Simultaneous diagonalizability of infinitely many endomorphisms in infinite dimension Let $V$ be a vector space and let $(f_i)_{i \in I}$ be a family of endomorphisms of $V$.

Question.
Suppose that each $f_i$ is diagonalizable and that $f_i$ commutes with $f_j$ for any two indices $i$ and $j$.
Is the family $(f_i)_{i \in I}$ simultaneously diagonalizable?

I’m using here the following notion of simultaneous diagonalizability.

The family $(f_i)_{i \in I}$ is simultaneously diagonalizable if we have the decomposition
$$
    V = \bigoplus_{(\lambda_i)_{i \in I}} \operatorname{Eig}( (f_i)_{i \in I}, (\lambda_i)_{i \in I} )
  $$
where $\operatorname{Eig}( (f_i)_{i \in I}, (\lambda_i)_{i \in I} ) = \{ v \in V \mid \text{$f_i(v) = \lambda_i v$ for every $i \in I$}\}$ is the common eigenspace of the family $(f_i)_{i \in I}$ for the eigenvalues $(\lambda_i)_{i \in I}$.
Equivalently, $V$ admits a basis consisting of common eigenvectors of the $f_i$.

It is (well-)known that the the answer to the question is yes if

*

*$I$ is finite;

*$V$ is finite-dimensional; or more generally

*if the linear subspace of $\operatorname{End}(V)$ spanned by the $f_i$ is finite-dimensional.

But I don’t know what happens in general if both $I$ is infinite and $V$ is infinite-dimensional.
 A: Surprisingly (to me anyway), this is false! Let $V$ be the vector space $k^X$ of functions on the Cantor set $X = \{ 0, 1 \}^{\mathbb{N}}$. $V$ is in fact a commutative $k$-algebra with respect to pointwise multiplication, so we can define commuting families of operators on $V$ via pointwise multiplication. Consider the countable family of functions given by each coordinate projection $p_i : \{ 0, 1 \}^{\mathbb{N}} \to \{ 0, 1 \} \subset k$; via pointwise multiplication these give a commuting family of endomorphisms $f_i$. Each $f_i$ is idempotent and hence diagonalizable; the only eigenvalues are $0$ and $1$, and $V$ splits up into a direct sum of the $0$-eigenspace and $1$-eigenspace, which consist of the kernel and image of $f_i$ (the functions supported on the sequences whose $i^{th}$ coordinate is $0$, and the functions supported on the sequences whose $i^{th}$ coordinate is $1$, respectively).
A similar argument shows that any finite subset of the $f_i$ are simultaneously diagonalizable. However, the $f_i$ are not simultaneously diagonalizable. The problem is that a simultaneous eigenvector for all of the $f_i$ must be a function $X \to k$ which is supported at a single sequence $x \in X$ in the Cantor set, whose coordinates are determined by the eigenvalues of the $f_i$ acting on this eigenvector. So the subspace of $V$ spanned by simultaneous eigenvectors for all of the $f_i$ consists of the functions with finite support, which is not all of $V$.

This counterexample comes out of thinking about what goes wrong if you try to push the usual argument for $I$ finite to the case that $I$ is infinite. The usual argument is to decompose into eigenspaces for $f_1$, say, then consider the action of $f_2$ on each eigenspace and decompose into eigenspaces for that action, and so forth. But if you try to repeat this "and so forth" argument countably many times you may find that you are "cutting up" the eigenspaces "too finely." This is very vague but trying to make it rigorous leads naturally to the above example, if we think of the minimal nontrivial amount of "cutting up" each new endomorphism can do as cutting the eigenspaces in "half." The Cantor set is, somewhat loosely speaking, the universal object which can be cut in half infinitely many times.

Edit, 6/13/22: Actually there's a much easier way of thinking about this counterexample. It just comes from wondering: what if instead of a direct sum decomposition we had a direct product decomposition
$$V \cong \prod_{(\lambda_i)_{i \in I}} \text{Eig}((f_i)_{i \in I}, (\lambda_i)_{i \in I})$$
and then trying to construct the minimal infinite such thing. One way to say this is, the desired statement can't possibly always hold because it's not stable under passing to dual spaces.
