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Notation: by the $d$'th diagonal of an $n \times n$ matrix $A$ I will denote the diagonal parallel to the main diagonal that starts in row 1, column $d$. I will extend this definition in the obvious way to non-positive values of $d$, e.g. the $1$'st diagonal is the main diagonal and the $0$'th diagonal is the one that runs from $A_{2, 1}$ to $A_{n, n-1}$.

Now consider the following $n \times n$ matrix $A$. Most entries are zero. Non-zero entries appear only on two diagonals: on the $0$th diagonal, all whose entries equal $-1$, and on the $d$'th diagonal for some $d \in \{1, \ldots, n\}$. The entries on the $d$'th diagonal I will denote $a_0, \ldots, a_{n-d}$ (read from top left to bottom right).

If you compute $\det A$ for a few examples you will notice something curious:

  • If $d \not| n$ then $\det A = 0$
  • If $d|n$ then $\det A = \prod_{d|k} a_k$

So divisibility by $d$ shows up twice. (Even if $d$ was never meant to stand for 'divisor' (but rather for 'diagonal').)

Examples: $$\begin{vmatrix} a_0 & 0 & 0 & 0 & 0 & 0 \\ -1 & a_1 & 0 & 0 & 0 & 0 \\ 0 & -1 & a_2 & 0 & 0 & 0 \\ 0 & 0 & -1 & a_3 & 0 & 0 \\ 0 & 0 & 0 & -1 & a_4 & 0 \\ 0 & 0 & 0 & 0 & -1 & a_5 \\ \end{vmatrix} = a_0a_1a_2a_3a_4a_5, \begin{vmatrix} 0 & a_0 & 0 & 0 & 0 & 0 \\ -1 & 0 & a_1 & 0 & 0 & 0 \\ 0 & -1 & 0 & a_2 & 0 & 0 \\ 0 & 0 & -1 & 0 & a_3 & 0 \\ 0 & 0 & 0 & -1 & 0 & a_4 \\ 0 & 0 & 0 & 0 & -1 & 0 \\ \end{vmatrix} = a_0a_2a_4,$$ $$\begin{vmatrix} 0 & 0 & a_0 & 0 & 0 & 0\\ -1 & 0 & 0 & a_1 & 0 & 0 \\ 0 & -1 & 0 & 0 & a_2 & 0 \\ 0 & 0 & -1 & 0 & 0 & a_3\\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 \\ \end{vmatrix} = a_0a_3, \begin{vmatrix} 0 & 0 & 0 & a_0 & 0 & 0 \\ -1 & 0 & 0 & 0 & a_1 & 0 \\ 0 & -1 & 0 & 0 & 0 & a_2 \\ 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 &0 & 0 & -1 & 0 \\ \end{vmatrix} = 0,$$ $$\begin{vmatrix} 0 & 0 & 0 & 0 & a_0 & 0 \\ -1 & 0 & 0 & 0 & 0 & a_1 \\ 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 &0 & 0 & -1 & 0 \\ \end{vmatrix} = 0, \begin{vmatrix} 0 & 0 & 0 & 0 & 0 & a_0 \\ -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 &0 & 0 & -1 & 0 \\ \end{vmatrix} = a_0.$$

I will sketch a non-informative proof of this fact below, but here I want to ask: is there a conceptual explanation for the appearance of integer division in these determinants, in terms of one of the geometric or combinatorial interpretations of determinant? (Even a proof by partioning the matrix into blocks would be interesting.)

For me the importance of the arithmetic properties of the indices came as a big surprise. After all they are only indices. It 'feels' weird that we can tell if the dimension of a vector space is a prime number by looking at the invertibility of certain geometric transformations of it. But maybe that is because I can only visualize low-dimensional spaces anyway.

Also: are there attractive generalizations?

Uninformative proof (sketch): Let $A'$ consist of the last $n-d$ rows and the last $n-d$ columns of $A$. Then $A'$ is of the same type as $A$ but smaller, so, proceeding by induction, we may assume that the result holds for $A'$. (Here we use that $d|(n-d)$ if and only if $d|n$.)

Next we note that the first $d-1$ columns (if they exist) only contain one $-1$ and otherwise $0$'s while the first row only contains $a_0$ and otherwise $0$s. So applying Laplace expansion to the first $d-1$ columns and the first row reveals that $\det A = a_0 \cdot \det A'$ and the result follows.

UPDATE: As an alternative to my proof above and Robert Israel's proof below one could also notice that this is a special case of the (less mysterious but more beautiful) observation in my newer question: Is this determinant identity known?. Although I would still be interested in an more geometric approach to these phenomena I am accepting Robert Israel's answer as to encourage future readers to extend any answer to this question they might have into an answer to the more general question Is this determinant identity known?.

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Vincent
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1 Answers1

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If $d=1$ it's trivial (the matrix is lower triangular, so the determinant is the product of the diagonal elements). So let's suppose $d > 1$. Think of what permutations $\sigma$ can have all the matrix elements $A_{i,\sigma(i)}$ nonzero (and therefore show up in the expansion of the determinant). Each $\sigma(i)$ is either $i-1$ (to get a $-1$) or $i-1+d$ (to get $a_{i-1}$). Thus in any case $\sigma(i) \equiv i - 1 \mod d$. Thus $$\sum_{i=1}^n (\sigma(i) - i) \equiv -n \mod d$$ But since $\sigma$ maps $\{1,\ldots,n\}$ one-to-one onto itself, the left side is $0$, i.e. $n$ must be divisible by $d$ in order to have any such permutations.

Now supposing $d \mid n$, say $n = d k$, what permutations do we get? Split up the $n$ rows into their classes mod $d$. First look at rows $1, 1+d, \ldots, 1+(k-1)d$. In row $1$ there is only the one nonzero entry ($a_0$ in column $d$), so we must have $\sigma(1) = d$. Then we can't have $\sigma(1+d) = d$ so we must have $\sigma(1+d) = 2d$, etc. Thus $\sigma(1+jd) = (j+1)d$, $j=0 \ldots k-1$. Thus we pick up the matrix elements $a_{jd}$, $j=0 \ldots k-1$.

Each of the other $d-1$ classes of rows includes one of the bottom $d-1$ rows, say row $n-i = dk-i$, $0 \le i \le d-2$, which has only one nonzero entry ($-1$ in column $n-i-1$), so $\sigma(dk-i) = n-i-1$. Then we can't have $\sigma(d(k-1)-i) = dk-i-1$, so $\sigma(d(k-1)-i) = d(k-1)-i-1$, etc. Thus we pick up only matrix elements of $-1$ for all these classes.

The rest is simply a matter of counting the number of $-1$'s and the sign of the permutation to make sure we have $\prod_j a_{jd}$ rather than $-\prod_j a_{jd}$.

Robert Israel
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