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The following problem appeared on a past exam at my institution.

Suppose $V$ is a finite-dimensional non-zero complex vector space and $\mathcal{F}\subseteq\mathrm{Lin}(V)$ is a family of commuting linear operators on $V$. Prove that there exists a non-zero vector $v\in V$ that is a simultaneous eigenvector of every linear operator in $\mathcal{F}$.

I already know how to prove this fact when $\mathcal{F}$ is finite or countably infinite. The proof proceeds as follows:

Let $(T_n)_{n\in\mathbb{N}}\in\mathcal{F}^\mathbb{N}$ be such that $F=\{T_n\in\mathrm{Lin}(V)|n\in\mathbb{N}\}$. Let $W_0\subseteq V$ be an eigenspace of $T_0$ corresponding to eigenvalue $\lambda_0\in\mathbb{C}$.

Now assume $n\in\mathbb{N}$ is such that for all $i\in\{1,\dots,n\}$, $W_i\subseteq W_{i-1}$ is an eigenspace of $T_i|_{W_{i-1}}$ corresponding to eigenvalue $\lambda_i\in\mathbb{C}$. We claim that $W_n$ is $T_{n+1}$-invariant. Given $v\in W_n$ we know that $v\in W_0$, so $$T_0(T_{n+1}(v))=T_{n+1}(T_0(v))=T_{n+1}(\lambda_0 v)=\lambda_0T_{n+1}(v)$$ proving $T_{n+1}(v)\in W_0$. If $T_{n+1}(v)\in W_k$ for some $k\in\mathbb{N}$, $k<n$, then $v\in W_{k+1}$ implies $$T_{k+1}|_{W_k}(T_{n+1}(v))=T_{n+1}(T_{k+1}|_{W_k}(v))=T_{n+1}(\lambda_{k+1} v)=\lambda_{k+1}T_{n+1}(v)$$ proving $T_{n+1}(v)\in W_{k+1}$. By induction, it follows that $T_{n+1}(v)\in W_n$, so $W_n$ is indeed $T_{n+1}$-invariant. Finally, let $W_{n+1}\subseteq W_n$ be an eigenspace of $T_{n+1}|_{W_n}$.

Proceeding recursively in this way for all $n\in\mathbb{N}$, we obtain a decreasing sequence of non-zero subspaces $(W_n)_{n\in\mathbb{N}}$ of $V$ such that $W_n$ consists of simultaneous eigenvectors of $T_0,\dots,T_n$ for all $n\in\mathbb{N}$. Set $$W:=\bigcap_{n\in\mathbb{N}}W_n$$ Since $0<\mathrm{Dim}(W_{n+1})\leq\mathrm{Dim}(W_n)$ for all $n\in\mathbb{N}$, there exists $k\in\mathbb{Z}^+$ and $N\in\mathbb{N}$ such that and $\mathrm{Dim}(W_n)=k$ for all $n\in\mathbb{N}\cap[N,\infty)$. Then $W=W_N$ is a non-zero subspace of $V$, so there exists a non-zero vector $v\in W$ that is a simultaneous eigenvector for every linear operator in $\mathcal{F}$.

Does anyone have any suggestions for how to proceed when $\mathcal{F}$ is uncountable? Can the above proof be adapted?

Adrian Derderian
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  • You can just work with a basis of $F$; this reduces you to the finite case. – Qiaochu Yuan Sep 03 '24 at 02:07
  • @QiaochuYuan How does finding a simultaneous eigenvector $v\in V$ for a basis of $\mathrm{Span}(\mathcal{F})$ imply that $v$ is also a simultaneous eigenvector for every linear operator in $\mathcal{F}$? It may be the case that $v$ does not correspond to the same eigenvalue for each element of our basis. – Adrian Derderian Sep 03 '24 at 02:40
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    @AdrianDerderian If $T_\alpha v = \lambda_\alpha v$ then $(\sum c_i T_{\alpha_i})v = (\sum c_i \lambda_{\alpha_i}) v$. – ronno Sep 03 '24 at 04:07
  • Alternatively you could choose a well-ordering of $\mathcal{F}$ and use transfinite induction; you already wrote down the arguments for successor and the limit ordinals in special cases. – ronno Sep 03 '24 at 04:10
  • I would organize the proof to use induction on $\dim V$. The base case is trivial. As is the case when all the operators are scalar. Otherwise, if there is a single operator that is not scalar, use one of its eigenspaces $W$ to bring the dimension down (commutation implies that other operators map $W$ to itself). – Jyrki Lahtonen Sep 03 '24 at 04:15
  • See for example this answer. That thread is generally about two matrices, but the particular answer is more general. – Jyrki Lahtonen Sep 03 '24 at 04:27
  • The point really is that when you do induction on $\dim V$, the number of commuting operators becomes irrelevant: $2$, $n$, $\aleph_0$, anything. Makes no difference. – Jyrki Lahtonen Sep 03 '24 at 09:21

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