Let $A$, $B$ be $n\times n$ matrices.
It seems $\det(A \otimes B - B \otimes A) = 0$.
Moreover it seems that the kernel of $A \otimes B - B \otimes A$ contains $n$ vectors.
Here is MatLab code to check:
n=4; a=randn(n,n);b=randn(n,n);svd(kron(a,b)-kron(b,a))
You will see that the are exactly n zeros in svd, so the rank is n^2-n, for generic matrices.
This should be very simple, but I cannot see why ..
[EDIT]. If matrices A,B commute and have joint eigenbasis $v_i$. Then $v_i\otimes v_i$ generate the kernel.
David Speyer provided correct answer which I accept. The answer given before is wrong. AoB-BoA (xoy+yox)=AxoBy+AyoBx -BxoAy- ByoAx is no zero by no means. Moreover if it would be true the kernel would be n(n-1)/2 dimensional. However it is n-dimensional typically as it can be seen by experiments. [END EDIT].
[EDIT 2 ]. Additional observation non-zero eigenvalues of $M= A \otimes B - B \otimes A$, comes in pairs: x and -x.
How to explain ?
[End EDIT]
[EDIT 2]
Here is some numertical example
n=2; a=diag(rand(n,1)), b=rand(n,n), m= kron(a,b)-kron(b,a), [v d]=eig(m), diag(d)
a =
0.4494 0 0 0.6596
b =
0.7532 0.0292 0.8047 0.7798
m =
0 0.0131 -0.0131 0 0.3617 -0.1464 0 -0.0192
-0.3617 0 0.1464 0.0192
0 -0.5308 0.5308 0
v =
0.0232 -0.0232 0.0531 -0.0007
-0.3305 -0.0711 0.0000 -0.1309
0.0711 0.3305 -0.0000 -0.1309
-0.9409 0.9409 0.9986 0.9827
d =
-0.2265 0 0 0
0 0.2265 0 0 0 0 0.0000 0 0 0 0 0.0000
ans =
-0.2265 0.2265 0.0000 0.0000
[end edit]
Motivation comes from this question on MO, c1=a , c2=b: