10
$\begingroup$

Let $M$ be any $n\times n$ matrix.

We define the usual cofactors: $C_{i,j}$ is $(-1)^{i+j}$ times the determinant of the submatrix obtained by deleting row $i$ and column $j$ of $M$.

We can write the determinant of $M$ using Laplace expansion along column $p$ as:

$$\det M = \sum_{q=1}^n M_{q,p} C_{q,p}$$

Now, for any $k, p\in \{1,...,n\}$ consider the sum:

$$W_{n,p}(k) = \sum_{q=1}^n {M_{q,p} C_{q,p}^2 \prod_{j\ne q}{C_{j,k}}}$$

Clearly $W_{n,p}(p)$ can always be factored, with $\det M$ as one factor:

$$W_{n,p}(p) = \sum_{q=1}^n{M_{q,p} C_{q,p}^2 \prod_{j\ne q}{C_{j,p}}} = \sum_{q=1}^n{M_{q,p} C_{q,p} \prod_{j=1}^n{C_{j,p}}} = \left(\det M\right) \prod_{j=1}^n{C_{j,p}}$$

But in all the specific cases I've examined, $\det M$ appears as a factor of $W_{n,p}(k)$ even when $k\ne p$. For example, with $n=3$, $p=1$ and $k=2$:

$$W_{3,1}(2)=\sum_{q=1}^3{M_{q,1} C_{q,1}^2 \prod_{j\ne q}{C_{j,2}}}=\\ \left(\det M\right)\left(M_{2,2} M_{2,3} M_{3,1}^2 M_{1,3}^2+M_{2,1} M_{2,3} M_{3,1} M_{3,2} M_{1,3}^2-M_{2,1} M_{2,2} M_{3,1} M_{3,3} M_{1,3}^2-M_{2,1}^2 M_{3,2} M_{3,3} M_{1,3}^2-M_{1,2} M_{2,3}^2 M_{3,1}^2 M_{1,3}+M_{1,2} M_{2,1}^2 M_{3,3}^2 M_{1,3}+M_{1,1} M_{2,1} M_{2,2} M_{3,3}^2 M_{1,3}-M_{1,1} M_{2,3}^2 M_{3,1} M_{3,2} M_{1,3}-M_{1,1} M_{1,2} M_{2,1} M_{2,3} M_{3,3}^2-M_{1,1}^2 M_{2,2} M_{2,3} M_{3,3}^2+M_{1,1} M_{1,2} M_{2,3}^2 M_{3,1} M_{3,3}+M_{1,1}^2 M_{2,3}^2 M_{3,2} M_{3,3}\right) $$

It might be worth noting that the second factor here cannot be written as a linear combination of products of the cofactors, with purely numeric coefficients. This is in stark contrast to the case $k=p$, when the quotient is simply a product of cofactors.

I am seeking a proof that $\det M$ always divides $W_{n,p}(k)$, and a general formula for the quotient in the non-trivial case $k\ne p$.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ Is your $S_n(k)$ divisible by $\sum_{q=1}^n C_{q,1}$ if you don't require $M$ to have its first column filled with $1$'s? $\endgroup$ Commented Aug 1, 2014 at 13:10
  • 2
    $\begingroup$ If $M$ has no special properties $S_n(k)$ doesn't factor at all for $k\ne 1$ (at least for the low values of $n$ I'm able to check by explicit computations). $\endgroup$ Commented Aug 1, 2014 at 13:48
  • 1
    $\begingroup$ I've now generalised the question to apply to a matrix $M$ with no special properties. $\endgroup$ Commented Aug 2, 2014 at 5:18
  • 1
    $\begingroup$ This suggests an identity of Dodgson (yes a.k.a. Lewis Carroll). I may be misremembering, but the word "condensation" pops into my head when I look at this. $\endgroup$ Commented Aug 2, 2014 at 5:49
  • 2
    $\begingroup$ The quotient given for $p=1$, $k=2$ can be written as $\sum M_{h,p}M_{i,3}M_{j,3}(M_{h,k}C_{h,k}-M_{h,p}C_{h,p})$ where the sum is over the three cyclic permutations $(h,i,j)=(1,2,3)$ or $(2,3,1)$ or $(3,1,2)$. It probably won't help for $n>3$ unless you can generalize the $M_{i,3}M_{j,3}$ terms, since they come from "the column(s) not including column $p$ or $k$". I'm sure you already know your quotient in the $n=2$ (non-trivial) case is the permanent. $\endgroup$ Commented Aug 2, 2014 at 15:02

