In an exercise we have to show that T: P(R) -> P(R) by T(p)(x) = p(x) - p(0): a) T is linear b) Is T one-to-one? c) Is T onto? I tried to sovle this exercise but I'm not sure whether it's correct.
In an exercise we have to show that T: P(R) -> P(R) by T(p)(x) = p(x) - p(0): a) T is linear b) Is T one-to-one? c) Is T onto? I tried to sovle this exercise but I'm not sure whether it's correct.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
In an exercise we have to show that T: P(R) -> P(R) by T(p)(x) = p(x) - p(0):
a) T is linear
b) Is T one-to-one?
c) Is T onto?
I tried to sovle this exercise but I'm not sure whether it's correct.
![E1:
T: P(R) -→ P(R) by T(p)(x) = p(x) - plo)
allet piqEP(IR) and letaEIR
Then Tip+q)(x) = P(x) +9(x) - [p(0)-9(0)]
= (P(x)-p(0)) + (9(0)-9(0))
I (p) (x) + T(g)(x)
= T (p+q) (x)
・・ T (ap)(x) = α (P)(x) - α(p)(0)
=αp(x-c)
=dT (P)(x)
DATE
11-4/17
b) Consider pig where
p(x) = 1 for all x
Q(x) = x-1. Then T(p) = P(x) = 1 but g(x) = 1-1=0.
Since 170 and so ker T7{0} ( 1€ kert)
Hence, I is not ^-^.
C) Let (ETR and let p(x) = 2 € P(R).
Then there does not exist a constant polynomial whose range
under I is 2
(e.g., TCC) = (-C=C₁ VCER)
So there does not exist p(x)ERR st. T(p)(x) = 2.
Hence, I is not onto.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F01b4eff1-1c92-4d86-9563-264f29556131%2F0d5bf443-7e60-42c6-acf4-993847c15c3d%2Fjvmwa2f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:E1:
T: P(R) -→ P(R) by T(p)(x) = p(x) - plo)
allet piqEP(IR) and letaEIR
Then Tip+q)(x) = P(x) +9(x) - [p(0)-9(0)]
= (P(x)-p(0)) + (9(0)-9(0))
I (p) (x) + T(g)(x)
= T (p+q) (x)
・・ T (ap)(x) = α (P)(x) - α(p)(0)
=αp(x-c)
=dT (P)(x)
DATE
11-4/17
b) Consider pig where
p(x) = 1 for all x
Q(x) = x-1. Then T(p) = P(x) = 1 but g(x) = 1-1=0.
Since 170 and so ker T7{0} ( 1€ kert)
Hence, I is not ^-^.
C) Let (ETR and let p(x) = 2 € P(R).
Then there does not exist a constant polynomial whose range
under I is 2
(e.g., TCC) = (-C=C₁ VCER)
So there does not exist p(x)ERR st. T(p)(x) = 2.
Hence, I is not onto.
Expert Solution
Step 1
Let us consider the transformation .
a) T is linear if .
b) T is one-one if then .
c) T is onto if For any there is thus, .
Step by step
Solved in 4 steps
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