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Let $A$ be an $n \times n$ matrix and assume that the sum of absolute values of all its entries equals to $1$. What is the maximal possible value of $\det(A)$?

My attempt: We know that $|a_{i,j}| \leq 1 \Rightarrow -1 \leq a_{i,j} \leq 1$ for all $1 \leq i,j \leq n$

Claim:Diagonal matrix has the largest determinant.

I don't know whether the claim holds for all matrices or not. If the claim is correct, then I have maximal determinant is $\frac{1}{n^n}$.

If the claim is not correct, can anyone provide me a counterexample?

UPDATE: From the comments below, clearly my claim is not correct (Thanks to Martigan's counterexample). But what if I reword the claim as follows:

Second Claim:The largest determinant of an $n \times n$ matrix can be attained by diagonal matrix.

From the second claim, clearly other type of matrix can also give the same determinant. But what I concern here is just diagonal matrix.

Idonknow
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    Claim is incorrect: take $a_{11}=a_{22}=0$ and $a_{12}=-a_{21}=\frac 12$. Det is still $\frac 14$ tough. – Martigan May 22 '15 at 12:01
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    look here: http://en.wikipedia.org/wiki/Hadamard%27s_inequality –  May 22 '15 at 12:03
  • In fact you can study diagonal matrix, for either $det(A)=0$ or $A$ can be made diagonal with unitary matrix (of $det=1$)... – Martigan May 22 '15 at 12:15
  • @martigan can you give an example for a $2\times 2$-matrix with the desired property and a determinant larger than $\frac{1}{4}$ ? – Peter May 22 '15 at 12:15
  • No, I was just saying that you can have non-diagonal matrices with same determinant as diagonal one. – Martigan May 22 '15 at 12:18
  • Take the one I provide first comment Peter. This is non-diagonal. – Martigan May 22 '15 at 12:21
  • OK, now I see it. – Peter May 22 '15 at 12:23
  • Since you can always make a matrix with the same determinant by permuting columns or rows (and possibly changing signs), clearly the maximum is not unique unless it occurred when all entries have the same absolute value. – David K May 22 '15 at 12:43

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Hint: $ ab+cd \leq |ab| + |cd| \leq (|a| + |c| ) ( |b| + |d| ) $

Applying the above concept, we have

$$ \det M = \sum_\sigma sgn(\sigma) \prod_i M_{i \sigma(i) } \leq \sum_{i=1}^n \prod_{j=1}^n | M_{i j } | \leq \prod_{j=1}^n ( \sum_{i=1}^n |M_{i j} | ) \leq \left( \frac{ \sum_{i,j} |M_{i j } | } { n} \right)^n = \frac{1}{n^n}.$$

Follow the above chain of inequalities to determine the $n!$ scenarios in which the maximum is attained.

Calvin Lin
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