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?

When you diagonalise a matrix over the complex numbers, you start by finding its eigenvalues, which may be complex. If the matrix is diagonalisable, then we can expect a basis of eigenvectors corresponding to these eigenvalues. If we compute the transformation $x \mapsto Ax$ in terms of this basis of eigenvectors, we get a complex diagonal matrix, similar to $A$.
To block-diagonalise $A$ over the reals, again find all the complex eigenvalues and a basis of eigenvectors. Suppose $\alpha + i \beta$ is a non-real eigenvalue, with corresponding eigenvector $v + i w$, where $v, w$ are vectors with real components. One can easily verify that $\alpha - i \beta$ is another eigenvalue with eigenvector $v - i w$. We can form a basis of eigenvectors such that, if $v + iw$ is in the basis (where $v$ and $w$ are real vectors), then so is $v - iw$.
Now, form a real basis like so: replace the conjugate eigenvector pairs $v \pm i w$ with the vectors $v, w$. If $v + iw$ has an eigenvalue $\alpha + i \beta$, then
\begin{align*}
Av &= A \left(\frac{(v + iw) + (v - iw)}{2}\right) \\
&= \frac{(\alpha + i\beta)(v + iw) + (\alpha - i\beta)(v - iw)}{2} \\
&= \frac{\alpha v + i \beta v + i \alpha w - \beta w + \alpha v - i \beta v - i \alpha w - \beta w}{2} \\
&= \alpha v - \beta w.
\end{align*}
Similar calculation reveals
$$Aw = \beta v + \alpha w.$$
So, when computing the matrix for $x \mapsto Ax$ with this basis, you'll find a block diagonal form, where each block is either a $1 \times 1$ block containing a real eigenvalue, or a $2 \times 2$ block of the form
$$\begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix},$$
where $\alpha \pm \beta i$ is a non-real eigenvalue of $A$.
