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