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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!

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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.

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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 $n$.

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

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