# Orthogonal matrices with eigenvalues equally spaced over unit circle?

We know that the eigenvalues of orthogonal matrices have norm 1, and thus they are all on the unit circle. However, I wonder if there is a way to construct a orthogonal matrix (with real entries) whose eigenvalue is equally spaced on unit circle (or as uniformly distributed as possible)? "A way to construct" means given an $$n$$, we can find such a $$n \times n$$ orthogonal matrix. Any help is appreciated!

The eigenvalues of the permutation matrix $$M = \pmatrix{0&\cdots & 0 & 0 & 1\\ 1&0\\ &1&0\\ &&\ddots & \ddots\\ &&&1&0}$$ will be all $$n$$th roots of unity, i.e. $$e^{2 \pi i k/n}$$ for $$i = 0,\dots,n-1$$. These are equally spaced over the unit circle.

You could make the eigenvalues be the $$n$$-th roots of unity, which are "evenly spaced" around the unit circle in $$\mathbb C$$. For instance, the matrix $$\begin{bmatrix}1&0&0\\0&\omega&0\\0&0&\omega^2\end{bmatrix}$$ works for $$n=3$$ where $$\omega=e^{2\pi i/3}$$. I leave it to you to generalize this to arbitrary $$n$$.

• Thanks, but sorry I forget to mention, I want the matrices to have real entries. – dave2d Jan 10 at 20:27
• Ah, well then yes my answer won't work. I'll leave it up just in case people are interested. – Dave Jan 10 at 20:27

Here is the easiest way to do it with real entries (I think). Given a real number $$\theta$$, the rotation matrix $$\begin{bmatrix}\cos \theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$$ has two complex eigenvalues, which have norm $$1$$ and argument $$\pm\theta$$. Using this, we can build the matrix we're after.

Take the complex unit circle, and distribute $$n$$ points along it so that

• They are evenly spaced
• The configuration is symmetric with respect to mirroring across the $$x$$-axis (i.e. complex conjugation)

Pair up the points in complex conjugate pairs, and for each pair, construct the corresponding rotation matrix as shown above (you can choose the sign of $$\theta$$ freely). Then take all those rotation matrices, and put them along the diagonal of an $$n\times n$$ matrix (with zeroes in all other entries).

If $$n$$ is odd, one of the points is $$1$$ (or $$-1$$), and doesn't have a pair mate. Just put a $$1$$ (or $$-1$$) on the diagonal of the $$n\times n$$ matrix instead of a rotation matrix when you get to that point.