Answer
$\lambda=2$
$\vec{V}=(1.5,1,0)r-(-1,0,1)s$
Work Step by Step
1. eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
10-\lambda &-12 &8 \\
0& 2-\lambda&0\\
-8&12 &-6-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2\\
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{vmatrix}
10-\lambda &-12 &8 \\
0& 2-\lambda&0\\
-8&12 &-6-\lambda
\end{vmatrix}=0$
$\lambda=2$
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2. eigenvectors
$\lambda_1=2$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
10-\lambda &-12 &8 \\
0& 2-\lambda&0\\
-8&12 &-6-\lambda
\end{bmatrix}$
= $\begin{bmatrix}
8&-12&8 \\
0 &0 &0\\
-8&12 &-8
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
1 & -1.5 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)
\]
let $r$ and $s$ is a free variable.
Then $x_1=1.5x_2-1x_3 = 1.5r-1s$
$\vec{V}=(1.5,1,0)r-(-1,0,1)s$
