**Conjecture** If $A$ and $B$ are two complex square matrices, then every eigenvector of $A\otimes B$ is of the form $x\otimes y$, where $x$ is an eigenvector of $A$ and $y$ is an eigenvector of $B$. 


Here, $A\otimes B$ denotes the [Kronecker Product](https://en.wikipedia.org/wiki/Kronecker_product) of two matrices. I would like to know if this conjecture is true.


**Motivation:** I know that the following is true:


**Theorem**
Let $A$ and $B$ be two complex square matrices. If $\lambda$ is
an eigenvalue of $A$ with corresponding eigenvector $x$ and $\mu$
is an eigenvector of $B$ with corresponding eigenvector $y$, then
$\lambda\mu$ is an eigenvalue of $A\otimes B$ with corresponding
eigenvector $x\otimes y$. Moreover, every eigenvalue of $A\otimes B$
arises as such a product. 


This is Theorem 4.2.12 in Horn and Johnson's "Topics in Matrix Analysis'',
or Theorem 13.12 [here](https://archive.siam.org/books/textbooks/OT91sample.pdf).
However, no statement is made that all eigenvectors arise in such
fashion. Does the conjecture follow from this theorem?



Eigenvectors of Kronecker Product
Counterexample: the matrix $I \times I$ has eigenvectors that are not in product form, since every vector is an eigenvector of it and not every vector can be written in product form.


