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This question is a follow-up to the previous question on symmetric matrices. Thanks to the responses by Christian Remling and Geoff Robinson to that question, the problem now becomes much more specific, as follows.

Suppose that a symmetric matrix $M\in\R^{n\times n}$ is orthogonal or, equivalently, satisfies the condition $M^2=I_n$, where $I_n$ is the $n\times n$ identity matrix. Suppose also that all the diagonal entries of $M$ are equal to one another. Is it then true that $M$ is of the form $aI_n+b\,\de\de^T$, where $\de$ belongs to the set (say $\De_n$) of all $n\times 1$ column matrices with all entries from the set $\{-1,1\}$; $a\in\{-1,1\}$; and $b\in\{0,-2a/n\}$?

This is true for $n\in\{2,3\}$.

Comment 1. Let $\mathcal M_n$ denote the set of all matrices $M$ satisfying the stated conditions, that is, the set of all symmetric orthogonal matrices $M\in\R^{n\times n}$ with constant diagonal entries.
The actual problem here is to show that for each $M\in\mathcal M_n$ all the off-diagonal entries $M_{ij}$ of $M$ with $i\ne j$ are of the form $c\de_i\de_j$ for some real $c$ and some $\de\in\De_n$; it is then easy to specify the appropriate $a$ and $b$, given in the above question. So, it is enough to show that \begin{equation} \prod_{\de\in\De_n}\Big(\sum_{1\le i<j\le n}\sum_{1\le k<\ell\le n}(M_{ij}\de_i\de_j-M_{k\ell}\de_k\de_\ell)^2\Big)=0, \end{equation} which is how the cases the cases of $n\in\{2,3\}$ were verified.

Comment 2. More generally, even without the condition on the diagonal entries of $M$, it is enough to show that $M-aI_n$ is of rank $1$ for some real $a$.

  • $\begingroup$ what is an $n\times n$ column matrix? $\endgroup$ Jul 20, 2018 at 15:35
  • 2
    $\begingroup$ @AbdelmalekAbdesselam : Thank you for spotting the typo. It is now corrected. $\endgroup$ Jul 20, 2018 at 15:36

1 Answer 1


No. A symmetric $M$ will satisfy $M^2=1$ if and only if the spectrum is contained in $\pm 1$, which is equivalent to $M=P-(1-P)=2P-1$ for some orthogonal projection $P$. Now you're asking if the extra condition that the diagonal is constant will give $P$ rank $1$ or $n-1$.

It's clear that this won't follow because we have such examples with $\textrm{rank }P=1$ for $n=2$ and can just take orthogonal sums of those for larger (even) $n$'s.

  • $\begingroup$ So, the cases of $n\in\{2,3\}$ are the only ones when the answer is yes. Turning to orthoprojectors is a nice idea. $\endgroup$ Jul 20, 2018 at 17:06

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