Suppose $V$ is finite-dimensional and $E$ is a subspace of $\mathscr L(V)$ Suppose $V$ is finite-dimensional and $E$ is a subspace of $\mathscr L(V)$ such that $ST\in E$ and $TS \in E$ for all $S \in \mathscr L(V)$ and all $T\in E$. Prove that $E = \{0\}$ or $E=\mathscr L(V)$. 
I have started the proof, but I get lost and am not sure how to finish out what I have: 
Suppose $v_1,\ldots,v_n$ is a basis of $V$. If $E=\{0\}$, we are done. Suppose $E\neq\{0\}$, then there exists a nonzero $T\in E$, which means there exists some $v_k\in\{v_1,\ldots,v_n\} $ such that $T(v_k)\neq0$. Let $a_1,\ldots,a_n\in \Bbb F$ such that $T(v_k)=a_1v_1+\cdots+a_nv_n\neq0$ meaning there exists some $a_l\in \{a_1,\ldots,a_n\}$ such that $a_l\neq0$. 
Clearly, I'll need to incorporate the fact that $ST$ and $TS$ are in $E$, and hopefully get to the point that $I\in E$. 
 A: You can choose the basis you like. Let $T\ne0$, $T\in E$. Suppose $T(v_1)\ne0$, so in particular $v_1\ne0$.
Complete $v_1$ to a basis $\{v_1,\dots,v_n\}$.
Since $w_1=T(v_1)\ne0$, you can complete it to a basis $\{w_1,\dots,w_n\}$. Consider the maps $R_i$ and $S_i$ defined by
$$
R_i(v_j)=\begin{cases}
v_1 & j=i\\[4px]
0 & j\ne i
\end{cases}
\qquad
S_i(w_j)=\begin{cases}
v_i & j=1 \\[4px]
0 & j>1
\end{cases}
$$
Then
$$
S_i T R_i (v_i)=S_i T(v_1)=S_i(w_1)=v_i
$$
and, for $j\ne i$,
$$
S_i T R_i (v_j)=S_iT(0)=0
$$
Hence, if $U=\sum_{i=1}^n S_iTR_i\in E$, we have
$$
U(v_j)=v_j
$$
for $j=1,\dots,n$. So $U$ is the identity and therefore $E=\mathscr{L}(V)$.
A: Here's one way to show that $I \in E$:
Let $T$ be a non-zero element of $E$.  Let $v_1$ be a vector such that $T(v_1) \neq 0$.  Extend $v_1$ into a basis $\{v_1,v_2,\dots,v_n\}$.
Select $S$ so that $ST$ is a the linear map satisfying
$$
ST(v_k) = \begin{cases}
v_1 & k=1\\
0 & k \neq 1
\end{cases}
$$
Call this map $T_1$; we have now shown that $T_1 \in E$. Now, let $P_j$ denote the linear map satisfying
$$
P_j(v_k) = \begin{cases}
v_1 & k=j\\
v_j & k=1\\
0 & k \notin \{j,1\}
\end{cases}
$$
Let $T_j$ denote the map $P_jT_1P_j$.  We have shown that $T_j \in E$ for all $j$.  Verify that
$$
T_1 + T_2 + \cdots + T_n = I
$$
