Invertibility of a Kronecker Product Prove that $A\otimes B$ is invertible if and only if $B\otimes A$ is invertible.
I don't have a clue where to start to be honest. I am not very familiar yet to the Kronecker Product so could you please help me with providing an easily understandable proof.
Thanks in advance.
 A: Hint: If $A$ is a $m\times m$ matrix and $B$ a $n\times n$ matrix then 
$$\det(A\otimes B)=(\det A)^n\cdot (\det B)^m.$$
A: The previous answers assume that $A$ and $B$ are square.  This is fair because    it is not generally possible for $A\otimes B$ to be invertible when $A$ and $B$ are rectangular.
In the more general setting where $A$ is $m\times n$ and $B$ is $r\times s$ (and $mr = ns$) it is still easy to see that $A\otimes B$ is equivalent to $B \otimes A$ up to permuting the rows and columns.
Both $A \otimes B$ and $B \otimes A$ contain exactly the same entries $a_{ij} b_{kl}$.  More explicitly, one places $a_{ij} b_{kl}$ in row $(i-1)r+k$ and column $(j-1)s+l$, and the other places it in row $(k-1)m+i$ and column $(l-1)n+j$.  Note that the row numbers are uniquely determined by $i,k$ and the column numbers determined independently by $j,l$.
A: Added: Assuming that $A$ and $B$ are square matrices and neither is $0\times 0$:

You can show that $A\otimes B$ is invertible if and only if $A$ and $B$ are invertible, from which this result follows.  
If $A$ and $B$ are invertible, do you know how to find $(A\otimes B)^{-1}$?
If $A$ is not invertible, then there exists $C\neq 0$ such that $CA=0$.  It follows that $(C\otimes I)(A\otimes B)=0$, which implies that $A\otimes B$ is not invertible.  
A: Do you know that if two matrices $A$ and $B$ represent two linear maps $a : V \to V'$ and $b : W \to W'$ (with respect to some chosen bases of the vector spaces $V$, $V'$, $W$ and $W'$), then the Kronecker product $A \otimes B$ represents the linear map $a \otimes b : V \otimes W \to V' \otimes W'$ (with respect to the appropriately defined bases: e.g., if our bases of $V$ and $W$ are $\left(e_i\right)_{i \in I}$ and $\left(f_j\right)_{j \in J}$, then for $V \otimes W$ we use the basis $\left(e_i \otimes f_j\right)_{\left(i,j\right)\in I\times J}$) ? If so, then you will realize that your question is equivalent to the following question: If the tensor product $a \otimes b$ of two linear maps $a$ and $b$ is invertible, then so is $b \otimes a$. And this follows from the following commutative diagram:
$
\begin{array}{ccc}
V \otimes W & \overset{\cong}{\longrightarrow} & W \otimes V \\
\downarrow a \otimes b & & \downarrow b \otimes a \\
V' \otimes W' & \overset{\cong}{\longrightarrow} & W' \otimes V'
\end{array}
$
where the horizontal arrows are canonical isomorphisms.
This is essentially an algebraic version of Erick Wong's proof.
A: Assume that $A\in M_{m,n},B\in M_{p,q}$. If we stack the matrices $X,Y$ row by row, then $A\bigotimes B:X\in M_{n,q}\rightarrow AXB^T,B\bigotimes A:Y\in M_{q,n}\rightarrow BYA^T$. Let $\phi:X\in M_{q,n}\rightarrow X^T,\psi:Y\in M_{p,m}\rightarrow Y^T$ ; one has $A\bigotimes B\circ\phi=\psi\circ B\bigotimes A$. 
Case 1. If $p=q,m=n$, that is, if $A,B$ are square matrices, then $A\bigotimes B$ and $B\bigotimes A$ are similar.
Case 2. Otherwise it is false. Yet $A\bigotimes B$ and $B\bigotimes A$ have same singular values $(\sigma_i\tau_j)_{i,j}$ where $(\sigma_i)_i, (\tau_j)_j$ are the singular values of $A,B$. Indeed, if one has the SVD $A=U_1D_1V_1,B=U_2D_2V_2$, then we deduce the SVD $A\bigotimes B=(U_1\bigotimes U_2)(D_1\bigotimes D_2)(V_1\bigotimes V_2)$ and $B\bigotimes A=(U_2\bigotimes U_1)(D_2\bigotimes D_1)(V_2\bigotimes V_1)$. 
A direct consequence is the following generalization of the result requested by the OP: $rank(A\bigotimes B)=rank(B\bigotimes A)=rank(A)rank(B)$.
