Ytukyg

Throughout, let $A$ and $B$ be complex $m times m$ and $n times n$ matrices respectively. By $A otimes B$, we mean the matrix formed from the Kronecker product of $A$ and $B$, and by $A oplus B$, we mean the matrix formed by the Kronecker sum of $A$ and $B$. Namely, $$A otimes I_n + I_m otimes B,$$ where $I_r$ is an $r times r$ identity matrix.

Let $C$ be an arbitrary $mn times mn$ matrix that commutes with $A oplus B$. What (if anything) can be said about $C$?

For example, one could write $C$ as
$$C=sum_{k=1}^msum_{l=1}^n (X_{kl} otimes e^n_{kl})=sum_{i=1}^msum_{j=1}^n (e^m_{ij} otimes Y_{ij}),$$
where $e^r_{ij}$ is the standard $r times r$ matrix with a 1 as the $(i,j)$-th entry and every other entry is 0, and $X_{kl}$ and $Y_{ij}$ are $m times m$ and $n times n$ matrices respectively. If $C$ commutes with $A oplus B$, then does every matrix $X_{kl}$ commute with $A$ and every matrix $Y_{ij}$ commute with $B$?

I have found no counterexamples thus far to the above, but I also fail to see why it might be true in general. If we write
$$C(A oplus B)=(A oplus B)C$$
then we can deduce from the mixed-multiplication property of Kronecker products that
$$sum_{k=1}^msum_{l=1}^n ((X_{kl}A-A X_{kl}) otimes e^n_{kl}) + sum_{i=1}^msum_{j=1}^n (e^m_{ij} otimes (Y_{ij}B-B Y_{ij}))=0,$$
but it doesn't seem clear to me at all that one might be able to deduce from this that $X_{kl}A-A X_{kl}=0$ and $Y_{ij}B-B Y_{ij}=0$ for any $i,j,k,l$.

Let $spectrum(A)=(lambda_i)_{ileq m},spectrum(B)=(mu_j)_{jleq n}$.

We consider the case when $A,B$ are generic (for example take random $A,B$). Then the $(lambda_i)$ (resp. the $(mu_j)$) are distinct.

Moreover $spectrum(Aoplus B)=(lambda_i+mu_j)_{i,j}$ has $mn$ distinct elements - Note that $Aotimes I$ and $Iotimes B$ commute-.

Then $C(Aoplus B)$ is a vector space of dimension $mn$ constituted by the polynomials in $Aoplus B$.

Finally, the $mn$ linearly independent matrices in the form $A^iotimes B^j,i< m,j<n$ constitute a basis of $C(Aoplus B)$.

EDIT. Answer to the OP. I think you did not understand one word of my post.

For i) A generic matrix $A=[a_{i,j}]$ is s.t. there are no algebraic relations between the $(a_{i,j})$. More precisely, the $(a_{i,j})$ are said to be parameters (they are mutually transcendental over $mathbb{C})$. You can simulate such a matrix by choosing it at random. Do this with your PC instead of writing pseudo counter examples; you will find in particular that for such matrices $A,B$, the $lambda_i+mu_j$ are distinct.

For ii). My friend, $C(Aoplus B)$ is the commutant of $Aoplus B$ (well-known notation) and, therefore, is a vector space. On the other hand, the commutant of a matrix that has distinct eigenvalues is constituted with the polynomials in this matrix.

For iii). Your counter-examples are only particular well-known cases (all that you write is absolutely standard and is not the object of my post). With probability $1$, the commutant of your matrix admits the $(A^iotimes B^j)$ as basis.

For iv). When one does not understand, one asks. I do not intend to waste any more time with your file.