Exercise 5.1.13

Answers

1.

Since the diagram is commutative, we know that

T (v) = ϕ_{β}^{- 1} (T (ϕ_{β} (v))) = ϕ_{β}^{- 1} (λ ϕ_{β} (v)) = λv .

2.

One part of this statement has been proven in the previous exercise. If

ϕ_{β}^{- 1} (y)

is an eigenvector of

T

corresponding to

λ

, we have

Ay = ϕ_{β} (T (ϕ_{β}^{- 1} (y))) = ϕ_{β} (λ ϕ_{β} (y)) = λy .

jephianlin

2011-06-27 00:00