# Special formula for the permanent of the sum of two matrices

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.

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 $$\mathrm{per}(A+B)=\mathrm{per}(B)+\sum_{k=1}^{m-1}\sum_{\alpha,\beta\in Q_{k,m}}\mathrm{per}A[\alpha|\beta]\cdot\mathrm{per}A(\alpha|\beta)+\mathrm{per}(A)$$
• Is there a typo and it should be per$B(\alpha | \beta)$ ? Jan 22, 2018 at 22:26
• 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 $\mathrm{per}(A+B)=\mathrm{per}(B)+\sum_{k=1}^{m-1}\sum_{\alpha,\beta\in Q_{k,m}}\mathrm{per}A[\alpha|\beta]\cdot\mathrm{per}B(\alpha|\beta)+\mathrm{per}(A)$ ? Jan 24, 2018 at 12:15