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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).

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  • $\begingroup$ No, take for $R$ the subalgebra of matrices preserving $U$. $\endgroup$ – abx Oct 23 '16 at 12:41
  • $\begingroup$ It can be shown that the $\mathbb{C}$-subalgebra $A$ generated by $R$ also satisfies the property that any 2 elements have a common eigenvector (because $R$ is Zariski-dense in $A$ and this property is Zariski-closed). Hence it is no restriction to assume that $R$ is a subalgebra. $\endgroup$ – YCor Oct 24 '16 at 20:30
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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 $\mathbf{C}^{n+k-1}\times\{0\}$.)

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