Testwiki:Reference desk/Archives/Mathematics/2021 October 18

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October 18

Counting permutations in the nullspace of a matrix

Given a matrix $M$ (size $m\times n$) and a matrix $N$ (size $n\times 1$, with entries in (0, 1)), is there a linear algebra type way to count the number of distinct permutations of $N$ such that $M*N_{permuted}=0$? For example, with

$M=\left[\begin{array}{cccccc}-1& -1& 0& 1& -1& 0\\ 1& 0& -1& -1& 0& 1\end{array}\right],N={\left[\begin{array}{cccccc}1& 1& 0& 0& 0& 0\end{array}\right]}^{⊺}$

then the answer would be 2 with ${\left[\begin{array}{cccccc}1& 0& 0& 1& 0& 0\end{array}\right]}^{⊺}$ and ${\left[\begin{array}{cccccc}0& 0& 1& 0& 0& 1\end{array}\right]}^{⊺}$. 174.73.241.150 (talk) 01:35, 18 October 2021 (UTC)

I don't see how to establish this in any formal or semi-formal way, but my mathematical instinct makes me expect that the operations of linear algebra will not help to count these permutations. If $M$ is a null matrix, and $w$ is the weight of $N$, the number of permutations for which the product is the null vector equals $(\genfrac{}{}{0ex}{}{n}{w}).$ I do not see a plausible way how that would be, for example, the permanent of some matrix that is a function of $M$ (which is null) and $N.$  --Lambiam 05:58, 18 October 2021 (UTC)

Yes, the number of permutations of N is finite so this seems to fall under the domain of discrete programming rather than linear algebra. I strongly suspect the problem can be stated in way that's NP-hard. For example if N is a 0-1 matrix then you essentially have a 0-1 programming problem; it seems unlikely that the special case we're dealing with here would make the problem significantly easier. --RDBury (talk) 11:07, 18 October 2021 (UTC)