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

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


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 $m_{n,j} = \sum_{k=1}^{n-1}a_im_{k,j}$ for some $a_k's$; furthermore since $M$ is integer-valued, upon scaling one can assume the $a_k$'s are integers.

Now $M_n = \begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots\\m_{1,n-1}&m_{2,n-1}&\ldots&m_{n-1,n-1}\end{pmatrix}$

Lets compare $M_{n-1}$ to $M_n$:

$M_{n-1} = \begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&m_{1,n}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&m_{2,n}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&m_{n,n-2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n-2}&m_{n,n}\end{pmatrix} =$

$\small\begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&a_1m_{1,1}+a_2m_{1,2}+\ldots a_{n-2}m_{1,n-2}+a_{n-1}m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&a_1m_{1,2}+a_2m_{2,2}+\ldots a_{n-2}m_{2,n-2}+a_{n-1}m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&a_1m_{1,n-2}+a_2m_{2,n-2}+\ldots a_{n-2}m_{n-2,n-2}+a_{n-1}m_{n-1,n-2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n-2}&a_1m_{1,n}+a_2m_{2,n}+\ldots a_{n-2}m_{n-2,n}+a_{n-1}m_{n-1,n}\end{pmatrix}$

Now since a determinant is unchanged by subtracting a multiple of one column from another, for each $i$ from 1 to $n-2$ subtract $a_i$ times the $i^{th}$ column from the last column, this leaves the following matrix with the same determinant as $M_{n-1}$:

$\begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&a_{n-1}m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&a_{n-1}m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&a_{n-1}m_{n-1,n-2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n-2}&a_{n-1}m_{n-1,n}\end{pmatrix}$

Now do the same thing along the rows of this matrix, giving the following matrix with the same determinant as $M_{n-1}$

$\begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&a_{n-1}m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&a_{n-1}m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&a_{n-1}m_{n-1,n-2}\\a_{n-1}m_{1,n-1}&a_{n-1}m_{2,n-1}&\ldots&a_{n-1}m_{n-2,n-1}&a_{n-1}^2m_{n-1,n-1}\end{pmatrix}$.

Note that this matrix is obtained from $M_n$ by multiplying the last row by $a_{n-1}$ and then multiplying the last column by $a_{n-1}$, hence one has $D_{n-1} = a_{n-1}^2D_n$. 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.

  • $\begingroup$ Do we even need that scaling? $\endgroup$ – Thierry Zell May 1 '11 at 23:25
  • $\begingroup$ @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

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