# 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?

• You've given us a specific example, but asked for a general procedure. In what way are you generalising? Do you want a general method for block-diagonalising a complex-diagonalisable real matrix? – Theo Bendit Mar 23 at 1:52
• Ideally yes. So that I can do this example and any other I encounter in the future... – SuperCiocia Mar 23 at 1:53

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 $$A$$.

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 $$\left( \begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right) \; \; ? \;$$

• Hey thanks. I am a physicists so I have little idea of what you just said. I can diagonalise $A$ all right. How do i find $\alpha$? – SuperCiocia Mar 23 at 1:54
• @SuperCiocia what are the eigenvalues of one of your two by two matrices with sine and cosine of some angle? – Will Jagy Mar 23 at 1:56
• $\cos (\theta )-i \sin (\theta )$ and $\cos (\theta )+i \sin (\theta )$ – SuperCiocia Mar 23 at 1:57
• @SuperCiocia alright, starting with your real matrix $R$ and desired diagonal complex matrix $C,$ what is a matrix $P$ with $P^{-1}R P = C \; ? \; \;$ Note that this also gives $PC P^{-1} = R \; , \; \;$ which is why I brought it up in this fashion – Will Jagy Mar 23 at 2:01
• I would make $P$ from the eigenvectors or $R$. Which gives me $\alpha$ complex exponentials? – SuperCiocia Mar 23 at 2:07