Exercise 5.1.23

Answers

If $T$ is diagonalizable, there is a basis $β$ consisting of eigenvectors. By the previouse exercise, we know that if $T (x) = λx$ ,

f (T) (v) = f (λ) v = 0 v = 0 .

This means that for each $v \in β$ , we have $f (T) (v) = 0$ . Since $β$ is a basis, $f (T) = T_{0}$ .

jephianlin

2011-06-27 00:00