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Can Somebody Check My Proofs Please

  1. Feb 6, 2010 #1
    Hi, would someone be able to check my proofs for me and tell me if they are right and if not what is wrong please?


    So for the first one I said let u=p(x) and v=b(x)
    and T(ku)=(kp)(x)=kp(5)x2=kTu
    So it is a linear transformation.

    For the second I said T(u+v)=p(x)+b(x)=x2p(1/x)+x2b(1/x)=Tp(x)+Tb(x)
    and T(ku)=x2kp(1/x)=k(x2p(1/x))=kTp(x)

    So it is also a linear transformation.
    For the third I said T(kp)(x)=xkp'(x)kp''(x)=k2xp'(x)p''(x) which does not equal KTp(x)

    So it is not a linear transformation

    Did I do these right?

    Thanks for any help :)
  2. jcsd
  3. Feb 6, 2010 #2
    for the first one, what would happen if you took A= x^2 and B=-x^2?
  4. Feb 6, 2010 #3


    Staff: Mentor

    You haven't used the given information that P2(x) is the space of polynomials of degree 2 or less. Every function in this space is of the form p(x) = ax2 + bx + c, for some constants a, b, and c.
  5. Feb 6, 2010 #4
    So am I suppose to write it out as T(ax2 + bx + c+dx2+ex+f)=[25(a+d)+5(b+e)+(c+f)]x2?

    I am a bit confused :confused:
  6. Feb 6, 2010 #5
    Hm I did it that way and I still get that it is a linear transformation.

    Well either way I get zero.
  7. Feb 6, 2010 #6


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    You have some problems with your notation. For instance, you have u=p(x) and v=b(x), so [itex]T(u+v) = T[(p+b)(x)] \ne p(x)+b(x)[/itex].
  8. Feb 6, 2010 #7
    Right I made that mistake, but I don`t think it affected the wrongness of my answer.
  9. Feb 6, 2010 #8
    T(u+v)=T[p+b](x)=(p+b)(5)x2. Does this not equal p(5)x2+b(5)x2? I don't understand why not.
  10. Feb 6, 2010 #9


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    I think your conclusions are right.
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