It seems like the following is probably true: Let $V,W$ be $K$-vector spaces. Let $\phi:V\rightarrow V$ and $\psi:W\rightarrow W$ be linear maps, then $u\in V\otimes W$ is an eigenvector of $\phi \otimes \psi$ with eigenvalue $\lambda$ iff there are $v\in V$, $w\in W$ and $\kappa,\mu\in K$ such that $u=v\otimes w$, $\lambda=\kappa \mu$ , and $v$ is an eigenvector of $\phi$ with eigenvalue $\kappa$ and $w$ is an eigenvector of $\psi$ with eigenvalue $\mu$.
The backwards implication is trivial, but I can't quite work out the forwards implication, so maybe it's false.
*In light of omnomnom's post, how about the following: for all $\lambda\in K$, there are $\kappa_i,\mu_i \in K$ such that $\lambda=\kappa_i\mu_i$ and $$\ker\left(\phi \otimes \psi -\lambda I\right)=\bigoplus_i\ker\left(\phi -\kappa_i I\right) \otimes \ker\left(\psi -\mu_i I\right)$$
Best Answer
Your statement is not necessarily true. For example: take $V = W = \Bbb C^2$ over $\Bbb C$, and $\phi = \psi = \operatorname{id}$. We then find that $$ u = \pmatrix{1\\0} \otimes \pmatrix{1\\0} + \pmatrix{0\\1} \otimes \pmatrix{0\\1} $$ is an eigenvector of $\phi \otimes \psi$ which cannot be written as a tensor product. That is, we can't write $u = v \otimes w$ for any (eigenvectors) $v,w$.