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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.

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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.

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  • $\begingroup$ While I see the idea behind it [see also the comments in my post], I don't see why this proves that the statement is wrong. Can you provide a specific counter-example? $\endgroup$ – Dominik Aug 26 '15 at 11:14
  • $\begingroup$ I agree with Dominik. $\endgroup$ – Ludwig Aug 26 '15 at 11:40
  • $\begingroup$ @Dominik Better? (At least for the current version of the statement.) $\endgroup$ – Algebraic Pavel Aug 26 '15 at 12:03
  • $\begingroup$ Yes, what a surprisingly simple counterexample. Thanks. $\endgroup$ – Dominik Aug 26 '15 at 12:07
  • $\begingroup$ @Algebraic Pavel: Thank you for the counterexample! I have only one more question. If, in your example, I replace the identity block of $C$ by $\lambda I$, $\lambda\in\mathbb{R}$, $\lambda\neq 0$, then I have that $v$ is an eigenvector of $C$ associated with eigenvalue $\lambda$. What about the case $\lambda=0$? In this case the counterexample does not work, since $C$ is of the form $\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\otimes D$. Do you have a counterexample also for the 0-eigenvalue case? $\endgroup$ – Ludwig Aug 26 '15 at 18:14
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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 $u_i, v_j$.

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  • $\begingroup$ I assume that $v_B$ is any vector in $\mathbb{R}^{d_B}$. In your example, if I choose $v_B=[0, 1]^\top$, then $v=v_A\otimes v_B$ is no more an eigenvector of $C$. $\endgroup$ – Ludwig Aug 26 '15 at 9:34
  • $\begingroup$ I see, then you should change the wording of your question to "assume that $v_A \otimes v_B$ is an eigenvector for all $v_b \in \mathbb{R}^{d_B}$". I suspect that the statement still remains false, because this only yields $d_B$ linearly independent eigenvectors, while $A$ is a $d_A \times d_A$ matrix. But I don't have a counterexample at the moment. $\endgroup$ – Dominik Aug 26 '15 at 10:05

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