# How to find the nilpotent submatrices of a symmetric, real matrix?

Given a symmetric, real $n \times n$-matrix $M$, is there a way to find all $m \times m$-submatrices ($1 < m < n$) that are nilpotent?

By the Cauchy interlacing theorem, I know that $M$ must have both negative and positive eigenvalues, which we will take for granted.

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Given your comment about interlacing, it seems your submatrices are principal submatrices, hence symmetric. In which case your question is at the level of a homework exercise, and is out of place on this site. – Chris Godsil Mar 4 '13 at 12:57
Sorry, I wasn't aware of a lower bound on the hardness of the problem. Shall I close the question? – Turion Mar 4 '13 at 14:09

If you consider arbitrary submatrices, the question seems to be quite vague. For finding in the sense of counting nilpotent symmetric matrices, it seems to be more interesting to consider other fields than $\mathbb{R}$. Indeed, let the field be $\mathbb{F}_q$, and let $a(n,k,q)$ be the number of nilpotent matrices of rank $k$ over $\mathbb{F}_q$. Then $$a(n,k,q)=\frac{\Phi_n(q)\Phi_{n-1}(q)}{\\Phi_{n-k}(q)\Phi_{n-k-1}(q)\Phi_k(q)}q^{\binom{k}{2}}$$ for $k \le n-1$, where $\Phi_r(q)=\prod_{1\le i\le r}(q^i-1)$. This is due to G. Lusztig. Then count the nilpotent, symmetric ones of rank $k$, see http://www.win.tue.nl/~aeb/math/symnilp.html .