Exercise 2.1.21

Answers

1.

To prove

T

is linear we can check

T (σ_{1} + σ_{2}) (n) = σ_{1} (n + 1) + σ_{2} (n + 1) = T (σ_{1}) (n) + T (σ_{2}) (n)

and

T (cσ) (n) = cσ (n + 1) = cT (σ) (n) .

And it’s similar to prove that $U$ is linear.

2.

It’s onto since for any

σ

V

. We may define another sequence

τ

such that

τ (0) = 0

and

τ (n + 1) = σ (n)

for all

n \geq 1

. Then we have

T (τ) = σ

. And it’s not one-to-one since we can define a new

σ_{0}

with

σ_{0} (0) = 1

and

σ_{0} (n) = 0

for all

n \geq 2

. Thus we have

σ_{0} \neq 0

but

T (σ_{0}) = 0

3.

T (σ) (n) = σ (n - 1) = 0

for all

n \geq 2

, we have

σ (n) = 0

for all

n \geq 1

. And let

σ_{0}

be the same sequence in the the previous exercise. We cannot find any sequence who maps to it.

jephianlin

2011-06-27 00:00