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Exercise 3.F.16

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Proof. Let $T_{1}, \dots, T_{n}$ be a basis of $L (V, W)$ . Define $R : L (V, W) \to L (W^{'}, V^{'})$ by

R T_{j} = T_{j}^{'}, for j = 1, \dots, n

Let $T \in L (V, W)$ and $ψ \in W^{'}$ . There are scalars $a_{1}, \dots, a_{n}$ such that $T = a_{1} T_{1} + \dots + a_{n} T_{n}$ . We have

\begin{aligned} (RT) ψ & = (R (a_{1} T_{1} + \dots + a_{n} T_{n})) ψ \\ = (a_{1} R T_{1} + \dots + a_{n} R T_{n}) ψ \\ = (a_{1} T_{1}^{'} + \dots + a_{n} T_{n}^{'}) ψ \\ = a_{1} T_{1}^{'} ψ + \dots + a_{n} T_{n}^{'} ψ \\ = ψ \circ (a_{1} T_{1}) + \dots + ψ \circ (a_{n} T_{n}) \\ = ψ \circ (a_{1} T_{1} + \dots + a_{n} T_{n}) \\ = ψ \circ T \\ = T^{'} (ψ) \end{aligned}

Hence $R$ satisfies the desired property of taking $T$ to $T^{'}$ . (I think showing the existence of $R$ like this wasn’t necessary since the question already presumes the existence of such map, but I kept it anyway)

We but need to prove that $\dim null R = 0$ , since $L (V, W)$ and $L (W^{'}, V^{'})$ have the same dimension. Suppose that $T \in null R$ . We have that $0 = RT = T^{'}$ . By Exercise 15, $T = 0$ , therefore $null R = 0$ as desired. □

celiopassos

2017-10-06 00:00

Exercise 3.F.16

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