Suppose $V$ is finite-dimensional and $\mathcal E$ is a subspace of $\mathcal L(V)$ such that $ST \in \mathcal E$ and $TS \in \mathcal E$ for all $S \in \mathcal L(V)$ and all $T \in \mathcal E$. Prove that $\mathcal E = \{0\}$ or $\mathcal E = \mathcal L(V)$.

Suppose $\mathcal E \ne \{0\}$, I'm going to show this implies $\mathcal E = \mathcal L(V)$.

Let $T$ be a nonzero map in $\mathcal E$ and let $v \in V$ be such that $Tv \ne 0$.
Let $v_1 = v$ and extend to a basis $v_1,\dots,v_n$ of $V$, let $A$ be the matrix of $T$ with respect to this basis.

Basically, we extract the nonzero component of $A$ by multiplying it by elementary matrices (row operations) to get

(In general more then 2d but latexifying matrices is annoying)

Then we add them to get that the identity is in $\mathcal E$ (subspace of $\mathcal E$ is closed under linear combinations), which allows us to show $\mathcal E = \mathcal L(V)$ because for any $S \in \mathcal L(V)$, $IS = S \in \mathcal E$.

todo: Provide details (I'm lazy right now)