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Exercise 3.B.24

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Proof. Suppose $null T_{1} \subset null T_{2}$ . Let $w_{1}, \dots, w_{n}$ be a basis of $range T_{1}$ and $v_{1}, \dots, v_{n} \in V$ an inverse image of this basis when $T_{1}$ is applied. We will prove that $V = null T_{1} + span (v_{1}, \dots, v_{n})$ .

Suppose $v \in V$ . There are $a_{1}, \dots, a_{n} \in 𝔽$ such that

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

Let $ν = a_{1} v_{1} + \dots + a_{n} v_{n}$ . We have that $0 = T_{1} v - T_{1} ν = T_{1} (v - ν)$ . Hence $v - ν \in null T_{1}$ . But $v = (v - ν) + ν$ , therefore $v \in null T_{1} + span (v_{1}, \dots, v_{n})$ , implying $V \subset null T_{1} + span (v_{1}, \dots, v_{n})$ . The inclusion in the other direction is clearly true, hence $V = null T_{1} + span (v_{1}, \dots, v_{n})$ .

Let $S \in L (W, W)$ be any linear map such that

S w_{j} = T_{2} v_{j}, for j = 1, \dots, n

Let $v \in V$ . There are $u \in null T_{1}$ and $c_{1}, \dots, c_{n} \in F$ such that

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

Thus

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

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

Conversely, suppose $T_{2} = S T_{1}$ . Let $u \in null T_{1}$ . Then

0 = S (0) = S T_{1} u = T_{2} u

Hence $u \in null T_{2}$ , as desired. □

celiopassos

2017-10-06 00:00

Exercise 3.B.24

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