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Exercise 5.B.5

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Proof. Note that, for non-negative $k$ , ${(ST S^{- 1})}^{k} = S T^{k} S^{- 1}$ , because the $S$ ’s and $S^{- 1}$ ’s cancel out (this can easily me proven by induction on $k$ ). Let $n = \deg p$ and $a_{0}, a_{1}, \dots, a_{n} \in 𝔽$ be the coefficients of $p$ . Then

\begin{aligned} p (ST S^{- 1}) & = \sum_{k = 0}^{n} a_{k} {(ST S^{- 1})}^{k} \\ = \sum_{k = 0}^{n} a_{k} S T^{k} S^{- 1} \\ = S (\sum_{k = 0}^{n} a_{k} T^{k}) S^{- 1} \\ = Sp (T) S^{- 1} \end{aligned}

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celiopassos

2017-10-06 00:00

Exercise 5.B.5

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