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Dear math stack exchange community,

I was told that in the paper

http://www.tandfonline.com/doi/abs/10.1080/03081088708817770

there was a formula for the permanent of the sum of two matrices $X$ and $Y$ via permanents of certain submatrices of $X$ and $Y$.

Unfortunately, I don't have access to this paper (and thus, to this formula) at the moment.

I would be very grateful. if somebody could give the formula here.

kjetil b halvorsen
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Bernhard Boehmler
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1 Answers1

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For all $k\leq n$, let $Q_{k,n}$ the set of all finite sequences of positive integer $\alpha=\{\alpha_{i}\}_{i=1}^{i=k}$ such that $$ 1\leq \alpha_1 < \alpha_2< \ldots <\alpha_{k-1}< \alpha_k\leq n. $$
If $A$ happens to be a $n\times n$ matrix and if $\alpha,\beta\in Q_{k,n}$ then $A(\alpha|\beta)$ denote a $(n-k)\times(n-k)$ submatrix submatrix of $A$ obtained by deleting the rows $\alpha$ and the columns $\beta$; Let $A[\alpha|\beta]$ the complement of $A(\alpha|\beta)$ in $A$.

THEOREM 1 (Expansion of the permanent of the sum of two matrices) Let $A$ and $B$ be arbitrary $m\times m$ matrices. Then $$ \mathrm{per}(A+B)=\mathrm{per}(B)+\sum_{k=1}^{m-1}\sum_{\alpha,\beta\in Q_{k,m}}\mathrm{per}A[\alpha|\beta]\cdot\mathrm{per}A(\alpha|\beta)+\mathrm{per}(A) $$

See also in wikipedia verbete on the permanents and its references.

Elias Costa
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  • Is there a typo and it should be per$B(\alpha | \beta)$ ? – Bernhard Boehmler Jan 22 '18 at 22:26
  • @BernhardBoehmler, No, there is no such type in the article. The notation of the article is just the one that was posted in the answer. – Elias Costa Jan 23 '18 at 11:22
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    So it is really $ \mathrm{per}(A+B)=\mathrm{per}(B)+\sum_{k=1}^{m-1}\sum_{\alpha,\beta\in Q_{k,m}}\mathrm{per}A[\alpha|\beta]\cdot\mathrm{per}A(\alpha|\beta)+\mathrm{per}(A) $ and not $ \mathrm{per}(A+B)=\mathrm{per}(B)+\sum_{k=1}^{m-1}\sum_{\alpha,\beta\in Q_{k,m}}\mathrm{per}A[\alpha|\beta]\cdot\mathrm{per}B(\alpha|\beta)+\mathrm{per}(A) $ ? – Bernhard Boehmler Jan 24 '18 at 12:15
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    There is in fact a typo in the above answer. It should be perB(α|β) rather than perA(α|β). – jkuck Jun 18 '18 at 19:51
  • @EliasCosta Is it possible for you to say what this complement means here? Maybe it is too much to ask, but working on this result in the 2 by 2 case would be nice. I can't seem to compute it – Chunna Dec 04 '23 at 18:08
  • @Chunna For positive integers $n$ and $k$ with $(1 \leq k \leq n), Q_{k, n}$ denotes the set $\left{\left(i_1, \ldots, i_k\right) / 1 \leq i_1<\ldots<i_k \leq n\right}$. For $\alpha, \beta \in Q_{k, n}$, let $A(\alpha / \beta)$ be the submatrix of $A$ obtained by deleting the rows indexed by $\alpha$ and columns indexed by $\beta$ and $A[\alpha / \beta]$ be the submatrix of $A$ with rows and columns indexed by $\alpha$ and $\beta$ respectively. – Elias Costa Dec 19 '23 at 21:50