Suppose $V$ is finite-dimensional and $E$ is a subspace of $\mathscr L(V)$

Question

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 $I\in E$.

Answer 1

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 $E=\mathscr{L}(V)$.

Answer 2

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 $$ T_1 + T_2 + \cdots + T_n = I $$