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Exercise 3.D.4

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Proof. For the implication in the forward direction, by _Exercise 3_ we only need to prove the existence of an injective linear map $S \in L (range T_{2}, W)$ such that $T_{1} = S T_{2}$ .

Let $w_{1}, \dots, w_{n}$ be a basis of $range T_{2}$ and $v_{1}, \dots, v_{n} \in V$ a reverse image of it when $T_{2}$ is applied. Because $T_{2} v_{1}, \dots, T_{2} v_{n}$ is linearly independent, it follows that $null T_{2} \cap span (v_{1}, \dots, v_{n}) = 0$ . Hence $null T_{2} + span (v_{1}, \dots, v_{n})$ is a direct sum. We will show that $V = null T_{2} \oplus span (v_{1}, \dots, v_{n})$ . Let $v \in V$ . There are $a_{1}, \dots, a_{n} \in 𝔽$ such that

\begin{aligned} T_{2} v & = a_{1} w_{1} + \dots + a_{n} w_{n} \\ = a_{1} T_{2} v_{1} + \dots + a_{n} T_{2} v_{n} \\ = T_{2} (a_{1} v_{1} + \dots + a_{n} v_{n}) \end{aligned}

Then

T_{2} (v - (a_{1} v_{1} + \dots + a_{n} v_{n})) = T_{2} v - T_{2} (a_{1} v_{1} + \dots + a_{n} v_{n}) = T_{2} v - T_{2} v = 0

Thus $v - (a_{1} v_{1} + \dots + a_{n} v_{n}) \in null T_{2}$ . But $v = (v - (a_{1} v_{1} + \dots + a_{n} v_{n})) + (a_{1} v_{1} + \dots + a_{n} v_{n})$ . Hence $v \in null T_{2} \oplus span (v_{1}, \dots, v_{n})$ and $V \subset null T_{2} \oplus span (v_{1}, \dots, v_{n})$ .

Clearly, $null T_{2} \oplus span (v_{1}, \dots, v_{n}) \subset V$ , therefore $V = null T_{2} \oplus span (v_{1}, \dots, v_{n})$ .

Define $S \in L (range T_{2}, W)$ by

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

Because $null T_{1} \oplus span (v_{1}, \dots, v_{n})$ is also a direct sum, it follows that $T_{1} v_{1}, \dots, T_{1} v_{n}$ is linearly independent. Thus $S$ is injective. Let $v \in V$ . There is $u \in null T_{1}$ and $a_{1}, \dots, a_{n} \in F$ such that

v = u + a_{1} v_{1} + \dots + a_{n} v_{n}

Thus

\begin{aligned} T_{1} v & = T_{1} (u + a_{1} v_{1} + \dots + a_{n} v_{n}) \\ = a_{1} T_{1} v_{1} + \dots + a_{n} T_{1} v_{n} \\ = a_{1} S w_{1} + \dots + a_{n} S w_{n} \\ = a_{1} S T_{2} v_{1} + \dots + a_{n} S T_{2} v_{n} \\ = S T_{2} (a_{1} v_{1} + \dots + a_{n} v_{n}) \\ = S T_{2} u + S T_{2} (a_{1} v_{1} + \dots + a_{n} v_{n}) \\ = S T_{2} v \end{aligned}

Hence $T_{1} = S T_{2}$ , as desired.

Conversely, suppose $S$ is an invertible operator on $W$ such that $T_{1} = S T_{2}$ .

Let $v \in null T_{1}$ . Then

0 = T_{1} v = S T_{2} v

Because $S$ is invertible, it follows that $T_{2} v = 0$ . Hence $v \in null T_{2}$ .

For the inclusion in the other direction, let $v \in null T_{2}$ . We have

0 = S T_{2} v = T_{1} v

Hence $v \in null T_{1}$ and, therefore $null T_{1} = null T_{2}$ . □

celiopassos

2017-10-06 00:00

Exercise 3.D.4

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