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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 ?

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How can I write matrices with latex ? – Peter Jun 16 '14 at 8:14
For the LaTeX question see my edit ... – martini Jun 16 '14 at 8:15
Simples counter-example: $ \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2}\\ 1/\sqrt{2} & -1\sqrt{2} \end{pmatrix} $ – Peter Franek Jun 16 '14 at 8:24
It is not reasonable to expect this as your property is invariant under an orthogonal change of basis. Thr characteristic polynomial is of degree 2 which tells you the eigenvalues, and since your matrix is symmetric, this tells you up to an orthogonal transformation it is a diagonal matrix with plus or minus ones on the diagonal. – George Shakan Jun 16 '14 at 8:26
A counterexample with rational values : $\frac{1}{25}$[[-7,24][24,7]] – Peter Jun 16 '14 at 8:40
up vote 13 down vote accepted

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.

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Very nice answer! Thank you very much! – Peter Jun 16 '14 at 8:27
In more geometric terms, such a matrix is an orthogonal reflection through a subspace. I.e., if $A$ is symmetric and orthogonal, then $P:=\frac12(A+I)$ is an orthogonal projection, and $A=2P-I$ is the reflection through the image of $P$. – Harald Hanche-Olsen Jun 16 '14 at 8:34

$A$ 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 $-1$.

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@ Urgje I am unable to understand how the exponential formula arrives. – Orca May 11 at 7:38

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