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

Question

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?

Answer 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.

Answer 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 $-1$.

Answer 3

You can construct orthogonal and symmetric matrices using a nice parametrization from Sanyal [1] and Mortari [2].

We want a matrix $R$ both orthogonal and symmetric, i.e.

$$ R^T R = I \ \ \text{and}\ \ R = R^T $$

which also means $R^2 = I$ and arbitrary powers are well-behaved

$$ R^k = \begin{cases} R \ \ \ \text{if} \ \ k \ \ \ \text{is odd} \\ I \ \ \ \text{if} \ \ k \ \ \ \text{is even} \end{cases} $$

And now the parametrization. One can construct such a matrix with a choice of $n$ orthogonal vectors $\{r_k\}_{k=1}^n$ and the desired number of positive eigenvalues $p \in [0,n]$

$$ R = \sum_{k=1}^p r_kr_k^T - \sum_{k=p+1}^n r_kr_k^T $$

They also point out that

if $p=n$, then $R=I$ whereas if $p=0$, then $R=-I$.

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1. Sanyal, A. K., Geometrical Transformations in Higher Dimensional Euclidean Spaces, Master’s thesis, Department of Aerospace Engineering, Texas A&M University, College Station, TX, May 2001.
2. Mortari, D.. (2004). Ortho-Skew and Ortho-Sym Matrix Trigonometry. Advances in the Astronautical Sciences. 115.

Answer 4

Can a matrix with the desired properties only contain the values -1,0 and 1 ?

For this part of your question every 3-D rotation matrix (it's orthogonal) about any axis ( defined by a unit vector $v$) by angle $\pi$ is symmetric.

You can generate plenty of them with Rodrigues' rotation formula which for a $\pi$ case takes simpler form $rot(v, \pi)= 2vv^T-I$ and they are not necessary consist only of $-1, 0, 1 $.