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Exercise 3.B.25

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Proof. Let $T = T_{1}$ and consider the same notatition from 3.22.

Suppose $range T_{1} \subset range T_{2}$ . Let $ν_{1}, \dots, ν_{n} \in V$ be an inverse image of $T_{1} v_{1}, \dots, T_{1} v_{n}$ when $T_{2}$ is applied (this inverse image exists because of the initial assumption). Define $S \in L (V, V)$ by

\begin{aligned} S u_{j} & = 0, for j = 1, \dots, m \\ S v_{j} & = ν_{j}, for j = 1, \dots, n \end{aligned}

Then

\begin{aligned} T_{1} v & = T_{1} (a_{1} u_{1} + \dots + a_{m} u_{m} + b_{1} v_{1} + \dots b_{n} v_{n}) \\ = b_{1} T_{1} v_{1} + \dots + b_{n} T_{1} v_{n} \\ = b_{1} T_{2} ν_{1} + \dots + b_{n} T_{2} ν_{n} \\ = b_{1} T_{2} S v_{1} + \dots + b_{n} T_{2} S v_{n} \\ = T_{2} S (b_{1} v_{1} + \dots + b_{n} v_{n}) \\ = T_{2} S (a_{1} u_{1} + \dots + a_{m} u_{m}) + T_{2} S (b_{1} v_{1} + \dots + b_{n} v_{n}) \\ = T_{2} Sv \end{aligned}

Hence $T_{1} = T_{2} S$ .

Conversely, suppose $T_{1} = T_{2} S$ . Let $w \in range T_{1}$ . There exists $v \in V$ such that $T_{1} v = w$ . But $T_{2} Sv = T_{1} v = w$ and, because $Sv \in V$ , it follows that $w \in range T_{2}$ , completing the proof. □

celiopassos

2017-10-06 00:00

Exercise 3.B.25

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