A question about matrix kernels and Kronecker products 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 $(*)$, belongs to the kernel of the symmetric matrix $C\in\mathbb{R}^{d\times d}$, with $d:=d_Ad_B$, for all $v_B\in\mathbb{R}^{d_B}$. Namely, $Cv=C (v_A\otimes v_B)=0$, for all $v_B\in\mathbb{R}^{d_B}$ and for fixed $v_A\in\mathbb{R}^{d_A}$. Is it true that $C$ has the form
$$
C=A\otimes B,
$$
where $A\in\mathbb{R}^{d_A\times d_A}$ and $B\in\mathbb{R}^{d_B\times d_B}$?
Thank you for your help.

No. Consider $v_A=(1,0,0)^T,\ v_B\in\mathbb R^2$ and
$$
C=\pmatrix{
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&1&0&0&1\\
0&0&0&1&1&0\\
0&0&0&1&1&0\\
0&0&1&0&0&1}.
$$
