Consider the matrix algebra $\mathcal{M}_n(\mathbb{C})$ (acting on $n$ dimensional space $V$) and let $R$ be subring of matrices of $\mathcal{M}_n(\mathbb{C})$.

Suppose that any two elements of $R$ have a common eigenvector. Does it follow that there is a common eigenvector for all $R$?

Note: the original question was the same under the additional assumption that any two elements have a common eigenvector in a fixed hyperplane $U$ of $V$. But it is equivalent to the previous question (which appears at first sight harder): if we have $R$ as above, we consider the subring $R'$ of operators in $V\oplus\mathbb{C}$ stabilizing $V$ and whose restriction to $V$ belongs to $R$. Then any two elements of $R'$ have a common eigenvector in the hyperplane $V$; so if the results holds under this hyperplane condition, then we deduce that $R'$ has a common eigenvector $w$. But if we consider the operator mapping $(0_V,1)$ to a nonzero vector in $V$ and $V$ to 0, its kernel is exactly $V$ and all its eigenvectors are in $V$; since it belongs to $R'$, we deduce that $w\in V$. So both questions are equivalent (and thus it is much more natural to formulate it with no reference to a hyperplane).