Exercise 6.6.7

Answers

1.

It comes from Theorem 6.25(c).

2.

T_{0} = T^{n} = \sum_{i = 1}^{k} λ_{i}^{n} T_{i} .

Now pick arbitrary eigenvector $v_{i}$ for the eigenvalue $λ_{i}$ . Then we have

0 = T (v_{i}) = (\sum_{i = 1}^{k} λ_{i}^{n} T_{i}) (v_{i}) = λ_{i}^{n} v_{i} .

This means that $λ_{i}^{n} = 0$ and so $λ_{i} = 0$ for all $i$ . Hence we know that

T = \sum_{i = 1}^{k} λ_{i} T_{i} = T_{0} .

3.

By the Corollary 4 after the Spectral Theorem, we know that

T_{i} = g_{i} (T)

for some polynomial

g_{i}

. This means that

U

commutes with each

T_{i}

U

commutes with

T

. Conversely, if

U

commutes with each

T_{i}

, we have

TU = (\sum_{i = 1}^{k} λ_{i} T_{i}) U = \sum_{i = 1}^{k} λ_{i} T_{i} U

= \sum_{i = 1}^{k} λ_{i} U T_{i} = U (\sum_{i = 1}^{k} λ_{i} T_{i}) = UT .

4.

Pick

U = \sum_{i = 1}^{k} λ_{i}^{\frac{1}{2}} T_{i},

where $λ_{i}^{\frac{1}{2}}$ is an arbitrary square root of $λ_{i}$ .

5.

Since

T

is a mapping from

V

V

T

is invertible if and only if

N (T) = {0}

. And

N (T) = {0}

is equivalent to that

0

is not an eigenvalue of

T

6.

If every eigenvalue of

T

1

0

. Then we have

T = 0 T_{0} + 1 T_{1} = T_{1}

, which is a projection. Conversely, if

T

is a projection on

W

along

W^{'}

, we may write any element in

V

u + v

such that

u \in W

and

v \in W^{'}

. And if

λ

is an eigenvalue, we have

u = T (u + v) = λ (u + v)

and so

(1 - λ) u = λv .

Then we know that the eigenvalue could only be $1$ or $0$ .

7.

It comes from the fact that

T^{∗} = \sum_{i = 1}^{k} \bar{λ_{i} T_{i}} .

jephianlin

2011-06-27 00:00