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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?

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    $\begingroup$ In the nondiagonalizable case, it's not even true that all eigenvectors lie in the span of products of eigenvectors. Take $A$ and $B$ to be the $2$-by-$2$ Jordan block with eigenvalue $1$. Then $e_1\otimes e_1$ and $e_1\otimes e_2-e_2\otimes e_1$ are both eigenvectors of $A\otimes B$. The problem of computing the extra eigenvectors can always be reduced to the case where $A$ and $B$ are Jordan blocks by Jordanizing both. $\endgroup$
    – MTyson
    Commented Dec 1, 2019 at 19:32
  • $\begingroup$ Thanks @MTyson! This is actually closer to what I had in mind. $\endgroup$
    – Henrique de Oliveira
    Commented Dec 5, 2019 at 16:51

1 Answer 1

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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.

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  • $\begingroup$ Ha! Good point. Now I feel silly. I'll see if there is a different way of phrasing the question... $\endgroup$
    – Henrique de Oliveira
    Commented Dec 1, 2019 at 17:55
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    $\begingroup$ if both matrices are diagonalizable, their tensor product does have a basis of eigenvectors of the form you indicate $\endgroup$
    – alesia
    Commented Dec 1, 2019 at 18:23

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