Eigenvector shared by two endomorhisms

Question

I am guessing if the following fact is true: Let be $V$ a finite vector space above a field $K$. Let $f, g$ be two endomorphisms of $V$ with $f g = g f$. We assume that both $f$ and $g$ have got at least one eigenvector. Then $f$ and $g$ have got one eigenvector in common. I have proved for a 2-dimensional vectorial space. Do you think that this is always true? Is it a known fact?

Answer 1

First we prove that since $f$ and $g$ commute and if $\lambda$ is an eigenvalue of $f$ then the associated eigenspace $E_\lambda (f)=\ker(f-\lambda\operatorname{id})$ is invariant by $g$: in fact, if $x\in E_\lambda(f)$ then $f(x)=\lambda x$ so $$f(g(x))=g(f(x))=g(\lambda x)=\lambda g(x)$$ and then $g(x)\in E_\lambda (f)$. Now let $g_1$ the restriction of $g$ on $E_\lambda(f)$ and let $v$ an eigenvector of $g_1$ associated to an eigenvalue $\mu$ of $g_1$ then clearly $v$ is a common eigenvector of $f$ and $g$.

Remark This result depends on the field $\Bbb K$ since the eigenvalues $\lambda$ or $\mu$ may not exist but the result is obviously true if $\Bbb K=\Bbb C$.