A Set of Commuting Operators Has a Common Eigenvector

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

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