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Exercise 5.C.12

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Proof. Let $v_{1}, v_{2}, v_{3}$ and $w_{1}, w_{2}, w_{3}$ be eigenvectors of $R$ and $T$ , respectively, corresponding to $2, 6, 7$ . Note that the $v$ ’s and $w$ ’s are both bases of $𝔽^{3}$ . Define $S \in L (𝔽^{3})$ by

\begin{aligned} S v_{1} & = w_{1} \\ S v_{2} & = w_{2} \\ S v_{3} & = w_{3} \end{aligned}

It is easy to see that $S$ is invertible. Thus

\begin{aligned} S^{- 1} TS (a_{1} v_{1} + a_{2} v_{2} + a_{3} v_{3}) & = S^{- 1} T (a_{1} w_{1} + a_{2} w_{2} + a_{3} w_{3}) \\ = S^{- 1} (2 a_{1} w_{1} + 6 a_{2} w_{2} + 7 a_{3} w_{3}) \\ = 2 a_{1} v_{1} + 6 a_{2} v_{2} + 7 a_{3} v_{3} \\ = R (a_{1} v_{1} + a_{2} v_{2} + a_{3} v_{3}) \end{aligned}

Therefore $R = S^{- 1} TS$ . □

celiopassos

2017-10-06 00:00

Exercise 5.C.12

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