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.
Best Answer
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.