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

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The discussion centers on finding the kernel and range of the linear transformation L(p(x)) = xp'(x) from P3 to P3. The initial assumption was that the kernel is {0} and the range is P3, but the correct answers are ker(L) = P1 and L(P3) = Span(x^2, x). Participants clarify that to find the kernel, one should set L(p(x)) to zero and solve for the coefficients of the polynomial, leading to the conclusion that the kernel includes polynomials of degree 1 or less. The confusion arises from misapplying the method of setting coefficients to zero instead of solving the resulting equations. Understanding the definitions of Pn and the implications of the transformation is crucial for accurately determining the kernel and range.
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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