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Exercise 8.C.10

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Proof. Let $λ_{1}, \dots, λ_{m}$ denote the distinct eigenvalues of $T$ and $d_{1}, \dots, d_{m}$ their corresponding multiplicities. Then

p (z) = {(z - λ_{1})}^{d_{1}} \dots {(z - λ_{m})}^{d_{m}} .

The eigenvalues of $T^{- 1}$ are $\frac{1}{λ_{1}}, \dots, \frac{1}{λ_{m}}$ and by Exercise 3 in section 8A they have multiplicities $d_{1}, \dots, d_{m}$ . Thus

\begin{aligned} q (z) & = {(z - \frac{1}{λ_{1}})}^{d_{1}} \dots {(z - \frac{1}{λ_{m}})}^{d_{m}} \\ = z^{d_{1}} {(1 - \frac{1}{λ_{1} z})}^{d_{1}} \dots z^{d_{m}} {(1 - \frac{1}{λ_{m} z})}^{d_{m}} \\ = z^{d_{1} + \dots + d_{m}} {(1 - \frac{1}{λ_{1} z})}^{d_{1}} \dots {(1 - \frac{1}{λ_{m} z})}^{d_{m}} \\ = z^{\dim V} {(1 - \frac{1}{λ_{1} z})}^{d_{1}} \dots {(1 - \frac{1}{λ_{m} z})}^{d_{m}} \\ = z^{\dim V} \frac{1}{λ_{1}^{d_{1}}} {(λ_{1} - \frac{1}{z})}^{d_{1}} \dots \frac{1}{λ_{m}^{d_{m}}} {(λ_{m} - \frac{1}{z})}^{d_{m}} \\ = z^{\dim V} \frac{1}{λ_{1}^{d_{1}} \dots λ_{m}^{d_{m}}} {(λ_{1} - \frac{1}{z})}^{d_{1}} \dots {(λ_{m} - \frac{1}{z})}^{d_{m}} \\ = z^{\dim V} \frac{{(- 1)}^{d_{1} + \dots + d_{m}}}{λ_{1}^{d_{1}} \dots λ_{m}^{d_{m}}} {(\frac{1}{z} - λ_{1})}^{d_{1}} \dots {(\frac{1}{z} - λ_{m})}^{d_{m}} \\ = z^{\dim V} \frac{{(- 1)}^{d_{1} + \dots + d_{m}}}{λ_{1}^{d_{1}} \dots λ_{m}^{d_{m}}} p (\frac{1}{z}) \\ = z^{\dim V} \frac{1}{p (0)} p (\frac{1}{z}) . \end{aligned}

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celiopassos

2017-10-06 00:00

Exercise 8.C.10

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