# Is there an easy way to compute the maximum isotropic subspace over finite fields?

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.

• 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. – Noam D. Elkies Feb 2 '19 at 0:08
• Quadratic forms are only equivalent to symmetric matrices in odd characteristic. – Chris Godsil Feb 2 '19 at 0:49
• See section 2 of web.archive.org/web/20051016064639/http://www.math.unicaen.fr/… – literature-searcher Feb 2 '19 at 3:21