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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.

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    $\begingroup$ Claim is incorrect: take $a_{11}=a_{22}=0$ and $a_{12}=-a_{21}=\frac 12$. Det is still $\frac 14$ tough. $\endgroup$ – Martigan May 22 '15 at 12:01
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    $\begingroup$ look here: en.wikipedia.org/wiki/Hadamard%27s_inequality $\endgroup$ – MotylaNogaTomkaMazura May 22 '15 at 12:03
  • $\begingroup$ In fact you can study diagonal matrix, for either $det(A)=0$ or $A$ can be made diagonal with unitary matrix (of $det=1$)... $\endgroup$ – Martigan May 22 '15 at 12:15
  • $\begingroup$ @martigan can you give an example for a $2\times 2$-matrix with the desired property and a determinant larger than $\frac{1}{4}$ ? $\endgroup$ – Peter May 22 '15 at 12:15
  • $\begingroup$ No, I was just saying that you can have non-diagonal matrices with same determinant as diagonal one. $\endgroup$ – Martigan May 22 '15 at 12:18
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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.

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