# Simultaneous Diagonalization of collection of linear operators

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?

## 1 Answer

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.