Exercise $14$, Section $3.A$ - Linear Algebra Done Right

Question

Exercise: Suppose $V$ is finite-dimensional with $\dim V \ge 2$. Prove that there exist $S,T \in L(V, V)$ such that $ST \ne TS$

$L(V, V)$ denotes the set of all linear maps from $V\to V$.

Proof: Define $T : V\to V$ by $T(a_1v_1+\dots+a_nv_n)=2a_1v_1+\dots+a_nv_n$ and define $S: V\to V$ by $S(a_1v_1+\dots+a_nv_n)=(a_1+\dots+a_n)v_1+\dots+(a_1+\dots+a_n)v_n$. Pick non-zero scalars $a_1,\dots,a_n\in F$.

We see that $(ST)v=S(2a_1v_1+\dots+a_nv_n)=(2a_1+\dots+a_n)v_1+\dots+(2a_1+\dots+a_n)v_n$ and

$(TS)v=T((a_1+\dots+a_n)v_1+\dots+(a_1+\dots+a_n)v_n)=(2a_1+\dots+2a_n)v_1+\dots+(a_1+\dots+a_n)v_n$.

This shows that $ST\ne TS$.

Is this solution correct?

Edit: Let $v_1,\dots,v_n$ be a basis of V.

Answer 1

Your proof is correct but not efficient. Since $\mathrm{dim}(V)=2$, we have $\exists B=\{\alpha_1,\alpha_2\}$ basis of $V$. Linear map is uniquely determined by it’s action on basis $B$. It’s easy to check $\exists !S,T\in L(V,V)$ such that $S(\alpha_1)=\alpha_2$, $S(\alpha_2)=\alpha_1$ and $T(\alpha_1)=\alpha_1$, $T(\alpha_2)=\alpha_1$. Then $S\circ T(\alpha_1)$ $=S(T(\alpha_1))$ $=S(\alpha_1)$ $=\alpha_2$ and $T\circ S(\alpha_1)$ $=T(S(\alpha_1))$ $=T(\alpha_2)$ $=\alpha_1$. Thus $S\circ T(\alpha_1)\neq T\circ S(\alpha_1)$. Hence $S\circ T\neq T\circ S$.