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?
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Best Answer
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 $$\dim W < \dim V.$$ 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 $$\dim V_{\lambda}^{1}(A) < \dim V.$$ 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!