Locally Finite Operator

Question

I didn't fully understand what a locally finite operator is, so it would be beneficial to see an example. And could you give me an example of an infinite vector space (I guessed that complex numbers field might be infinite but a field is always 1 dimensional)? Thank you!

Answer 1

Let $V$ be a vector space. An operator $T: V \to V$ is called locally finite if we can write $V = \bigcup_{i \in I} U_i$ for some collection of subspaces $\{U_i\}_{i \in I}$ such that each $U_i$ is both finite-dimensional and invariant under $T$, meaning $T U_i \subseteq U_i$. Here are some examples:

1. If $V$ is finite-dimensional, then every $T: V \to V$ is locally finite.
2. If $V$ has countable basis $v_1, v_2, \ldots$, then the identity operator is locally finite, we may take $U_i = \mathrm{span}\langle v_1, \ldots, v_i \rangle$.
3. If $V$ has countable basis $v_1, v_2, \ldots$, then define the "left-shift" operator $T: V \to V$ as $Tv_1 = 0$, and $Tv_i = v_{i-1}$ for $i > 1$. Then $T$ is locally finite, by taking $U_i = \mathrm{span}\langle v_1, \ldots, v_i \rangle$.

An example of an operator which is not locally finite would be a right-shift operator, which arises when you consider $V = \mathbb{C}[t]$, the vector space of polynomials in one variable, and $T: V \to V$ acting by multiplication by $T$, so for example $T(t^2 + 2t) = t^3 + 2t^2$. This operator has no (nonzero) finite-dimensional invariant subspace $U$, since if $p(t) \in U$ then so is $t p(t), t^2 p(t)$, and so on.