Let $X$ and $Y$ be $n\times n$ real matrices and $S$ be a diagonal matrix with $\pm 1$ on the diagonal. If $$ \det(SX + Y) = 0 $$ for all choices of $S$, then $Y$ is singular. One way to see this is to pick $S$ uniformly at random, and then use multi-linearity of the determinant and the permutation expansion to get that $$ \mathbf{E}_S\left[ \det(SX + Y)\right] = \det(Y) $$ This sort of thing seems like it must be extremely well-known and should be in a book (or an exercise in a book) about random matrices.

My question is which book?

  • $\begingroup$ Is ${\bf{E}}_S$ the expected value? $\endgroup$
    – Karlo
    Commented Aug 23, 2017 at 10:54
  • $\begingroup$ This seems like a really, really difficult way to find out that $Y$ is singular. If you can compute determinants, just go and compute the one of $Y$. Furthermore, $Y$ might still be singular even if the determinant is not always zero. Thus I don't really see a benefit of this result yet, except maybe as an exercise for people learning the determinant. If you can find a practical use for this criterion, it might point you towards the right literature. $\endgroup$
    – Dirk
    Commented Aug 23, 2017 at 10:55
  • $\begingroup$ @DirkLiebhold: Um... the use is when this is the information you have about $Y$. I already have a use for this. That's not the question. $\endgroup$
    – Louis
    Commented Aug 23, 2017 at 11:13
  • 1
    $\begingroup$ @Karlo: Yes, with respect to $S$. $\endgroup$
    – Louis
    Commented Aug 23, 2017 at 11:14
  • $\begingroup$ @Louis, your condition implies that $Y$ AND $X$ are singular. $\endgroup$
    – user91684
    Commented Sep 19, 2017 at 11:11


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