1 Answer 1

Reset to default
2
$\begingroup$

This is a proof that $\det M$ divides $W_{n,p}(k)$. I can assume that the entries of $M$ are variables of polynomial ring $\mathbb{C}[M_{11},\ldots,M_{nn}]$; then, it suffices to prove that points from the (irreducible) variety $\det M=0$ satisfy $W_{n,p}(k)=0$.

Indeed, let $M'$ be a generic point on this variety; denote by $v^i$ the $i$th column vector of $M'$. Since $M'$ is generic, there is $l\notin\{p,k\}$ such that $v^l=\sum_{t\neq l} \lambda_t v^t$ with all $\lambda_t$ nonzero. Then, the ratio of cofactors $C_{qp}/C_{qk}$ equals $\alpha=\lambda_p/\lambda_k$ and, in particular, it does not depend on $q$. We conclude $$W_{n,p}(k)=\sum\limits_{q=1}^{n}\,M'_{qp}C_{qp}^2\prod\limits_{j\neq q}C_{jk}=\sum\limits_{q=1}^{n}\,M'_{qp}C_{qp}\alpha\prod\limits_{j=1}^nC_{jk}=\alpha\prod\limits_{j=1}^nC_{jk}\cdot\left(\sum\limits_{q=1}^{n}\,M'_{qp}C_{qp}\right)=0.$$

Improve this answer
$\endgroup$
3
  • $\begingroup$ Nice work. Here is a minor variation on this argument, which is purely algebraic (completely avoiding any mention of varieties and generic points): Let $\Delta = \det M$. For any $u < v$ and $a < b$, we have $C_{ua} C_{vb} \equiv C_{ub} C_{vd} \mod \Delta$ (indeed, the Desnanot-Jacobi identity yields that $C_{ua} C_{vb} - C_{ub} C_{vd} = \pm \Delta m$, where $m$ is a certain $\left(n-2\right)\times\left(n-2\right)$-minor of $M$). We shall call this the 2-by-2 congruence. Now we can easily ... $\endgroup$ Commented Aug 18, 2019 at 13:31
  • $\begingroup$ ... see that $C_{qp} \prod\limits_{j\neq q} C_{jk} \equiv C_{1p} \prod\limits_{j\neq 1} C_{jk} \mod \Delta$ for any $n$, $p$, $k$ and $q$ (indeed, for $q = 1$ this is obvious, whereas for $q > 1$ it follows from the 2-by-2 congruence $C_{qp} C_{1k} \equiv C_{1p} C_{qk} \mod \Delta$). Substituting this congruence into the definition of $W_{n,p}\left(k\right)$, we ... $\endgroup$ Commented Aug 18, 2019 at 13:34
  • $\begingroup$ ... find $W_{n,p}\left(k\right) \equiv \sum\limits_{q=1}^n M_{qp} C_{qp} C_{1p} \prod\limits_{j\neq 1} C_{jk} = \underbrace{\left(\sum\limits_{q=1}^n M_{qp} C_{qp}\right)}_{= \det M = \Delta \equiv 0 \mod \Delta} C_{1p} \prod\limits_{j\neq 1} C_{jk} \equiv 0 \mod \Delta$, qed. $\endgroup$ Commented Aug 18, 2019 at 13:34

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.