Orthogonal matrices with eigenvalues equally spaced over unit circle?

Question

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!

Answer 1

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.

Answer 2

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 $n$.

Answer 3

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.