6
$\begingroup$

$\newcommand{\R}{\mathbb{R}}$

The question is

Is there a constructive (say, parametric) description of the set (say $M_n$) of all symmetric matrices $A\in\R^{n\times n}$ such that all the diagonal entries of $A$ are $0$ and the matrix $(A-tI_n)^2$ is diagonal for some real $t$?

Here, of course, $I_n$ is the $n\times n$ identity matrix.

E.g., if $A=I_n-\frac1n\,1_n1_n^T$, then all the diagonal entries of $A$ are $0$ and
$(A-\frac12\,I_n)^2=\frac14\,I_n$, so that $A\in M_n$. Here, as usual, $1_n1_n^T$ is the matrix in $\R^{n\times n}$ all of whose entries are $1$.

Even a necessary condition for a matrix $A$ to be in the set $M_n$ could be useful, if it is close enough to sufficiency.

$\endgroup$
  • $\begingroup$ @ChristianRemling : Thank you for your comment. I have now edited the question, trying to express better what I need. $\endgroup$ – Iosif Pinelis Jul 19 '18 at 4:52
  • 2
    $\begingroup$ I think my previous comment still in principle outlines a procedure: Fix a diagonal matrix $D$. Then each eigenspace of $D$ is also invariant under $A$ if $(A-t)^2=D$. On such a space (with ev $d$, let's say), $A$ will have the desired property if and only if its spectrum is contained in $t\pm \sqrt{d}$. $\endgroup$ – Christian Remling Jul 19 '18 at 5:27
  • $\begingroup$ Of course, this gives various conditions that have to be satisfied. For example, $D$ can have at most one simple eigenvalue ($=t^2$), and the ev's of any part of $A$ must sum to zero. We probably also want a statement that if I have eigenvalues for $A$ as above with sum zero, then there will be an $A$ with these ev's. $\endgroup$ – Christian Remling Jul 19 '18 at 5:29
  • $\begingroup$ other examples of such matrices are symmetric conference matrices, i.e. a square matrix $A$ of order $n$ with zero diagonal and $\pm 1$ off-diagonal, such that $A^2 = (n-1) I_n$. Such matrices exist for many values of $n$ for example when $n\equiv 2 \ (mod \ 4)$ and $n−1$ is a prime power. $\endgroup$ – Mahdi Jul 19 '18 at 11:41
  • $\begingroup$ @AlexandreEremenko : I have re-checked the example, and it seems all right. Indeed, $(11^T)^2=n11^T$, whence $4(A-\frac12\,I)^2=(I-\frac2n\,11^T)^2=I-\frac4n\,11^T+\frac4{n^2}\,n11^T=I$. $\endgroup$ – Iosif Pinelis Jul 19 '18 at 13:33
6
$\begingroup$

Partial answer: As Christian Remling noted in remarks, it suffices to deal with the case that $(A-tI_{n})^{2} = \lambda I_{n}.$ Since $A$ is symmetric, we can only have $\lambda = 0$ when $A = tI_{n}$ for some $t$. But since the diagonal entries of $A$ are all $0,$ this only happens when $A = 0.$ Hence we may suppose that $\lambda \neq 0.$ Then $A = tI_{n} + \sqrt{\lambda}T$ for some matrix $T$ with $T^{2} = I_{n}.$ If $T = I_{n}$ we again see that $A$ is a scalar matrix, so the zero matrix. Hence the only non-zero possibilities for $A$ arise when $T^{2} = I \neq T.$ Now trace $T$ = $r-s$ where $T$ has $r$ eigenvalues $1$ and $s$ eigenvalues $-1$ Also, since $tI_{n} + \sqrt{\lambda}T$ has all diagonal entries zero, it follows that all diagonal entries of $T$ must be equal. Hence each diagonal entry of $T$ is $\frac{r-s}{n}$ and $t + \frac{(r-s)\sqrt{\lambda}}{n} = 0.$

Now $r = s$ gives $t = 0,$ so that $A = \sqrt{\lambda}T.$ If $r \neq s,$ then $\lambda = \frac{n^{2}t^{2}}{r-s}.$

Hence the problem is reduced in essence to determining which symmetric matrices $T$ of order two have all diagonal entries equal. But if $T$ is such a matrix, then $\frac{I+T}{2}$ and $\frac{I-T}{2}$ are mutually orthogonal idempotent symmetric matrices, each with all diagonal entries equal.

Now the problem is reduced to finding symmetric idempotent matrices $E$ with all diagonal entries equal (for we may take $T = 2E-I$ if we find such an $E$). If $E$ has rank $m,$ then note that each diagonal entry of $E$ is $\frac{m}{n}.$

We may obtain such an $E$ or each positive divisor $d$ of $n$: take $E$ to be the direct sum (in the obvious sense) of $n/d$ copies of $\frac{J_{d}}{d},$ where $J_{d}$ is the $d \times d$ matrix with all entries $1.$

At the moment, I don't see how to determine all possibilities for such an $E$. Later edit: (...but someone else might). One observation which might turn out to be relevant is that if $E$ is such an idempotent matrix, and $X$ is a "signed permutation matrix", that is, a matrix with exactly one non-zero entry in each row and each column, that entry being $\pm 1$), then $XEX^{t}$ is another such matrix. I don't know if there are other orthogonal matrices leaving this set of idempotents invariant.

Even later edit: Note that if $E,F$ are mutually orthogonal symmetric idempotent matrices with all diagonal matrices with all entries equal, then $E+F$ is symmetric idempotent with all diagonal entries equal.

This allows us to produce symmetric idempotent matrices with all diagonal entries equal of every rank for some values of $n$. For example, when $n = 2^{r},$ we may consider $Y = \frac{1}{\sqrt{n}} X,$ where $X$ is the character table of an elementary Abelian $2$-group of order $2^{r}.$

Let $u_{i}$ be the $i$-th column of $Y$. Then $E_{i} = u_{i}u_{i}^{t}$ is a symmetric idempotent matrix of rank $1$ with all diagonal entries $\frac{1}{n}.$ For $j \neq i,$ it is easy to see that $E_{i}E_{j} = E_{j}E_{i} = 0.$

Hence for $1 \leq k \leq n,$ we may take the sum of $k$ distinct $E_{i}$'s to get an idempotent symmetric matrix with all diagonal entries $\frac{k}{n}.$

$\endgroup$
  • $\begingroup$ Thank you very much for your answer, +1. This is a nice reduction. I'd also need a constructive description of the $E$'s, to be fed into an optimization problem. Perhaps, this is impossible, though. $\endgroup$ – Iosif Pinelis Jul 19 '18 at 13:23
  • $\begingroup$ I think the question of determining when there can be $n$ such mutually orthogonal symmetric idempotent $n \times n$ matrices (with equal main diagonal entries) is closely related to the existence of $n \times n$ Hadamard matrices, as the last example in my answer illustrates. $\endgroup$ – Geoff Robinson Jul 26 '18 at 8:54

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.