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



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



Subring of matrix algebra with common eigenvectors
In an attempt to prove this, I finally got the following counterexample. Fix $2\le n<k$ (for instance $(n,k)=(2,3)$, in size 5). Consider the subalgebra $R$ of $(n+k)$-square matrices of the form 
$$M(A,B,t)=\begin{pmatrix}A & B\\ 0 & tI\_k\end{pmatrix},\quad A\in M\_n(\mathbf{C}),\quad B\in M\_{n,k}(\mathbf{C}).$$
For $x=M(A,B,t)\in R$, write $t=t(x)$. Then the rank of $x-t(x)I$ is at most $n$; in other words, $\mathrm{codim}(\mathrm{Ker}(x-t(x)I)\le n$. Hence for any two $x,y\in R$,
$$\mathrm{codim}(\mathrm{Ker}(x-t(x)I)\cap \mathrm{codim}(\mathrm{Ker}(y-t(y)I)\le 2n<n+k,$$
so any two elements in $R$ have a common eigenvector. But $R$ has no common eigenvector: if $w=(u,v)\in\mathbf{C}^n\oplus\mathbf{C}^k$ where a common eigenvector, that $w$ is a common eigenvector for all $M(A,0,0)$ would imply $u=0$ (because $n\ge 2$) and that $w$ is a common eigenvector to all $M(0,B,0)$ would imply $v=0$.


On the other hand, the result is true under some further assumptions, e.g., if $V$ is an irreducible complex $R$-module (clear since then $R$ generates the whole algebra of matrices), or more generally if it is a semisimple module (= direct sum of irreducible submodules).


(Remark related to the original formulation of the question: in the above construction, if $k\ge n+2$ then the same proof shows that any two elements have a common eigenvector in the hyperplane $\mathbf{C}^{n+k-1}\times\{0\}$.)


