Let $R$ be a unitary commutative semi-ring and let $M_n(R)$ be the semi-ring of $n \times n$ matrices with coefficients in $R$. Let's call $I$ the identity matrix in $M_n(R)$ (it exists because $R$ is unitary).

Is it true that for any $A$ and $B$ in $M_n(R)$ such that $AB=I$ we have $BA=I$?


This is proven in Inversion of Matrices over a Commutative Semiring by Reutenauer and Straubing. The proofs aren't especially short though (as you request in the comments). The first two paragraphs give good context for their paper though:

It is a well-known consequence of the elementary theory of vector spaces that if $A$ and $B$ are $n$-by-$n$ matrices over a field (or even a skew field) such that $AB = 1$, then $BA = 1$. This result remains true for matrices over a commutative ring, however, it is not, in general, true for matrices over noncommutative rings.

In this paper we show that if $A$ and $B$ are $n$-by-$n$ matrices over a commutative semiring, then the equation $AB = 1$ implies $BA = 1$. We give two proofs: one algebraic in nature, the other more combinatorial. Both proofs use a generalization of the familiar product law for determinants over a commutative semiring.

It's worth mentioning that "semi-ring" for them requires having a multiplicative identity, so they do prove the result you desire.

  • $\begingroup$ Thank you very much for this answer! I should have mention it in the main post, but my principal motivation was to avoid rings such that I cannot use determinant and see if it is still true, and your answer also give an answer to this motivation : as a generalization of determinant seems to be required in "Both proves". $\endgroup$ – jcdornano Aug 3 '18 at 6:04
  • $\begingroup$ @jcdornano If you don't want to use rings (to the degree that determinants are undefined), then how would you expect matrix multiplication to be defined? $\endgroup$ – Arthur Aug 3 '18 at 6:06
  • $\begingroup$ @Arthur : Matrix multiplication over a semi-ring works the usual way, but determinant cannot be defined for any matrix when the underlying additive monoïde of the ring is not a group $\endgroup$ – jcdornano Aug 3 '18 at 6:11
  • 1
    $\begingroup$ @jcdornano I forgot the minus sign in the determinant definition, you're right. $\endgroup$ – Arthur Aug 3 '18 at 6:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.