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This question is a follow-up to the [previous question on symmetric matrices](https://mathoverflow.net/questions/306346/a-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 $a$. 



Symmetric orthogonal matrices with constant diagonal entries
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.


