I recently read an article on generalized inverses and Green's relations (by X.Mary). The framework is semigroups, but obviously it has a lot of application within matrix theory. In the article mention is made of the "trace product" - it is defined as:

if $a,b \in S$, then $ab$ is a trace product if $ab \in \mathcal{R}_a \cap \mathcal{L}_b$.

Here $\mathcal{R}_a$ and $\mathcal{L_b}$ are the relevant classes from Green's relations. So I traced (no pun intended) this product back to two articles, one by Rees and one by Miller and Clifford (Regular D-classes in semigroups). So here the explanation is given of a semigroup $D^0$, constructed from a $D$ class and called the trace of $D$. The definition of the binary operation of this semigroup then includes this trace product. So this semigroup is used to prove that $D$ is partially isomorphic to a regular matrix semigroup.

I am not so familiar with the abstract algebra of semigroups - does anyone know if this trace product mentioned here bears any relation to the trace of a matrix? or perhaps trace of a matrix product? as it is usually referred to in the context of matrix theory. Perhaps in a semigroup of matrices (say square matrices of the same size) you get some classes as per Green's definitions that can be characterized in terms of the trace of a matrix, and it is related to this trace product (it's a long shot I know...)?

NB: You can read it online if you register - and the section on the trace construction is on page 276/277.


1 Answer 1


I believe that there is no connection with the trace in matrix theory. I think they are viewing $D^0$ as being a trace (in the sense of remnant) of the semigroup $S$.

For the semigroup of $n\times n$ matrices over a field the $D$-classes consist of matrices with a common rank $r$. So $AB$ is a trace product iff the ranks of $A,B,AB$ are the same. I don't see any relation with trace except in the special case $A,B,AB$ all are idempotents, in which case the product is a trace product iff they all three have the same trace.

  • $\begingroup$ Thank you, yes, that is what I suspected as well... Ok so am I correct in then saying that the D-classes correspond to matrix equivalence (for matrix $A$ there exist invertible $P$ and $Q$ so that $PAQ=B$). Does the $R$ and $L$ classes correspond to row equivalence and column equivalence? $\endgroup$ Commented Apr 26, 2014 at 20:00
  • $\begingroup$ Yes. You are correct. $\endgroup$ Commented Apr 26, 2014 at 21:04

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