Suppose we have an $n \times n$ matrix and we know there are exactly $N$ zero entries. The determinant is the sum of $n!$ terms, each formed by selecting exactly one entry from each row and column and multiplying them. If a given term comes from selecting one or more zeroes then that term becomes zero. This raises the question of how many of the terms can be nonzero.

Intuitively I imagine the maximum numbers are achieved if we take a matrix of all $1$s and start putting in zeroes, starting with entries $(1,2), (1,3), \ldots, (1,n)$ so the only nonzero entry in that row is at $(1,1)$, and then moving to the next row and putting zeroes in $(2,1), (2,3), \ldots, (2,n)$ so the only nonzero entry in that row is at $(2,2)$, and so on until we're filling in zeroes on the last row at $(n,1), (n,2), \ldots, (n,n-1)$. At this stage we have the identity matrix and there is exactly one nonzero term.

I imagine this has already been proved somewhere but cannot find a good reference. Could someone please provide a proof, or better yet, a book where this is proved as part of some wider theory? Ideally by something more elegant than induction?


1 Answer 1


Yes, the maximum number of zeros that can occur without making the determinant zero is $N=n(n-1)$. If there are more than $n(n-1)$ zeros then there are fewer than $n$ non-zero entries and so there is at least one column that contains only zero entries, and so the determinant is zero.

A related question is how to maximize the number of non-zero terms in the determinant when $N=n$ i.e. when the $n \times n$ matrix contains exactly $n$ zero terms.

The value of each term in the determinant sum is the product of the entries along the main diagonal in one of the $n!$ matrices that result from a permutation of the columns (or rows) of the original matrix. This ignores a factor of $\pm 1$, but for the purposes of counting non-zero terms this is irrelevant.

If we have one zero in each row and in each column (e.g. the $n$ zeros lie along a diagonal) then a zero term will occur if the permutation of columns puts any column in the position where its zero is on the main diagonal. On the other hand, a non-zero term will occur if each column is in one of the other $n-1$ positions where its zero is off the main diagonal.

Of the $n!$ permutations of columns there are $!n$ (sub-factorial $n$) in which no zero lies on the main diagonal where

$!n = n! \sum_{i=0}^{i=n} \frac{(-1)^i}{i!}$

because this is the same as the counting the derangements of $n$.

I believe this maximises the number of non-zero terms and minimizes the number of zero terms. If two zeros lie on the same row or column then I think the number of zero terms increases.


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