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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.

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1 Answer 1

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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.

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  • $\begingroup$ Is there a typo and it should be per$B(\alpha | \beta)$ ? $\endgroup$ Jan 22, 2018 at 22:26
  • $\begingroup$ @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. $\endgroup$ Jan 23, 2018 at 11:22
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    $\begingroup$ 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) $ ? $\endgroup$ Jan 24, 2018 at 12:15
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    $\begingroup$ There is in fact a typo in the above answer. It should be perB(α|β) rather than perA(α|β). $\endgroup$
    – jkuck
    Jun 18, 2018 at 19:51

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