TITLE: Subspace of $L(V)$
QUESTION [6 upvotes]: Suppose $V$ is finite dimensional and $E$ is a subspace of $L(V)$ such that $ST\in E$ and $TS \in E$ for all $S \in L(V)$ and $T\in E$. Prove that $E=\{0\}$ or $E =L(V)$.
When $E$ is non-trivial, I want to prove that $1 \in E$. Then we can deduce that $L(V) \subset E$ and we are done. However, I don't know how to use the fact '$ST\in E$ and $TS \in E$'.
To me, $L(V)$ is a ring and clearly $E$ is an ideal of ring. Still, why $1 \in E?$

REPLY [3 votes]: Let $v_1,\ldots,v_n$ be a basis for $V$. Let $0\neq T\in E$. So exists $ v\in V$ such that $T(v)\neq 0$. 
Let $R_i\in L(V)$ be such that $R_i(v_i)=v$ and $R_{i}(v_j)=0$, for $1\leq i\neq j\leq n$. 
Let $S_i\in L(V)$ be such that $S_i(T(v))=v_i$, for $1\leq i\leq n$.
Thus, $\sum_{i=1}^nS_iTR_i(v_j)=S_jTR_j(v_j)=S_jT(v)=v_j$ for $1\leq j\leq n$. Thus, $\sum_{i=1}^nS_iTR_i=Id$ and since $S_iTR_i\in E$, we obtain $Id\in E$.
If $S\in L(V)$ then $Id.S=S\in E$.