Diagonalizability of transpose (dual) operator for infinite-dimension vector spaces (purely algebraic version)

Question

I'm writing some notes on Linear Algebra and thought of the following question:

What is the relation between diagonalizability of a linear operator and its dual.

Here are my definitions: A linear operator $T\colon V\to V$ (on a vector space $V$ over a field $F$) is diagonalizable if $V=\sum_{\lambda\in F}\ker(T-\lambda)$. (This is equivalent to the usual definitions, such as existence of an eigenbasis; AC is assumed). The dual of a vector space $V$ is the vector space $V^*$ of linear maps $V\to F$ (functionals); The dual of the operator $T$ is the map $T^*\colon V^*\to V^*$ given by $T^*f = f\circ T$. The spectrum of $T$ is the set of eigenvalues: $Spec(T)=\left\{\lambda\in F\mid \ker(T-\lambda)\neq 0\right\}$.

I could prove the following:

Theorem: If $T$ is diagonalizable and $Spec(T)$ is finite, then $T^*$ is diagonalizable. Moreover, $Spec(T)=Spec(T^*)$.

Here's a sketch of the proof: Suppose $T$ is diagonalizable. An eigenvector $f$ of $T^*$ associated to an eigenvalue $\lambda$ is a function $f\colon V\to F$ such that $T^* f(v)=\lambda f(v)$ for every eigenvector $v$ of $T$; equivalently, $\lambda_vf(v)=\lambda f(v)$ for every eigenpair $(v,\lambda_v)$ of $T$. This is equivalent to saying that $\sum_{\mu\neq\lambda}\ker(T-\mu)\subseteq\ker(f)$. This also implies $Spec(T^*)\subseteq Spec(T)$.

So given any function $f\in V^*$, we decompose $V=\oplus_{i=1}^n\ker(T-\lambda_i)$, since $Spec(T)$ is finite and $T$. Let $f_i$ be the operator which is given as $f$ on $\ker(T-\lambda_i)$ and $0$ the other summands. By definition, we have, for any $i$, $\sum_{j\neq i}\ker(T-\lambda_j)\subseteq\ker(f_i)$, so $f_i$ is an eigenvector of $T^*$ (associated to $\lambda_i$) and $f=\sum_i f_i$. This same type of argument yields the inclusion $Spec(T)\subseteq Spec(T^*)$.

Now, I have a conjecture which I cannot prove:

Conjecture: If $T$ is not diagonalizable or $Spec(T)$ is infinite, then $T^*$ is not diagonalizable.

I tried to show that by checking where the proof in the theorem above goes awry without success. Any help is much appreciated.

Answer 1

This is true. We need to repeatedly use the following key lemma:

Key lemma: If $T^{\ast}$ is diagonalizable, it remains diagonalizable after quotienting by any invariant subspace of $V^{\ast}$. Taking the contrapositive, if $T^{\ast}$ is not diagonalizable on some such quotient then it is not diagonalizable.

We'll also need to use the fact that the action of $T$ equips both $V$ and $V^{\ast}$ with the structure of a $K[T]$-module, where $K$ is the underlying field, and that in terms of the module structure, a linear operator $X$ acts diagonalizably iff the induced $K[X]$-module is a direct sum of modules of the form $K[X]/(X - \lambda)$ (this is a slightly stronger condition than the condition that the module is semisimple).

The argument splits into a few cases. We need to use the axiom of choice below, although it is heavily disguised, in order to conclude that if $f : U \to W$ is an injection then $f^{\ast} : W^{\ast} \to U^{\ast}$ is a surjection, so that nontrivial $K[T]$-submodules of $V$ do in fact correspond to nontrivial $K[T]$-quotient modules of $V^{\ast}$.

Case 1: $T$ has either 1) a generalized eigenvector which is not an eigenvector, or 2) an eigenvector over a field extension of $F$ whose eigenvalue is not in $F$. Then $V$ contains $F[T]/f(T)$ as a submodule where either $f(T) = (T - \lambda)^2$ or $f(T)$ is irreducible of degree greater than $1$ over $F[T]$. It follows that $V^{\ast}$ also contains $F[T]/f(T)$ as a quotient module, hence $T^{\ast}$ does not act diagonalizably on this quotient.

Case 2: $T$ has infinite spectrum. Then $V$ contains $\bigoplus_{\lambda} F[T]/(T - \lambda)$ as a submodule, where the sum runs over some infinite set of distinct eigenvalues. It follows that $V^{\ast}$ admits $\prod_{\lambda} F[T]/(T - \lambda)$ as a quotient module, so it suffices to show that $T^{\ast}$ acting on such an infinite product is not diagonalizable.

But it's clear that the direct sum of the eigenspaces of $\prod_{\lambda} F[T]/(T - \lambda)$ is $\sum_{\lambda} F[T]/(T - \lambda)$ and that this is a proper submodule, since for example it does not contain $\prod_{\lambda} 1$.

Case 3: $T$ has a torsion-free vector (a vector $v \in V$ generating a submodule isomorphic to $F[T]$; in particular, a vector which cannot be written as a sum of eigenvectors). Then $V$ contains $F[T]$ as a submodule. It follows that $V^{\ast}$ admits $F[T]^{\ast}$ as a quotient module, so it suffices to show that $T^{\ast}$ acting on $F[T]^{\ast}$ is not diagonalizable.

$F[T]^{\ast}$, as an $F[T]$-module, can be explicitly described as exponential (divided power) power series

$$F \langle \langle S \rangle \rangle = \left\{ \sum a_k \frac{S^k}{k!} : a_k \in K \right\}$$

(the factorial divisions are just notation, this works in any characteristic) with $T^{\ast}$ acting by the derivative $\frac{\partial}{\partial S}$. The isomorphism with $F[T]^{\ast}$ sends $\frac{S^k}{k!}$ to the linear functional $e_k : F[T] \to F$ which extracts the $k^{th}$ coefficient of a polynomial; this corresponds to thinking of $\frac{S^k}{k!}$ as acting by $\frac{1}{k!} \frac{\partial^k}{\partial T^k}$ together with evaluation at $0$. It follows from this identification that $T^{\ast}(e_k) = e_{k-1}$, hence the isomorphism with the derivative.

This means the eigenvectors of $T^{\ast}$ acting on $F[T]^{\ast}$ are given by the exponentials

$$e^{\lambda S} = \sum \lambda^k \frac{S^k}{k!}$$

(again, the factorial divisions are just notation, this works in any characteristic), so it suffices to show that these eigenvectors span a proper submodule. In turn it suffices to show that there exists an exponential power series $f$ which does not satisfy a linear homogeneous constant-coefficient differential equation $p \left( \frac{\partial}{\partial S} \right) f = 0$. This can be done with a cardinality argument but it's more fun to be explicit; over any field $F$ we can take a lacunary power series such as

$$f(S) = \sum_{n \ge 0} \frac{S^{n^2}}{(n^2)!}.$$

We can show that $f$ does not satisfy a differential equation of the form $p \left( \frac{\partial}{\partial S} \right) f = 0$ where $\deg p = d$ by looking at what happens once the terms of $f$ get farther apart than $d$; the recurrence is no longer able to cancel them, so $p \left( \frac{\partial}{\partial S} \right) f \neq 0$ from that point on.

Corollary: If $T$ has infinite spectrum or is not diagonalizable, then $T^{\ast}$ is not diagonalizable.