TITLE: Eigenvectors of Kronecker Product
QUESTION [8 upvotes]: 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 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.
However, no statement is made that all eigenvectors arise in such
fashion. Does the conjecture follow from this theorem?

REPLY [8 votes]: 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.