How can I block diagonalise this matrix? I have this matrix:
$$A = \left(
\begin{array}{cccc}
 0 & 0 & 0 & -1 \\
 1 & 0 & 0 & -1 \\
 0 & 1 & 0 & -1 \\
 0 & 0 & 1 & -1 \\
\end{array}
\right)$$
and I would like to show that it can be block diagonalised into :
$$ B = \left(
\begin{array}{cccc}
 \cos 2\pi/5 & -\sin 2\pi/5 & 0 & 0 \\
 \sin 2\pi/5 & \cos 2\pi/5 & 0 & 0 \\
 0 & 0 & \cos 4\pi/5 & -\sin 4\pi/5 \\
 0 & 0 & \sin 4\pi/5 & \cos 4\pi/5\\
\end{array}
\right) = \left(
\begin{array}{cccc}
 \frac{1}{4} \left(-1+\sqrt{5}\right) & -\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}} & 0 & 0 \\
 \sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}} & \frac{1}{4} \left(-1+\sqrt{5}\right) & 0 & 0 \\
 0 & 0 & \frac{1}{4} \left(-1-\sqrt{5}\right) & -\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}} \\
 0 & 0 & \sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}} & \frac{1}{4} \left(-1-\sqrt{5}\right) \\
\end{array}
\right)$$
What is the general procedure?

it is already a companion matrix (both that form and its transpose are used, depends on circumstances). The characteristic polynomial is $$x^4 + x^3 + x^2 + x + 1 = \frac{x^5-1}{x-1}$$
With four distinct eigenvalues (complex) it diagonalizes. Next, you need to figure out how to take a specific diagonal matrix, with complex $\alpha$ such that $|\alpha| = 1,$
$$
\left(
\begin{array}{rr}
\alpha & 0 \\
0 & \bar{\alpha}
\end{array}
\right)
$$
and send it back to one of your two by two blocks with sine and cosine.
To reverse the job:  with some real angle $\theta,$ exactly how do you diagonalize
$$
\left(
\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}
\right) \; \; ? \;
$$
