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Exercise 5.C.6

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Proof. Let $λ_{1}, \dots, λ_{m}$ denote the distinct eigenvalues of $T$ , $v_{1}, \dots, v_{n}$ corresponding eigenvectors and $α_{1}, \dots, α_{n}$ the corresponding eigenvalues of $S$ . Then

\begin{aligned} ST v_{j} & = S (λ_{j} v_{j}) \\ = α_{j} λ_{j} v_{j} \\ = α_{j} T v_{j} \\ = T (α_{j} T v_{j}) \\ = TS v_{j} \end{aligned}

for $j = 1, \dots, n$ . Thus, by 3.5, $ST = TS$ . □

celiopassos

2017-10-06 00:00

Exercise 5.C.6

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