To Find Eigenvalues Find the eigenvalues of the $6\times 6$ matrix
$$\left[\begin{matrix} 
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
\end{matrix}\right]$$
The options  are $1, -1, i, -i$
It is a real symmetric matrix and the eigenvalues of a real symmetric matrix are real. Hence $i$ and $-i$ can't be its  eigenvalues. Then what else we can say?
Is there any easy way to find it?

We can also change basis to get $$I_3 \otimes C_2 = \left(\begin{array}{ccc}C_2&0&0\\0&C_2&0\\0&0&C_2\end{array}\right)$$ where $C_2 = \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$ is a representation for the generating element of the cyclic 2-group. For this small matrix it is easy to find value 1,-1 and vectors $\left(\begin{array}{c}1\\1\end{array}\right)$ and $\left(\begin{array}{r}1\\-1\end{array}\right)$. Please note that the eigenvectors are different from my other answer above, but this is OK (why?)
