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).

  • $\begingroup$ No, take for $R$ the subalgebra of matrices preserving $U$. $\endgroup$
    – abx
    Oct 23, 2016 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, 2016 at 20:30

1 Answer 1


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 $\mathbf{C}^{n+k-1}\times\{0\}$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.