Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & 1 & 0\\
-4 & 2-\lambda & 0\\
17 & -11 & -2 - \lambda\end{bmatrix}
\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix}$
$\begin{bmatrix}
1-\lambda & 1 & 0\\
-4 & 2-\lambda & 0\\
17 & -11 & -2 - \lambda\end{bmatrix}=0$
$\lambda_1=\lambda_2=3, \lambda_3=-2$
2. Find eigenvectors:
For $\lambda=3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
1-\lambda & 1 & 0\\
-4 & 2-\lambda & 0\\
17 & -11 & -2 - \lambda\end{bmatrix}
=\begin{bmatrix}
-2 & 1 & 0\\
-4 & -1 & 0\\
17 & -11 & -5
\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0\end{bmatrix} \\$
We obtain reduced row echelon form:
$S=\begin{bmatrix}
-2 & 1 & 0\\
-4 & -1 & 0\\
17 & -11 & -5
\end{bmatrix} \approx \begin{bmatrix}
1 & -3 & -5\\
0 & 1 & 2\\
0 & 0 & 0
\end{bmatrix}$
$\lambda =3$ has multiplicity $2$, but it contains only one unpivot column. Hence, $A$ is not diagonalizable.
