Exercise 3.A.14

Answers

Proof. Let $v_{1}, \dots, v_{n}$ be a basis of $V$ . Define $S, T \in L (V, V)$ such that

\begin{aligned} S (a_{1} v_{1} + \dots + a_{n} v_{n}) & = a_{n} v_{1} + \dots + a_{1} v_{n} \\ T (a_{1} v_{1} + \dots + a_{n} v_{n}) & = a_{1} v_{1} + \dots + a_{n - 1} v_{n - 1} . \end{aligned}

By 3.5 $S$ and $T$ are well defined linear maps. But

\begin{aligned} ST v_{n} & = S 0 = 0 \\ TS v_{n} & = T v_{1} = v_{1} \end{aligned} .

Hence $ST \neq TS$ . □

celiopassos

2017-10-06 00:00