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!
Orthogonal matrices with eigenvalues equally spaced over unit circle
eigenvalues-eigenvectorslinear algebramatricesorthonormal
Best Answer
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.