An example where $\text{perrank} (A)= \frac{\text{rank} (A)}{2}$

Question

It is given to me, without a proof that when $A= \begin{bmatrix} I_n & I_n \\ I_n & -I_n \end{bmatrix} $, $$\text{perrank} (A)= \frac{\text{rank} (A)}{2}$$ where $I_n$ is the identity matrix of degree $n$. In order to prove this, it is easy to see that $\text{rank}(A)=2n$ since the determinant of $A$ is $(-2)^n$, hence non-zero. So now I need to show that $\text{perrank(A)}=n$. Since $\text{per}(I_n)=1\neq 0$ it is only left to show that for any $k\times k$ submatrix $B$ of $A$ where $k>n$, $\text{per}(B)=0$. This is where I am stuck.

Answer 1

I came up with a solution since no one else did:

Lemma. If $M\in M_m(\mathbb{R})$ has a $s\times t$ zero submatrix such that $s+t=m+1$ , then $\text{per}(M)=0$.
proof: Define $N=[n_{ij}]$ like this: $n_{ij}=1 \Leftrightarrow m_{ij}\neq 0$ and $n_{ij}=0 \Leftrightarrow m_{ij}= 0$. Since the zeros of the two matrices are the same, $N$ also has such $s\times t$ zero submatrix. so the ones of $N$ are all covered by the other $n-s$ rows and $n-t$ columns. Thus the minimum covering lines are less than or equal to $m-s+m-t=2m-(s+t)=m-1$. By Konig's theorem, the minimum covering is equal to the maximum number of $1$s that are not in the same row and same column, thus this maximum is $m-1$. So if we take any $n_{1\pi(1)},\dots ,n_{m\pi(m)}$ where $\pi\in S_m$, at least one of them is zero, and since the zeros of $M,N$ are the same, this also happens for any $m_{1\pi(1)},\dots ,m_{m\pi(m)}$, meaning that $\text{per}(A)=0$.

Going back to the original problem now, let's look at any $(n+k)\times(n+k)$ submatrices as a "choice of $n+k$ columns and rows" from A. Since we are choosing $n+k>n$ columns, there exists $1\leq i\leq n$ such that both the $i$th and $n+i$th column are chosen. they look like this: $$ \begin{bmatrix} 0 & 0 \\ \vdots & \vdots \\ 0 & 0 \\ 1 & 1 \\ 0 & 0 \\ \vdots & \vdots \\ 0 & 0 \\ 1 & -1 \\ 0 & 0 \\ \vdots & \vdots \\ 0 & 0 \end{bmatrix} $$ Now we know that $n+k$ rows are also chosen, if the row which contains $\begin{bmatrix} 1 & -1 \end{bmatrix}$ is not chosen then at least $n+k-1$ rows containing $\begin{bmatrix} 0 & 0 \end{bmatrix}$ are chosen, giving us a $(n+k-1)\times 2$ zero submatrix which using the lemma, would mean that the permanent of the $(n+k)\times(n+k)$ submatrix is zero. On the other hand if $\begin{bmatrix} 1 & -1 \end{bmatrix}$ is chosen, when we are computing the permanent of this submatrix, there are two non-zero options in the row containing $1$ and $-1$, and these two cancel each other, giving us a zero permanent again, so we are done!!