Let $M$ be a symmetric square matrix with integer coefficients and $M_k$ the matrix obtained by deleting the kth line and kth column. If det(M)=0 does it follow that $\det(M_kM_j)$ is a square?

8$\begingroup$ I would like to know what led you to consider this question. Who knows, this may even suggest an angle of attack. $\endgroup$ – Thierry Zell May 1 '11 at 21:31

$\begingroup$ I guess the answer is yes for $3\times 3$ matrices! $\endgroup$ – Andrés E. Caicedo May 1 '11 at 21:35

3$\begingroup$ Well, 2 by 2 anyway. $\endgroup$ – Will Jagy May 1 '11 at 22:10

2$\begingroup$ The Kirchhoff matrix tree theorem uses the fact that det($M_i$)=det($M_j$), so it's true for Laplacian matrices, at least. (en.wikipedia.org/wiki/Kirchhoff%27s_matrix_tree_theorem) $\endgroup$ – Peter Shor May 1 '11 at 22:18
Yes one gets the square of the determinant of the matrix where one deletes row j and column k. That follows from Dodgson's condensation formula for determinants.

14$\begingroup$ (Also known as Lewis Carroll, inventor of cats) $\endgroup$ – Mariano SuárezÁlvarez May 1 '11 at 23:09


2$\begingroup$ My guess is the Cheshire Cat in Alice in Wonderland. $$ $$ en.wikipedia.org/wiki/Cheshire_Cat $\endgroup$ – Will Jagy May 2 '11 at 1:53

Abdelmalek's answer was posted while I was finishing up this, but I decided to post it anyway since it gives insight into why the statement is indeed true.
Let $M = (m_{i,j})\in Mat^{n\times n}(\mathbb{Z})$ and define $D_i := Det(M_i)$
Since det(M) = 0, one has that some row of M is linearly dependent on the others, WLOG say the last row so that $m_{n,j} = \sum_{k=1}^{n1}a_im_{k,j}$ for some $a_k's$; furthermore since $M$ is integervalued, upon scaling one can assume the $a_k$'s are integers.
Now $M_n = \begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n1}\\ \vdots&\vdots&\ddots&\vdots\\m_{1,n1}&m_{2,n1}&\ldots&m_{n1,n1}\end{pmatrix}$
Lets compare $M_{n1}$ to $M_n$:
$M_{n1} = \begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n2}&m_{1,n}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n2}&m_{2,n}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n2}&m_{2,n2}&\ldots&m_{n2,n2}&m_{n,n2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n2}&m_{n,n}\end{pmatrix} =$
$\small\begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n2}&a_1m_{1,1}+a_2m_{1,2}+\ldots a_{n2}m_{1,n2}+a_{n1}m_{1,n1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n2}&a_1m_{1,2}+a_2m_{2,2}+\ldots a_{n2}m_{2,n2}+a_{n1}m_{2,n1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n2}&m_{2,n2}&\ldots&m_{n2,n2}&a_1m_{1,n2}+a_2m_{2,n2}+\ldots a_{n2}m_{n2,n2}+a_{n1}m_{n1,n2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n2}&a_1m_{1,n}+a_2m_{2,n}+\ldots a_{n2}m_{n2,n}+a_{n1}m_{n1,n}\end{pmatrix}$
Now since a determinant is unchanged by subtracting a multiple of one column from another, for each $i$ from 1 to $n2$ subtract $a_i$ times the $i^{th}$ column from the last column, this leaves the following matrix with the same determinant as $M_{n1}$:
$\begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n2}&a_{n1}m_{1,n1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n2}&a_{n1}m_{2,n1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n2}&m_{2,n2}&\ldots&m_{n2,n2}&a_{n1}m_{n1,n2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n2}&a_{n1}m_{n1,n}\end{pmatrix}$
Now do the same thing along the rows of this matrix, giving the following matrix with the same determinant as $M_{n1}$
$\begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n2}&a_{n1}m_{1,n1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n2}&a_{n1}m_{2,n1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n2}&m_{2,n2}&\ldots&m_{n2,n2}&a_{n1}m_{n1,n2}\\a_{n1}m_{1,n1}&a_{n1}m_{2,n1}&\ldots&a_{n1}m_{n2,n1}&a_{n1}^2m_{n1,n1}\end{pmatrix}$.
Note that this matrix is obtained from $M_n$ by multiplying the last row by $a_{n1}$ and then multiplying the last column by $a_{n1}$, hence one has $D_{n1} = a_{n1}^2D_n$. A similar argument holds for the other $D_k$, thus $det(M_jM_k) = D_jD_k = a_{j}^2a_{k}^2D_n^2$ is indeed always a square.


$\begingroup$ @Thierry: I think you are right; the scaling is probably superfluous; I just mentioned it to make absolutely sure the $a_k$'s are integers for the conclusion in the final line. $\endgroup$ – ARupinski May 1 '11 at 23:27