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With $\Bbb K$ a commutative rings there a way to characterize $A,B\in\Bbb K^{n\times n}$ with $$Per(AB)=Per(A)Per(B)?$$

John provides a reasonable concept coverage multiplicative property.

How about if I seek $Per(A+B)=Per(A)+Per(B)$ instead of $Per(AB)=Per(A)Per(B)$?

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  • $\begingroup$ please don't delete a question and then repost it as new --- mathoverflow.net/questions/252987/… $\endgroup$
    – Carlo Beenakker
    Commented Oct 27, 2016 at 19:16
  • $\begingroup$ Actually that problem was two different problems. I am looking for only multiplicative property. I can reincarnate the other problem. It was downvoted once though. $\endgroup$
    – user94040
    Commented Oct 27, 2016 at 19:22

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Searching the literature brings up a concept known as permanent groups. A permanent group is a group of nonsingular matrices on which the permanent is multiplicative. In was conjectured by M. Marcus in Permanents that the set nonsingular matrices of the form $PD$, where $P$ is a permutation matrix and $D$ is a diagonal matrix, is a maximal permanent group (see Conjecture 12). This conjecture is proven over the complex numbers by Beasley in Maximal groups on which the permanent is multiplicative. Further work has been done by Beasley and Cummings in the papers titled Permanent Groups, Permanent Groups II, and Permanent Semigroups, etc.

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  • $\begingroup$ how about if I seek $Per(A+B)=Per(A)+Per(B)$? $\endgroup$
    – user94040
    Commented Oct 27, 2016 at 21:50
  • $\begingroup$ Beasley's result does not solve the whole multiplicative problem. He considers only groups containing the permutation matrices. $\endgroup$
    – Denis Serre
    Commented Mar 10, 2017 at 9:16
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Trivially, the additive group $G$ of matrices with vanishing last row satisfies $Per(A+B)=Per(A)+Per(B)=0$ for all $A,B\in G$.

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  • $\begingroup$ Yes, by vanishing I mean that each entry in that row is zero. Yes, it is quite trivial! $\endgroup$
    – Zach Teitler
    Commented Mar 10, 2017 at 3:14

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