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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

$M={\begin{bmatrix}-1&-1&0&1&-1&0\\1&0&-1&-1&0&1\\\end{bmatrix}},N={\begin{bmatrix}1&1&0&0&0&0\end{bmatrix}}^{\intercal }$

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

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

\textstyle {\binom {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)[reply]

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)[reply]