# Matrices Commuting with a Kronecker Sum

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 $$C=\sum_{k=1}^m\sum_{l=1}^n (X_{kl} \otimes e^n_{kl})=\sum_{i=1}^m\sum_{j=1}^n (e^m_{ij} \otimes Y_{ij}),$$ 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 $$\sum_{k=1}^m\sum_{l=1}^n ((X_{kl}A-A X_{kl}) \otimes e^n_{kl}) + \sum_{i=1}^m\sum_{j=1}^n (e^m_{ij} \otimes (Y_{ij}B-B Y_{ij}))=0,$$ 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 $$spectrum(A)=(\lambda_i)_{i\leq m},spectrum(B)=(\mu_j)_{j\leq n}$$.

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 $$spectrum(A\oplus B)=(\lambda_i+\mu_j)_{i,j}$$ has $$mn$$ distinct elements - Note that $$A\otimes I$$ and $$I\otimes B$$ commute-.

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

Finally, the $$mn$$ linearly independent matrices in the form $$A^i\otimes B^j,i< m,j constitute a basis of $$C(A\oplus 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, $$C(A\oplus B)$$ is the commutant of $$A\oplus B$$ (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 $$(A^i\otimes B^j)$$ as basis.

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

• Thank you for your response. It is very much appreciated! I have a few questions/comments: 1) Why do you say that the $(\lambda_i)$ and $(\mu_i)$ are distinct? This need not be true in general. A trivial example is if we take $A=B$, but it need not even be true for two matrices that are not similar. For example take $A=J_{1,3}$ and $B$ to be a direct sum of $J_{1,2}$ and $J_{1,1}$, where $J_{\lambda,r}$ is a Jordan block of size $r$ corresponding to an eigenvalue $\lambda$. In this example, $A$ is not similar to $B$, and yet they have the same spectra. – Iteraf Dec 9 '18 at 22:57
• 2) To be clear here, you are considering the matrix $C(A \oplus B)$ itself as a vector space? What exactly do you mean by 'constituted by the polynomials in $A \oplus B$'? What polynomials and what precisely is meant here by 'constituted'? – Iteraf Dec 9 '18 at 22:58
• 3)I fail to see how $A^i \otimes B^j$ can in generality be a basis for $C(A \otimes B)$ whilst the latter has dimension $mn$. For example take $A=B=e_{13}^3$ (so $A^2=B^2=0$). Similarly take $C=e_{19}^9$. Then $C(A \oplus B)=0$, and thus is 0-dimensional if considered as a (matrix) vector space, whereas $A \otimes B=e_{19}^9$ and is thus 1-dimensional. – Iteraf Dec 9 '18 at 22:58
• 4) I'm not sure how your response as a whole moves towards answering questions I had about the properties of the commutative matrix $C$ (or the matrices $X_{kl}$ and $Y_{ij}$). – Iteraf Dec 9 '18 at 22:58
• My apologies if you took offence at my questions and comments. It was genuinely an attempt to understand and to clarify some points of confusion, and it was not my intention to come across in any other way. With (4), this was intended as a prompt for further clarification, but I appreciate I was not clear here. – Iteraf Dec 10 '18 at 1:43