What can be said about a matrix which is both symmetric and orthogonal?

I tried to find matrices A, which are both orthogonal and symmetric, this means $A=A^{-1}=A^T$. I only found very special examples like I, -I or the matrix $$\begin{pmatrix} 0 &0& -1\\ 0& -1& 0\\ -1& 0& 0 \end{pmatrix} $$ Can a matrix with the desired properties only contain the values -1,0 and 1 ? Which matrices of a given size have the desired property ?

Solution 1:

For your first question, the answer is no. Every real Householder reflection matrix is a symmetric orthogonal matrix, but its entries can be quite arbitrary.

In general, if $A$ is symmetric, it is orthogonally diagonalisable and all its eigenvalues are real. If it is also orthogonal, its eigenvalues must be 1 or -1. It follows that every symmetric orthogonal matrix is of the form $QDQ^\top$, where $Q$ is a real orthogonal matrix and $D$ is a diagonal matrix whose diagonal entries are 1 or -1.

Solution 2:

$A$ is orthogonal and symmetric, so $A=A^{-1}$ and $A=A^{T}$. More general, let $A$ be a unitary and self-adjoint operator with discrete spectrum in a separable Hilbert space. Then $A=\exp [iW]$ with $W$ self-adjoint and $A=A^{\ast }=\exp [-iW]$. Thus $W=\sum_{n}\lambda _{n}P_{n}$ with $\lambda _{n}\in \mathbb{R}$ and the $P_{n}$ are orthogonal projectors, $\lambda _{m}\neq \lambda _{n}$, $m\neq n$ and $P_{m}P_{n}=\delta _{mn}P_{m}$ . Now \begin{equation*} A=\sum_{n}\exp [i\lambda _{n}]P_{n}=A^{\ast }=\sum_{n}\exp [-i\lambda _{n}]P_{n}, \end{equation*} so $\exp [2i\lambda _{n}]=1$ leading to $\lambda _{n}=k_{n}\pi $, $k_{n}\in \mathbb{Z}$, which is either $+1$ or $-1$.