About a problem of linear transformation $ ST-TS=I $

Question

The problem is, $S$ & $T$ are two linear operator i.e. belongs to $$L(V)$$ such that $ST-TS=I_n$ then prove that $L(V)$ is infinite dimensional vector space.

Now, I have showed that for all $n\in \mathbb{N}$ $S^{n+1}T-TS^{n+1}=(n+1)S^n$ by induction procedure. And I have showed that $\forall n\in\mathbb{N}$ $S^n\neq 0$, I mean zero linear operator.

How then I prove that $L(V)$ is infinite dimensional. Obviously I have to find a linearly independent set of $n$ many linear operators of for each $n\in\mathbb{N}$. But how construct it. May be $S^n$ are linearly independent to each other; not proved yet. Any help will be appreciated.

Answer 1

If $V$ is a finite-dimensional vector space, you can show that such a thing cannot happen by taking traces on both sides.

In cases of positive characteristic, there exists a counterexample: over $\mathbb{F}_{2}$, we have $ST-TS = I_{2}$ where $$ S = \begin{pmatrix} 0&1 \\0 & 0\end{pmatrix}, \quad T = \begin{pmatrix} 0 & 0 \\ 1& 0\end{pmatrix}. $$

Answer 2

By induction, we find that

$$ S^nT - TS^n = \begin{cases} nS^{n-1}, & n \geq 1 \\ 0, & n = 0 \end{cases} $$

We claim the following:

Claim. $p(S) \neq 0$ for any polynomial $p \in F[x]$.

In view of the Cayley-Hamilton theorem, this is enough to argue that $V$ is not finite-dimensional.

Proof of Claim. Assume otherwise and let $m(x)$ be the minimal polynomial of $S$. (This is simply the generator of the principal ideal $\{p(x) \in F[x] : p(S) = 0 \}$.) If we write $m(x) = \sum_{k} a_k x^k$, then

$$ 0 = m(S)T - Tm(S) = \sum_{k\geq 0} a_k (S^k T - TS^k) = \sum_{k\geq 1} k a_k S^{k-1} = m'(S). $$

This contradicts the minimality of $m$.