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Kernel and range

  1. Nov 14, 2004 #1
    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!
  2. jcsd
  3. Nov 14, 2004 #2

    matt grime

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    Pn is the polys of degree n right? Take an arbitrary degree three poly:


    and apply L to it.

    What has been killed and what is left?
  4. Nov 14, 2004 #3
    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!
  5. Nov 14, 2004 #4


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    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?
  6. Nov 15, 2004 #5


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    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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