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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.

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Aspirin
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1 Answers1

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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!

Aspirin
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    Maybe I'm missing something, but don't you only need the "second" case? If some $A$ is not a multiple of the identity, then let $\lambda$ be an eigenvalue of $A$. Then $0 < V_{\lambda}^1 < V$. Hence $V_{\lambda}^1$ is a non-zero space invariant under all $A_i$'s, so by induction, it contains a common (non-zero) eigenvector of all the $A_i$'s. – Ted Apr 14 '12 at 19:06