Simultaneous Diagonalization of collection of linear operators

Question

This is problem 5E from the 4th edition of Linear Algebra Done Right:

Suppose $\mathcal{E}$ is a subset of $\mathcal{L}(V)$ and every element of $\mathcal{E}$ is diagonalizable. Prove that there exists a basis of $V$ with respect to which every element of $\mathcal{E}$ has a diagonal matrix if and only if every pair of elements of $\mathcal{E}$ commutes.

I'm having some trouble extending the result to $|\mathcal{E}|$ infinite and I'd like some help.

My attempt: The if direction is trivial so I omit it. We first show the result for $|\mathcal{E}|$ finite. The case $|\mathcal{E}|=2$ is easy to show: if $S,T \in \mathcal{E}$, note $$V=E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m,T)$$ and so since $S|_{E(\lambda_j, T)}$ is invariant under $E(\lambda_j,T)$ we can get bases of each $E(\lambda_j,T)$ that diagonalize each $S|_{E(\lambda_j,T)}$ and we get our result. Inducting on $|\mathcal{E}|=p$, for all $S \in \mathcal{E}/\{T\}$, we note since those $S$ commute pairwise, we can find a basis of each $E(\lambda_j,T)$ that diagonalizes each $S$ simultaneously and combining we again get a basis.

Now where I'm getting stuck with the infinite case. Let $\mathcal{U} = \text{span}_{e \in \mathcal{E}}(e)$. We have $\mathcal{E} \subset \mathcal{U}$ and $\mathcal{U}$ is finite dimensional.

I'm pretty sure $\mathcal{E}$ (being a spanning set of $\mathcal{U}$) must contain a finite basis $\phi_1,\cdots,\phi_m$ of $\mathcal{U}$ which would finish the proof since then $\{\phi_1,\cdots,\phi_m\} \subset \mathcal{E} \subset \mathcal{L}(V)$ has finite cardinality and the elements pairwise commute we can find a basis of $V$, say $v_1,\cdots,v_n$ for which each $\phi_j$ has a diagonal matrix. Then, for any $T \in \mathcal{E}$, we have since the $\phi_j$ span $\mathcal{U}$ that $T=a_1\phi_1+ \cdots + a_m\phi_m$ for some $a_j \in \mathbb{F}$ and thus the matrix of $T$ is diagonal with respect to the $v_j$.

However, I'm not 100% sure if this is true: does an infinite spanning set of a finite vector space always contain a finite basis?

Answer 1

Yes, your reasoning is correct. If $X$ is a spanning set for some finite-dimensional vector space $V$, choose some basis $(e_1,\dots,e_n)$ of $V$. Then each $e_i$ is a linear combination of a finite subset $X_i\subset X$ of $X$, so if you take $X'=X_1\cup\dots\cup X_n$ you get a finite subset of $X$ which generates $V$ (since all $e_i$ are combinations of elements of $X'$). This should allow you to apply your standard results about finite generating subsets containing bases.