eigen vector that orthogonal to each other, symmetry matrix symmetry matrix $\left(\begin{array}{ccc} 0 & 1 & 1 \\
      1 & 0 & 1 \\
      1 & 1 & 0  \end{array}\right) $
one of  eigen value is $\lambda_1=2$ and one of eigen vector is $x_1=\left(\begin{array}{ccc} \frac{1}{\sqrt3} \\
       \frac{1}{\sqrt3}  \\
       \frac{1}{\sqrt3}  \end{array}\right) $
then i found the two other eigen value is $\lambda_2=-1 $ & $\lambda_3=-1$
but the question want eigen vector that is orthogonal to each other and has magnitude $
|x_2|=|x_3|=1$ 
here how can i find the two other eigen vector that orthogonal to each other?
using usual computation $x=-y-z, i cant find the vector that is orthogonal to each other and with magnitude 1.
using trial and error?
 A: \begin{eqnarray*}
\begin{bmatrix}
   \frac{2}{\sqrt{6}}  \\    \frac{-1}{\sqrt{6}}  \\    \frac{-1}{\sqrt{6}}  \\
\end{bmatrix},
\begin{bmatrix}
   0  \\    \frac{1}{\sqrt{2}}  \\    \frac{-1}{\sqrt{2}}  \\
\end{bmatrix}
\end{eqnarray*}
Edit: Note that
\begin{eqnarray*}
\begin{bmatrix}
   2 \\    -1 \\    -1  \\
\end{bmatrix},
\begin{bmatrix}
   -1  \\    2  \\    -1  \\
\end{bmatrix},
\begin{bmatrix}
   -1  \\    -1 \\    2 \\
\end{bmatrix}
\end{eqnarray*}
are eigenvectors with eigenvalue $-1$ and between them they span a space of dimension $2$, so we need to find a linear combination that is orthogonal the first one 
\begin{eqnarray*}
\begin{bmatrix}
  2  \\    -1  \\    -1  \\
\end{bmatrix} \cdot ( \alpha
\begin{bmatrix}
   2 \\    -1  \\    -1  \\
\end{bmatrix}+ \beta
\begin{bmatrix}
   -1  \\    2  \\    -1  \\
\end{bmatrix}+ \gamma
\begin{bmatrix}
   -1  \\    -1  \\    2  \\
\end{bmatrix} ) .
\end{eqnarray*}
This gives $ 6 \alpha -3 \beta -3 \gamma=0$ and $ \alpha=1, \beta=-2, \gamma=0$ will do, and now you just need to normalise.
