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Exercise 3.B.28

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Proof. Define $S_{j} \in L (range T, 𝔽)$ such that $S_{j} (w_{j}) = 1$ and $S_{j} (w_{k}) = 0$ for $k \neq j$ . $S_{j}$ is well defined by 3.5 and we have that

S_{j} (c_{1} w_{1} + \dots + c_{m} w_{m}) = c_{j}

Now define $φ_{j} = S_{j} T$ , for $j = 1, \dots, m$ . Note that $φ_{j}$ is also linear, since it is the product of linear maps. Thus

\begin{aligned} Tv & = a_{1} w_{1} + \dots + a_{m} w_{m} \\ = (S_{1} Tv) w_{1} + \dots + (S_{m} Tv) w_{m} \\ = φ_{1} (v) w_{1} + \dots + φ_{m} (v) w_{m} \end{aligned}

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celiopassos

2017-10-06 00:00

Exercise 3.B.28

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