Eigenvectors and Kronecker product

Question

Let us define $$ v:=v_A\otimes v_B\quad (*) $$ where $v_A$ is a fixed vector in $\mathbb{R}^{d_A}$, $v_B$ is any vector in $\mathbb{R}^{d_B}$ and $\otimes$ denotes the Kronecker product. To rule out trivial cases assume $d_A,d_B>1$.

My question: Suppose that $v$, defined as in $(*)$, is an eigenvector of the symmetric matrix $C\in\mathbb{R}^{d\times d}$, with $d:=d_Ad_B$, for all $v_B\in\mathbb{R}^{d_B}$. Is it true that $C$ has the form $$ C=A\otimes I_{d_B}, $$ where $A\in\mathbb{R}^{d_A\times d_A}$ and $I_{d_B}$ denotes the identity matrix of dimension $d_B$?

Thank you for your help.

Answer 1

Let $$ C=\pmatrix{I&0\\0&D}, $$ where the identity $I$ and an arbitrary symmetric $D$ have the same dimension greater than one. For $v_A=[1,0]^T$ and any nonzero $v_B$ (of the same dimension as $I$ and $D$), $v=v_A\otimes v_B$ is an eigenvector of $C$. The matrix $C$ does not need to have a Kronecker product form since $D$ is arbitrary symmetric.

Answer 2

This is in general false. Consider $v_A = v_B = (1, 1)^t$ and $C = \begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1\end{pmatrix}$. Then $v_A \otimes v_B = (1, 1, 1, 1)^t$ and $C (v_A \otimes v_B) = 10 (v_A \otimes v_B)$, but clearly $C$ is not of the form $A \otimes I_2$.

In general, a single eigenvector doesn't say much about the structure of a matrix. I suppose a version of this statement might be true if there exists a whole eigenbasis of the form $u_i \otimes v_j$ with suitable vectors $u_i, v_j$.