vectorized product of two matrices, with one symmetric For each generic non-symmetric square matrix $A$, it is well known the relation with its transpose using the commutation matrix:
$K^{(n,n)} {\rm vec}(A) = {\rm vec}(A^T)$, and of course ${\rm vec}(A) = (K^{(n,n)})^{-1} {\rm vec}(A^T)$.
Given a symmetrix matrix $S$, I was wondering if there is a way in which I can use the kronecker product and vector operator properties to write something like this:
${\rm vec}(A^TS) = X {\rm vec}(AS)$ 
where $X$ is this particular "commutation matrix".
Thanks for the help.
 A: Well, you have the rule ${\rm vec}\{A X B\} = (B^T \otimes A) \cdot {\rm vec}\{X\}$.
Applied to your problem, you can write ${\rm vec}\{A^T S\} = (S^T \otimes I) \cdot {\rm vec}\{A^T\} = (S^T \otimes I) \cdot K_{n,n}^T \cdot {\rm vec}\{A\}$.
Also, you have ${\rm vec}\{A S\} = (S^T \otimes I) \cdot {\rm vec}\{A\}$.
Therefore, if $S$ is invertible, you can write ${\rm vec}\{A^T S\} = (S^T \otimes I) \cdot K_{n,n}^T \cdot (S^{-T} \otimes I) \cdot {\rm vec}\{A S\}$.
Interestingly, this can be rewritten further as ${\rm vec}\{A^T S\} = K_{n,n}^T \cdot (I \otimes S^T )  \cdot (S^{-T} \otimes I) \cdot {\rm vec}\{A S\} = K_{n,n}^T \cdot (S^{-T} \otimes S^T )\cdot {\rm vec}\{A S\}$ (assuming $A$ is square). 
So in that sense, we can say that a matrix $X$ exists that satisfies ${\rm vec}\{A^T S\} = X\cdot {\rm vec}\{A S\}$. However, since $X=K_{n,n}^T \cdot (S^{-T} \otimes S^T )$, this matrix will depend on $S$. Also note that we haven't used the symmetry of $S$ yet. This will not help much, other than letting a few transpose signs disappear.
*edit: Btw, this question is also equivalent to asking how to convert ${\rm vec}(S A)$ into  ${\rm vec}(A S^T)$. The answer is ${\rm vec}(S A) = (S^{-1} \otimes S) \cdot {\rm vec}(A S^T)$.
