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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:

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This question is a bit more subtle than one might expect. Unless I'm mistaken, it is essential that $V$ is finite dimensional. As an infinite dimensional counterexample consider the space of univariate polynomials $V=k[x]$, when $V\otimes V$ can be viewed as bivariate polynomials $k[x,y]$. If $A$ is multiplication by $x$ and $B$ is multiplication by $(x+1)$, then $A\otimes B$ (resp. $B\otimes A$) is multiplication by $x(y+1)$ (resp. $(x+1)y$). Therefore $A\otimes B- B\otimes A$ is multiplication by $$x(y+1)-(x+1)y=x-y,$$ which is an injective mapping (but not surjective)... – Jyrki Lahtonen Jun 8 '12 at 10:14
...(cont.) Therefore a proof of this claim will need to use the assumption that $V$ is f.d. – Jyrki Lahtonen Jun 8 '12 at 10:15
up vote 8 down vote accepted

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.

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Great, thank you ! the only thing Av= 1/lambda w... Do you have guesses about x and -x in Spec(M) ? – Alexander Chervov Jun 8 '12 at 12:24
Thanks for the correction! – David Speyer Jun 8 '12 at 12:26
+1 The argument involving the second symmetric and exterior powers is just so... – Jyrki Lahtonen Jun 8 '12 at 15:00

Edit: There is apperently a mistake. I will deal with this later today/on the weekend.

As a map of vector spaces $A\otimes B-B\otimes A$ is nothing else but the map $$V\otimes V\to V\otimes V$$ which sends $$x\otimes y\mapsto A(x)\otimes B(y)-A(y)\otimes B(x)$$ Clearly this map has non-trivial kernel since $x\otimes y+y\otimes x$ is sent to zero for all $x,y$. Therefore the determinant is zero.

For your second question: This can't be quite right. Take for example $A=B$ and the kernel has dimension $n^2$. A lower bound for the dimension of the kernel is given by $\binom n2$, as this is the dimension of the subspace spanned by the vectors above.

I would assume that you should get this lower bound $\binom n2$ most of the time since you take random matrices which are dense and non-singular with probability 1.

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Thank you! shame on me, how can I miss it – Alexander Chervov May 17 '12 at 13:41
Double shame on me. First this did not come to my mind, second I did not notice this is wrong. – Alexander Chervov Jun 8 '12 at 6:28
hmm? What is wrong? – Simon Markett Jun 8 '12 at 7:34
the second displayed equation is wrong, in fact $x\otimes y \mapsto Ax\otimes By - Bx\otimes Ay$ – m_t_ Jun 8 '12 at 8:10

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