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


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

1.9993    1.0000    1.0273    0.9734
0.9994    1.9999    1.0116    0.9891
0.9737    0.9888    1.9988    1.0386
1.0276    1.0113    0.9622    1.9988

(was done in matlab)

  • $\begingroup$ I just want to point out that your reasoning is incorrect: eigenvectors of $A^TA$ need not be eigenvectors of $A$. For instance, take $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. $\endgroup$ – pre-kidney Jan 14 '15 at 3:41
  • $\begingroup$ Yes you are right. We also need that the multiplicity of the eigenvalue is 1. I have changed it. $\endgroup$ – user35593 Jan 14 '15 at 10:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.