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Let $A$ and $B$ be two matrices of order $n$ over a finite subset of integers $S$ such that $A$ and $B$ are positive-definite, nonsingular and symmetric.

I am interested in proprieties about $A$, $B$ that would allow me to conclude that $det(A) \not = det(B),$ without actually computing the determinant.

Motivation: I would like to bound (from bellow) the number of such matrices (having some additional structure) such that they have mutually different determinants.

I assume this problem is hard in general, but any pointers to relevant literature would be appreciated.

Thanks!

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  • $\begingroup$ det A = det B is equivalent to the statement that log(A/B) is a traceless matrix; unlikely that this can be determined without an actual computation $\endgroup$
    – Carlo Beenakker
    Sep 23, 2011 at 13:21
  • $\begingroup$ Your counting problem reminds me a bit of the determinant spectrum problem indiana.edu/~maxdet/spectrum.html for which a little is known for $\pm1$ matrices. I think you want more even information though. $\endgroup$
    – j.c.
    Sep 23, 2011 at 17:34
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    $\begingroup$ I don't really understand the motivation: presumably the number of matrices with pairwise different determinants equals the number of possible determinants?! $\endgroup$
    – Igor Rivin
    Sep 23, 2011 at 19:57

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I am not sure about the relevance of positive definiteness, and symmetry, other than tightening some bounds, but to compute the determinant of an integer matrix the best way I know is to compute it modulo enough primes where the product of the primes is bigger than the bound on the determinant (for general matrices it's the Hadamard bound, in your case you can do a little better), and then reconstitute the whole determinant by chinese remaindering -- that actually takes most of the time in general. However, if you just want to check that the determinants are the same, the last step is not necessary (as long as the modular results are all the same), so you win.

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