Maximizing the permanent of a $\pm 1$ matrix

Question

I am trying to figure out when exactly the permanent of an invertible matrix with $\pm1$ entries is maximal.

Assume that it is known that for a general $\pm1$ matrix, the permanent is maximized when all the entries are $1$ (or $-1$, if the order is even).

Related to my previous post, I thought that the permanent of an $\pm1$ matrix (call it $A$) gets larger the "closer" it is the an all $1$'s matrix (in the sense that the number of entries which $A$ differs from an all $1$'s matrix is minimal.)

I am not sure if this is actually the case, and even if it is, how can I prove it?

For example, Let $n=6$, and consider the matrix of all $1$'s:

\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \end{pmatrix}

Therefore, the largest permanent of a $6\times6$ invertible $\pm1$ matrix will be achieved (not uniquely) by

\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 &-1 & 1 & 1 & 1 & 1 \\ 1 & 1 &-1 & 1 & 1 & 1 \\ 1 & 1 & 1 &-1 & 1 & 1 \\ 1 & 1 & 1 & 1 &-1 & 1 \\ 1 & 1 & 1 & 1 & 1 &-1 \end{pmatrix}

which is invertible, and was obtained by changing the least number of entries of the all $1$'s matrix. How can I prove it (assuming this is true)?

Answer 1

For those who are interested, I have found the answer in this paper. A simple application of corollary 8.9 in page 43 gives us the following theorem:

For any invertible $\pm1$ matrix $A \in M_n (\pm1)$, where $n \neq 4$ and $n \geq 2$,

$$\left| \operatorname{per} (A) \right|\leq \left| \operatorname{per} (D_{(n,n-1)})\right|$$

where

$$ D_{(n,n-1)} = \begin{bmatrix} -1 & 1 & 1 & \dots & 1 \\ 1 & -1 & 1 & \dots & 1 \\ 1 & 1 & -1 & \dots & 1 \\ \vdots & \vdots &\vdots & \ddots & \vdots \\ 1 & 1 & 1 & \dots & 1 \end{bmatrix}$$

Matrix of $1$'s with $-1$'s on the diagonal except the last entry which is $+1$. Furthermore, this bound is sharp.