6
$\begingroup$

Given a quadratic form (or a symmetric $n \times n$ matrix $A$), an isotropic subspace is a subspace $U$ such that $$U^t A U=0,$$

If I am not mistaken, when the matrix is over reals, the maximum dimension of an isotropic subspace is given by the Witt index, that is, the minimum of $n_{\ge 0}(A)$ and $n_{\le 0}(A)$, the number of non-negative and non-positive eigenvalues of $A$ respectively.

My first question is, when the matrix is over a finite field $GF(p^k)$, is there a simple expression or a relatively easy way to compute the Witt index?

The second question is: to compute the rank of a symmetric matrix over a finite field, is there a way other than the Gaussian elimination? For example, symmetric matrices over reals are diagonalizable so one can check how many non-zero real roots its characteristic polynomial has. But that is normally false for finite fields.

Improve this question
$\endgroup$
3
  • $\begingroup$ Re the final paragraph: in general a method such as Gaussian elimination is still faster than computing the characteristic polynomial; the fact that symmetric matrices are diagonalizable is important for many reason, but actual computation of the rank isn't usually one of these reasons. $\endgroup$
    – Noam D. Elkies
    Commented Feb 2, 2019 at 0:08
  • 2
    $\begingroup$ Quadratic forms are only equivalent to symmetric matrices in odd characteristic. $\endgroup$
    – Chris Godsil
    Commented Feb 2, 2019 at 0:49
  • $\begingroup$ See section 2 of web.archive.org/web/20051016064639/http://www.math.unicaen.fr/… $\endgroup$
    – literature-searcher
    Commented Feb 2, 2019 at 3:21

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .