Finding Kernel and Range for Linear Transformation L(p(x)) = xp'(x) in P3

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Homework Help Overview

The discussion revolves around finding the kernel and range of the linear transformation L(p(x)) = xp'(x) within the space of polynomials P3. Participants are exploring the definitions and implications of kernel and range in the context of linear transformations applied to polynomials.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to apply the transformation to specific polynomial forms and analyze the results. Questions are raised about the definitions of kernel and range, particularly regarding the conditions under which polynomials belong to these sets.

Discussion Status

The discussion is active, with participants sharing their interpretations of polynomial degrees and the application of the transformation. Some guidance is offered regarding the correct method to find the kernel, emphasizing the need to set L(p(x)) to zero rather than individual coefficients. Multiple interpretations of the definitions and processes are being explored.

Contextual Notes

There is some confusion regarding the definitions of Pn and the implications for the kernel and range, as well as the assumptions made about polynomial coefficients. Participants are questioning the approach to determining the kernel and range based on their previous learning experiences.

loli12
Hi, does anyone know how to figure out the kernal and range for this linear transformation from P3 into P3 : L(p(x)) = xp'(x)?
I thought ker(L)= {0} and range is P3. But the correct answer is ker (L) = P1, L(P3) = Span (x^2, x). Can someone explain to me how exactly do we fine the kernel and range for this? Thanks!
 
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Pn is the polys of degree n right? Take an arbitrary degree three poly:

a+bx+cx^2+dx^3

and apply L to it.

What has been killed and what is left?
 
Well, my book defined Pn to be polynomials of less than n degree.
if I apply the L to ax^2 + bx + c, then i get L(p(x)) = 2ax^2 + bx + 0
i remembered from one of the ques that my teacher did, she set all those coefficients equal to 0 to find the kernal, if i do the same thing, i get a=b=c=0, so, shouldn't i get ker(L)= {0}? I don't understnad how to get P1 for that. Can you explain in a more detailed way. Thanks a lot!
 
loli12 said:
Well, my book defined Pn to be polynomials of less than n degree.
if I apply the L to ax^2 + bx + c, then i get L(p(x)) = 2ax^2 + bx + 0
i remembered from one of the ques that my teacher did, she set all those coefficients equal to 0 to find the kernal, if i do the same thing, i get a=b=c=0,

For the kernel, you don't set the coefficients to zero, you set L(p(x)) to zero and try to find what p(x)'s will satisfy this. If L(ax^2+bx+c)=0, then 2ax^2 + bx + 0=0. What choices of a,b,c will satisfy this?
 
loli12 said:
Well, my book defined Pn to be polynomials of less than n degree.
if I apply the L to ax^2 + bx + c, then i get L(p(x)) = 2ax^2 + bx + 0
i remembered from one of the ques that my teacher did, she set all those coefficients equal to 0 to find the kernal, if i do the same thing, i get a=b=c=0, so, shouldn't i get ker(L)= {0}? I don't understnad how to get P1 for that. Can you explain in a more detailed way. Thanks a lot!

If p= ax2+ bx+ c, then p'= 2ax+ b so Li(p)= 2ax2+ bx . Setting "all those coeficients equal to 0" gives 2ax= 0 and b= 0.
How did you get c=0?

What is P1?
 

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