Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

up vote 1 down vote accepted

Every real symmetric matrix is similar to a diagonal matrix by the spectral theorem. The only nilpotent matrix which is similar to a diagonal matrix is the zero matrix. Hence if you mean principal submatrices then the answer is trivial, as Chris Godsil has pointed out.

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 .

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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