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Note: This question is a direct follow-up to this this post.

I was considering how the permanent of $1_n1_n^{\top}-2I_n$ will be changed if, Instead of $1_n1_n^{\top}-2I_n$ (matrix of ones with $-1$ on the diagonal), we would change the sign of one of the elements of the diagonal to positive. That is, we start from:
$$ A = \begin{bmatrix} -1 & 1 & 1 & \dots & 1 \\ 1 & -1 & 1 & \dots & 1 \\ 1 & 1 & -1 & \dots & 1 \\ \vdots & \vdots &\vdots & \ddots & \vdots \\ 1 & 1 & 1 & \dots & -1 \end{bmatrix} $$

with permanent given by

$$\operatorname{per}(A) = n!\sum_{m=0}^n\frac{(-2)^m}{m!} $$

And change it to $$A'= \begin{bmatrix} 1 & 1 & 1 & \dots & 1 \\ 1 & -1 & 1 & \dots & 1 \\ 1 & 1 & -1 & \dots & 1 \\ \vdots & \vdots &\vdots & \ddots & \vdots \\ 1 & 1 & 1 & \dots & -1 \end{bmatrix} $$ (Changed $a_{11}$ from $-1$ to $1$, but it could have been any other diagonal entry, it does not matter.)
What would be the permanent of $A'$?

I apologize if this question is trivial, but I tried using Laplace expansion (for permanents) and did not get anywhere

Rodrigo de Azevedo
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Noam
  • 381

2 Answers2

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  1. Since the first row of $A'$ equals the first row of $A$ plus $(2,0,\dots,0)$, multilinearity in that row gives $$ \operatorname{per}(A') = \operatorname{per}(A) + \operatorname{per}(B), $$ where $B$ is obtained from $A$ by replacing its first row with $(2,0,\dots,0)$.

  2. In $B$, any nonzero term in the permanent must use the entry $B_{1,1}=2$, so only permutations with $\pi(1)=1$ contribute. Hence $$ \operatorname{per}(B) =2\;\operatorname{per}\bigl(A_{[2:n],[2:n]}\bigr) =2\,(n-1)!\sum_{m=0}^{n-1}\frac{(-2)^m}{m!}. $$

  3. Using $$ \operatorname{per}(A)=n!\sum_{m=0}^n\frac{(-2)^m}{m!}. $$ we get $$ \operatorname{per}(A') = n!\sum_{m=0}^n\frac{(-2)^m}{m!} \;+\; 2\,(n-1)!\sum_{m=0}^{n-1}\frac{(-2)^m}{m!}, $$ which one can equivalently rewrite as $$ \operatorname{per}(A')=(n+2)(n-1)!\sum_{m=0}^{n-1}\frac{(-2)^m}{m!} \;+\;(-2)^n. $$

    If I’ve made any mistakes anywhere, please correct me.

kabenyuk
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1

Let ${\bf A}_n := {\bf 1}_n {\bf 1}_n^{\top} - 2 {\bf I}_n$ and ${\bf A}_n^\prime := {\bf A}_n + 2 \, {\bf e}_1 {\bf e}_1^{\top}$. From the Laplace expansion,

$$ \begin{aligned} \operatorname{perm} \left( \, {\bf A}_n \right) &= (-1) \operatorname{perm} \left( \, {\bf A}_{n-1} \right) + \text{other terms} \\ \operatorname{perm} \left( \, {\bf A}_n^\prime \right) &= (+1) \operatorname{perm} \left( \, {\bf A}_{n-1} \right) + \text{other terms} \end{aligned} $$

And, thus, assuming that $\operatorname{perm} \left( \, {\bf A}_n \right) = n!\sum\limits_{m=0}^n\frac{(-2)^m}{m!} $, we have

$$ \operatorname{perm} \left( \, {\bf A}_n^\prime \right) = \operatorname{perm} \left( \, {\bf A}_n \right) + 2 \, \operatorname{perm} \left( \, {\bf A}_{n-1} \right) = \cdots = \color{blue}{(n+2)(n-1)! \left(\sum_{m=0}^{n-1} \frac{(-2)^m}{m!}\right) + (-2)^n}$$

as obtained by kabenyuk via other means.

Rodrigo de Azevedo
  • 23,223