Is a normal matrix satisfying $A^TA=...$ circulant?

Question

Let $A=\{a_{ij}\}$ be a normal matrix such that $a_{ij}\geq 0$ with equality iff $i=j$. Suppose that $$ A^TA=\begin{pmatrix} a & b & \cdots & b\\ b & a & \ddots & \vdots\\ \vdots & \ddots & a & b\\ b & \cdots & b & a\\ \end{pmatrix},\ where\ b>0. $$ Does it follow that $A$ is a circulant matrix?

Note: There is a partial classification of non-negative normal matrices posted here, which seems like it can be used to attack this problem.

There is a geometric interpretation as well: both the set of rows and the set of columns of $A$ form equidistant sets of vectors on a sphere, and basic geometry appears to severely restrict the possibilities.

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Reposted from math.SE

Answer 1

NO.

Let $A$ satisfy the assumptions. If $P$ is a permutation matrix, then $B:=PA$ satisfies the assumptions too: on the one hand, we have $B^TB=A^TP^TPA=A^TA$. On the other hand (remark that the matrix in the question is permutation-invariant) $$BB^T=PAA^TP^T=PA^TAP^T=A^TA=B^TB.$$

If the claim is true, we find that $B=PA$ is circulant for every permutation matrix $P$. We deduce that $a_{ij}$ not only depends upon $j-i$ (modulo $n$), but also depends only upon $j-\sigma(i)$, for every $\sigma\in{\frak S}_n$. We conclude that $a_{ij}$ is constant.

Therefore every matrix $A$ satisfying the assumptions, such that $a_{ij}$ is not constant, provides a counter-example of the form $PA$ for some (many) permutation matrices $P$.

Answer 2

No consider for example $$A=\begin{pmatrix}0 & 1 & 1+\sqrt{2} & \sqrt{2}\\1+\sqrt{2} & 0 & \sqrt{2} & 1\\1 & \sqrt{2} & 0 &1+\sqrt{2}\\\sqrt{2} & 1+\sqrt{2} & 1 & 0 \end{pmatrix}.$$