -4 & -12
\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0\end{bmatrix} \\$
We obtain reduced row echelon form:
$\begin{bmatrix}
4 & 8 & 16\\
4 & 0 & 8\\
-4 & -4 & -12
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 & 4\\
0 & 1 & 1\\
0 & 0 & 0
\end{bmatrix}$
It has one unpivot column. (1)
For $\lambda=-3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
5-\lambda & 8 & 16 \\
4 & 1-\lambda & 8\\
-4 & -4 & -11 - \lambda\end{bmatrix}
=\begin{bmatrix}
8 & 8 & 16\\
4 & 4 & 8\\
-4 & -4 & -8
\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0\end{bmatrix} \\$
We obtain reduced row echelon form:
$\begin{bmatrix}
8 & 8 & 16\\
4 & 4 & 8\\
-4 & -4 & -8
\end{bmatrix} \approx \begin{bmatrix}
1 & 1 & 2\\
0 & 0& 0\\
0 & 0 & 0
\end{bmatrix}$
It has two unpivot columns (2)
From (1) and (2), $A$, is diagonalizable.
The canonical Jordan form is
$J=\begin{bmatrix} 1 & 0 & 0\\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{bmatrix} $
