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Exercise 3.B.21

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Proof. Suppose $T$ is surjective. Let $w_{1}, \dots, w_{m}$ be a basis of $W$ . There are $v_{1}, \dots, v_{m} \in V$ such that

T v_{j} = w_{j}, for j = 1, \dots, m

Define $S \in L (W, V)$ by

S w_{j} = v_{j}, for j = 1, \dots, m

Now suppose $w \in W$ . We have

w = a_{1} w_{1} + \dots + a_{m} w_{m}

For some scalars $a_{1}, \dots, a_{m}$ . Therefore

\begin{aligned} TSw & = TS (a_{1} w_{1} + \dots + a_{m} w_{m}) \\ = a_{1} TS w_{1} + \dots + a_{m} TS w_{m} \\ = a_{1} T v_{1} + \dots + a_{m} T v_{m} \\ = a_{1} w_{1} + \dots + a_{m} w_{m} \\ = w \end{aligned}

Thus $TS$ is the identity map on $W$ .

Conversely, suppose $TS$ is the identity map on $W$ . Let $w \in W$ , we have $TSw = w$ . Since $Sw \in V$ , it follows that there is a vector in $V$ that $T$ maps to $W$ . Since $w$ was arbitrary, we have that $T$ is surjective. □

celiopassos

2017-10-06 00:00

Exercise 3.B.21

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