Is there any risk to transform to $(B^{T} \otimes A)\operatorname{vec}(X)=\operatorname{vec}(C) $ for solving $AXB=C$ for X To solve the equation $AXB=C$ for X, we can use the property of vec operator and kronecker product to transform to $(B^{T}\otimes A)\operatorname{vec}(X)=\operatorname{vec}(C)$, where $\operatorname{vec}(\cdot)$ is the vectorization operator.
Is there any risk to do like this, since $B^{T}\otimes A$ is a much larger matrix than A or B? 
Thank you.
 A: We assume that the unknown $X\in M_{m,n}$. When $A\in M_{m,m},B\in M_{n,n}$ are full matrices, $B^T\otimes A$ is a full $mn\times mn$ matrix; then inversion of such a matrix has complexity $O(m^3n^3)$. As we will see, using this method (when $A,B$ are large matrices), is a very bad idea.
There are special methods to solve equations in the form $AXB+CXD=E$ where $A,C\in M_{m,m},B,D\in M_{n,n},E\in M_{m,n}$ are known and $X\in M_{m,n}$ is unknown. cf. 
i)   http://www.maths.lth.se/na/courses/NUM115/NUM115-09/sylvester.pdf
uses an algorithm that is an extension of the Bartels–Stewart method and the Hessenberg-Schur method. Originally the Bartels –Stewart algorithm was used to solve the Sylvester equation.
ii)    http://www.dm.unibo.it/~simoncin/matrixeq.pdf
That is important, is that the previous algorithms have complexity $O(n^3+m^3)$ that is much smaller than the complexity of the Kronecker product method. For example, solving a Lyapounov equation ($AX+XA^T=B$ with $n\times n$ matrices) -with Bartels–Stewart- has the same complexity ($\approx 40 n^3$) as finding eigenvalues and eigenvectors of a $n\times n$ matrix.
A: No, I don't think there is any risk algebraically (although there are obvious computational issues)- in fact, it's one of the benifits of the $\text{vec}$ operator. But you need to be aware that when you are done, you will have $\vec{X}$ not $X$
Check out the bottom of page 6: https://tminka.github.io/papers/matrix/minka-matrix.pdf
In that document (which is a great introduction to matrix calculus), they give the example of using the vec operator to solve the Lyapunov Equation $AX + XB = C$ for $X$.
