A Set of Commuting Operators Has a Common Eigenvector 
Let $A_1, A_2,\ldots$ be a (possibly infinite) set of linear operators on a finite-dimensional complex vector space $V$ such that $$A_iA_j=A_jA_i$$ for all $i$ and $j$. How to prove that all operators $A_1, A_2,\ldots$ have common eigenvector?

Thanks a lot.
 A: Lemma: let
$$V_{\lambda}^{m}(A) = \{v\in V|(A - \lambda E)^{m}v = 0\}$$
For any $A\in\{A_1, A_2, \ldots\}$. Then for any $B\in\{A_1, A_2, \ldots\}$
$$B(V_{\lambda}^{m}(A))\subset V_{\lambda}^{m}(A).$$
Proof: if $v\in V_{\lambda}^{m}(A)$ then $(A - \lambda E)^{m}Bv = B(A - \lambda E)^{m}v = 0.$ QED
Induction on the dimension:
Consider two special cases:
First: there is operator $A\in\{A_1, A_2, \ldots\}$ with tho different eigenvalues $\lambda_1$ and $\lambda_2$. Then subspace
$$W = V_{\lambda_1}^{n}(A) \subsetneq V$$
is invariant under the action of any operator $A_i$ (by lemma) and
$$\dim W &lt \dim V.$$
So by induction $A_1, A_2, \ldots$ have common eigenvector in $W$.
Second: all eigenvalues of any operator $A\in\{A_1, A_2, \ldots\}$ are the same. This case splits into two:
1. There is operator $A\in\{A_1, A_2, \ldots\}$ such that
$$V_{\lambda}^{1}(A) \neq V.$$
Then $V_{\lambda}^{1}(A)$ is invariant under the action of any operator $A_i$ and
$$\dim V_{\lambda}^{1}(A) &lt \dim V.$$
So by induction $A_1, A_2, \ldots$ have common eigenvector in $V_{\lambda}^{1}(A)$.
2. For any $A\in\{A_1, A_2, \ldots\}$
$$V_{\lambda}^{1}(A) = V.$$
Then any vector of $V$ is common eigenvector of operators $A_1, A_2, \ldots$. QED
Can you check my solution?
Thanks a lot!
