5
$\begingroup$

I'm reading up on linear algebra over semirings, and I'm wondering why people seem to stop short of showing an equivalence between linear transformations between free modules and matrices.

It seems clear to me that over commutative semirings, the usual developments one does for commutative rings go through: a linear transformation $f: R^m\to R^n$ between free left semimodules is such that $f(x+y)=f(x)+f(y)$ and $f(ax)=af(x)$, and then any such $f$ determines a matrix in $R^{n\times m}$ in the usual way; conversely, any such matrix induces a linear transformation.

Now if $R$ is a non-commutative ring, then it seems to be known that the above fails in general, but holds if one works with free bimodules. Similarly, I am quite sure that things go through for free bisemimodules over non-commutative semirings.

Now, two questions:

  1. Did I say something wrong above?

  2. Why can't I seem to find those things anywhere? Golan's Semirings and affine equations seems like an obvious place to look; he does matrix semirings and linear transformations, but doesn't seem to connect them. Droste et.al.'s Handbook has matrices, but no linear transformations; likewise for all the Bloom/Ésik/Kuich work I know.

(I'm a bit of an amateur in this area, so forgive me if this is stupid or I missed some literature.)

$\endgroup$
1
2
$\begingroup$

What I said was mostly right; except that there's no need for the bimodule structure on the $R^n$:

Let $R$ be a semiring and, for $n,m\ge 0$, $R^n$ and $R^m$ the free bimodules over $R$. Say that a function $f:R^m\to R^n$ is left-linear if $f(x+y)=f(x)+f(y)$ and $f(ax)=af(x)$ for all $x,y\in R^m$ and all $a\in R$.

One can easily show that a function $f:R^m\to R^n$ is left-linear iff there is a matrix $M\in R^{m\times n}$ such that $f$ is right-multiplication by $M$, such that $f(x)=xM$ for all $x$. Only the structure of $R^m$ as a right semimodule is needed for this.

Similarly one can define a notion of right linearity, which is then equivalent to left matrix multiplication and uses only the left semimodule structure.

As a last point, if a function is both left and right linear, then its two matrices are transposes of each other and all their elements are in the center of $R$.

$\endgroup$
1
  • $\begingroup$ My previous answer to my own question was wrong, so I have deleted it and replaced it with a better one. $\endgroup$ Mar 30 '20 at 14:07

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.