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Given an $n \times n$ matrix $(c_{ij})$ with entries in $\mathbb{R}$ and such that $c_{ij} \leq B$ for some $B > 0$, then it is obvious that the determinant, which we call $\Delta$, is at most $B^n$. However it is frequently the case that this bound is not reached.

My question is, can there be an substantial improvements over the estimate $|\Delta| \leq B^n$? It seems that the upper bound is not reachable. Is there an example of a matrix where the upper bound is close to being obtained?

Thank you for any insights.

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Scale down $B$ to 1 and use en.wikipedia.org/wiki/Hadamard%27s_maximal_determinant_problem –  Federico Poloni Apr 18 '12 at 21:30
    
As pointed out by @Joe Silverman, in general the Hadamard bound is your friend. But if your matrices have special structure, that bound is not going to be sharp. –  Igor Rivin Apr 18 '12 at 21:42
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1 Answer

Presumably you mean to say that $|c_{ij}|\le B$, since if you just say $c_{ij}<B$, then the $c_{ij}$ could be very negative.

But even then your bound $B^n$ is false. Here's a counterexample: $B=\begin{pmatrix} 2&1\\-1&2\\\end{pmatrix}$, then entries are bounded by $2$, but $\Delta=5>2^2$. The most trivial estimate is that $\Delta$ is a sum of $n!$ terms, each of which is at most $B^n$, so the trivial bound is $n!B^n$. One can, of course, do much better. Hadamard's inequality gives $|\Delta|\le n^{n/2}B^n$.

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Furthermore, for many n which are multiples of 4, the bound is sharp. I don't know how far away from Omega((BBn)^(n/2)) reality is in the maximal case using only real coefficients, but that is the belief backed up by a lot of evidence. Gerhard "Ask Me About Binary Matrices" Paseman, 2012.04.18 –  Gerhard Paseman Apr 18 '12 at 23:17
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