Are certain normal matrices circulant? (Part 2)

Let $\mathcal{F}$ denote the family of real normal matrices $A$ such that $A^TA=\begin{pmatrix} a & b \\ b & \ddots \end{pmatrix}$, for $b>0$.

As a user observed in the solution of Part 1 of this question, $\mathcal{F}$ is closed under left and right multiplication by permutation matrices. Partition $\mathcal{F}$ into equivalence classes, by writing $A\sim B$ if $B$ is obtained from $A$ by a sequence of row and column permutations.

Question 1: Does every equivalence class of $\mathcal{F}$ contain a circulant matrix?

Question 2: If so, can the circulant matrix be chosen such that $a_{ij}\geq 0$ with equality iff $i=j$?

Part 1 of this question

Original question on math.SE

Literature

There is a natural geometric reformulation of this problem:

Describe equidistant configurations of $n$ points on an $(n-1)$-sphere, subject to positivity constraints.

As $A\in \mathbb{R}^{n \times n}$ is normal we have $A=Q*S*Q^H$ with $D\in \mathbb{C}^{n\times n}$ diagonal and $Q\in \mathbb{C}^{n\times n}$ unitary. Let $e=(1\;1\dots 1)^T$, which is an eigenvector of $A^TA=Q*S^H*S*Q^H$ with multiplicity 1 and therefore also of $A$. Let $E$ be the matrix with every entry equal to 1. Note that there exist $\alpha_1, \alpha_2\in \mathbb{R}$ such that $\alpha_1 A+\alpha_2 E$ is orthogonal. Hence we can write $A=cE+dR$ with $R$ orthogonal. Vice versa for any $R$ orthogonal with $Re=e$ we can construct a normal $A$ with the desired properties. You can now simply take a generic $R$ to convince yourself that your conjecture is wrong, e.g. take the matrix
• I just want to point out that your reasoning is incorrect: eigenvectors of $A^TA$ need not be eigenvectors of $A$. For instance, take $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. – pre-kidney Jan 14 '15 at 3:41