Exercise 5.C.10

Answers

Proof. If $0$ is an eigenvalue of $T$ , then

\begin{aligned} \dim range T & = \dim V - \dim null T \\ = \dim V - \dim E (0, T) \\ \geq \dim (E (0, T) \oplus E (λ_{1}, T) \oplus \dots \oplus E (λ_{m}, T)) - \dim E (0, T) \\ = \dim E (0, T) + \dim E (λ_{1}, T) + \dots + \dim E (λ_{m}, T) - \dim E (0, T) \\ = \dim E (λ_{1}, T) + \dots + \dim E (λ_{m}, T) \end{aligned}

Where the third line follows from 5.38.

If $0$ is not an eigenvalue of $T$ , just disregard $E (0, T)$ and $\dim E (0, T)$ in each line. □

celiopassos

2017-10-06 00:00