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Hadamard's maximal determinant problem asks for the largest determinant of a matrix with elements equal to 1 or −1. But what would be the case when the elements are equal to $\pm1\text{ or }0$?

I've tried to look for some reference, but all the reference I found were about elements being equal to 0 or -1 at best.

Anyone knows something about the case for elements being equal to $\pm1\text{ or }0$? Thanks in advance.

Monkey
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2 Answers2

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Actually Hadamard's determinant inequality is more general and also applies to your question. Thanks to @hardmath for pointing out the reference here.

The determinant of any $n\times n$ complex matrix $A$ with columns $a_i$ obeys

$$ | \det(A) |\leq \prod_{1\leq i\leq n} \Vert a_i \Vert, $$ where the norm is the Euclidean norm.

It happens that certain $\pm 1$ entry Hadamard matrices achieve this bound. If you introduce a single zero in such a matrix without making it singular, you can compare the resulting determinant to this upperbound.

Since the proof is via the arithmetic-geometric mean inequality, and achieved for the case of equal norm columns, introducing this zero is unlikely to achieve the upperbound.

In your case of entries from $\{0,\pm 1\}$, the squareroot of the Hamming weight of each row would appear in the right hand side of the bound.

kodlu
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    It is true that Hadamard's original result (1893), posed for square matrices with complex entries of magnitude at most one, applies to the subset of matrices asked about here. This survey of the literature explains how that led to interest in determinants of $\pm 1$ entry matrices. – hardmath Aug 22 '22 at 02:22
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The maximum (absolute) determinant of an $n\times n$ (real) matrices with entries in $\{0,\pm 1\}$ is the same as the maximum determinant when their entries are restricted to $\{\pm 1\}$.

This is noted without proof in the OEIS A003432 sequence, "largest determinant of a (real) $\{0,1\}$-matrix of order $n$."

Suppose M = (m(i,j)) is an n X n matrix of real numbers. Let
a(n) = max det M subject to m(i,j) = 0 or 1 [this sequence],
g(n) = max det M subject to m(i,j) = -1 or 1 A003433,
h(n) = max det M subject to m(i,j) = -1, 0 or 1 A003433,
F(n) = max det M subject to 0 <= m(i,j) <= 1 [this sequence],
G(n) = max det M subject to -1 <= m(i,j) <= 1 A003433.

Then a(n) = F(n), g(n) = h(n) = G(n), g(n) = 2^(n-1)*a(n-1). Thus all five problems are equivalent.

Apparently the proof is a simple matter of considering the cofactor expansion of the matrix determinant. An article "The Hadamard Maximum Determinant Problem," by Joel Brenner (American Math Monthly Vol. 79, No. 6 (Jun. - Jul., 1972), pp. 626-630) available online through JSTOR merely says this:

Consider any real $n\times n$ matrix $A = (a_{ij})$ with $|a_{ij}|\le 1$. Expanding the determinant of any such $A$ by minors along successive rows it is apparent that $\det A$ is dominated by the determinant of a $(-1,1)$ matrix; i.e., a matrix all of whose entries are either $-1$ or $1$. Since there are finitely many such matrices the maximum determinant problem has a solution for each $n$.

We could elaborate on this by asking for the $\{0,\pm 1\}$-matrix, among those attaining the maximum determinant, having fewest zero entries. If there were a zero entry, then by the cofactor of that entry being either non-negative or non-positive, we could vary the entry from $0$ to either $1$ or $-1$ respectively to attain the maximum determinant with fewer zero entries.

hardmath
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