TITLE: $\det(A \otimes B - B \otimes A) = 0$ why? Why $rk(M) = n^2-n$ ? Why  x and -x in Spec(M) ?  
QUESTION [5 upvotes]: 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:
https://mathoverflow.net/questions/97036/relaxing-commutativity-for-c1-c2-find-q1-q2-1-c1-c2q1c2-q2c1-2-q1-q20/

REPLY [8 votes]: Here is a proof for generic $A$ and $B$. Let $v_1$, ..., $v_n$ be the eigenvectors of $A^{-1} B$, and $\lambda_1$, ..., $\lambda_n$ the eigenvalues. Defining $w_i = B v_i$, we have $A^{-1} w_i = \lambda_i v_i$ or $A v_i = \lambda_i^{-1} w_i$. 
So 
$$(A \otimes B - B \otimes A) (v_i \otimes v_i) = (\lambda_i^{-1} w_i \otimes w_i) - (w_i \otimes \lambda_i^{-1} w_i) = 0.$$
It is not clear to me exactly what happens if $A$ is not invertible, or if $A^{-1} B$ is not diagonalizable, or if some of the $\lambda_i$ are zero. However, rank can only go down under specialization, so the kernel of $A \otimes B - B \otimes A$ is always at least $n$ dimensional.

Another proof: WLOG $2$ is invertible in the base field. Then, $M$ carries $\mathrm{Sym}^2 V$ to $\bigwedge^2 V$ (both regarded as subspaces of $V \otimes V$) and vice versa. Since $\dim \mathrm{Sym}^2(V) = \dim \bigwedge^2 V + n$, the map from $\mathrm{Sym}^2 V$ to $\bigwedge^2 V$ must have kernel of dimension at least $n$.

Regarding the symmetry of the spectrum, let $\sigma$ be the map $V \otimes V \to V \otimes V$ which switches the tensor factors. Then $\sigma M = - M \sigma$, so $M$ is conjugate to $-M$. So $\mathrm{Spec}(M) = - \mathrm{Spec}(M)$, as desired.