Suppose $a,b$ are two matrices (arbitrary for now), and I have a function defined on a space of matrices, $T(x) = a x b$. This function is a linear and bounded transform on the a finite dimensional vector space of matrices, so can be represented as a matrix. Say $x$ is $m$ by $n$, then you can write $x$ as column vector of $mn$ entries (row major ordering, lets say), and work out the corresponding matrix representation for $T$, call it $M$.


Now suppose all matrices here are $n$ by $n$. So on the one hand, this is an expensive way to represent $T$, as it would take on the order of $n^6$ operations to apply in this way, vs $2 n^3$, from the definition (two $n$ by $n$ matrix-matrix mults, vs one $n^2$ by $n^2$).


So it seems that, in the space of arbitrary $n^2$ by $n^2$ matrices, there is a subset which can be represented as $x \mapsto a x b$, for some $n$ by $n$ matrices $a,b$.


So I'm wondering if there's generally a name for maps of this form, $x \mapsto a x b$, or if anyone generally has any comments. I know this looks like change of basis, but I'm thinking more generally than that. This may be a silly or ill-posed question, in which case I won't be offended if you say so :).


I'm asking because this shows up in Lagrange interpolation of functions of two (real) variables, and I'd like to know what to call the (c++) function which evaluates the transform, right now I'm calling it 'lagrange\_tensor', but I'm interested generally.


Also it may be nicer to work with the transpose of $b$, so $T = x \mapsto a x b^t$. And if either $a$ or $b$ is the identity, then the matrix $M$ above has many zeros, so that's one way to see why it's an expensive representation. As a side note, the set of operators of this form is not a vector space, as $axb + cxd \not = (a+c)x(b+d).$ Actually, this probably means it's not very interesting and I just answered my own question...


thanks



naming for the map $T = x \mapsto a x b$
I'd go for the obvious *two-sided multiplication operator*. A Google search shows this has been used indeed. 


