Answer
See below
Work Step by Step
Given
$\begin{bmatrix}
1-\lambda & -1 & 1\\
-1 & -1-\lambda & 1 \\
1 & 1 & 1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\lambda_1=2,\lambda_2=-1$
a) For $\lambda=2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-1 & -1 & 1\\
-1 & -3 & 1 \\
1 & 1 & -1
\end{bmatrix}$
Let $r,s$ be free variables.
$\vec{V}=r(-1,1,0)+s(1,0,1) \\
E_1=\{(-1,1,0);(1,0,1)\}
\rightarrow dim(E_1)=2$
Let $v_1=u_1=(1,0,1)\\
v_2=u_2=(-1,1,0)$
Use the Gram-Schmidt procedure:
$u_2=v_2-\frac{}{||u_1||^2} u_1=\frac{1}{2}(-1,2,1)$
An orthogonal basis for $E_1=\{(1,0,1);(-1,2,1)\}$
b) $\lambda=-1$
$B=\begin{bmatrix}
2 & -1 & 1\\
-1 & 0 & 1 \\
1 & 1 & 2
\end{bmatrix}$
Let $t$ be free variables.
$\vec{V}=t(-1,-1,1) \\
E_2=\{(-1,-1,1)\}
\rightarrow dim(E_1)=1$
Let $v=(-1,-1,1)$
Obtain $=0=$
The vectors in $E_1$ are orthogonal to the vectors in
$E_2 $
