Answer
$\lambda=5$
$\vec{V}=(1,0,0)r+(0,1,0)s+(0,0,1)t$
Work Step by Step
1. eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
5-\lambda &0 &0 \\
0& 5-\lambda&0\\
0&0 &5-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2\\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$
$\begin{vmatrix}
5-\lambda &0 &0 \\
0& 5-\lambda&0\\
0&0 &5-\lambda
\end{vmatrix}=0$
$\lambda=5$
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2. eigenvectors
$\lambda=5$
let $B=A-\lambda I$
$B= \begin{bmatrix}
5-\lambda &0 &0 \\
0& 5-\lambda&0\\
0&0 &5-\lambda
\end{bmatrix}$
= $\begin{bmatrix}
0&0 &0 \\
0& 0&0\\
0&0 &0
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
0&0 &0 & 0\\
0& 0&0& 0\\
0&0 &0 & 0
\end{array}\right)
\]
let $r, s , t$ are a free variables.
$\vec{V}=(1,0,0)r+(0,1,0)s+(0,0,1)t$
