# 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 $B\in\mathbb{R}^{d_B\times d_B}$?

No. Consider $v_A=(1,0,0)^T,\ v_B\in\mathbb R^2$ and $$C=\pmatrix{ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&1&0&0&1\\ 0&0&0&1&1&0\\ 0&0&0&1&1&0\\ 0&0&1&0&0&1}.$$
• Thank you for the counterexample. If $v_A$ and $v_B$ have the same dimension, i.e. $d_A=d_B$, is the answer still no? – Ludwig Aug 27 '15 at 11:21
• @Jacquard The answer is still no if $d_A=d_B\ge3$. E.g. suppose $d_A=d_B=3$ and $v_A=(1,0,0)^T$. Take a random real symmetric $C$ whose first three rows/columns are zero. It's almost always not a Kronecker product of two $3\times3$ matrices because its nonzero $3\times3$ sub-blocks are not scalar multiples of each other. – user1551 Aug 27 '15 at 11:30
• A last question: in case $d_A=d_B=2$, then is the answer positive? – Ludwig Sep 4 '15 at 8:27
• @Jacquard (Presumably $v_A\ne0$.) Yes. If $d_A=d_B=2$, by a change of basis (via an invertible matrix $P$ on $\mathbb R^2$ and $P\otimes P$ on $\mathbb R^4$), we may assume that $v_A=(1,0)^T$. So, the condition that $C(v_A\otimes v_B)=0$ for every $v_B$ means that the first two columns of $C$ are zero. As $C$ is symmetric, its first two rows are zero too. Therefore, $C$ is a Kronecker product of $A=\pmatrix{0&0\\ 0&1}$ and some $2\times2$ matrix $B$. – user1551 Sep 4 '15 at 9:31