Problem: Suppose $V$ is finite-dimensional with $\dim V \geq 2$. Prove that there exist $S, T \in \mathcal{L} (V,V)$ such that $ST \neq TS$, where $\mathcal{L} (V,V) $ is the set of all linear maps from $V$ to $V$.
Solution: Let $\dim V = n\geq 2$. Let $\mathcal{B} = \left\{ v_1, v_2, \ldots, v_n \right\} $ be a basis for $V$. Define $S, T \in \mathcal{L} (V,V)$ by
\[
S(v_i) =
\begin{cases}
v_1 & i = 1 \\
0 & \text{ otherwise},
\end{cases}
\]
and
\[
T(v_i) =
\begin{cases}
v_2, & i = 1 ;\\
0, & \text{ otherwise}.
\end{cases}
\]
Then,
\[
ST(v_1) = S(v_2) = 0 \quad TS(v_1) = T(v_1) = v_2.
\]
Thus,
\[
ST \neq TS.
\]