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Linear Transformations

  1. Oct 19, 2005 #1


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    Can anyone at least tell me how to get started on this problem I have?

    Determine whether the following are linear transforatmions from P2 to P3.
    L(p(x)) = xp(x)

    I understand when it's in vector form but not really picking up on the polynomial part of this.
  2. jcsd
  3. Oct 19, 2005 #2
    Let me start by giving you the definition of a linear transformation, in case you didnt already know.

    A function L: R^n--->R^m is called a linear transformation or linear map if it satisfies

    i) T(u+v)= T(u) + T(v) for all u,v in R^n
    ii) T(cv)= cT(v) for all v in R^n, and scalar c

    How does that help?
  4. Oct 19, 2005 #3
    Sorry...I seemed to have used notation wrongly. I'm use to calling T my transformation, so I wrote my definition in such a way. I really mean..

    i) L(u+v)= L(u) + L(v) for all u,v in R^n
    ii) L(cv)= cL(v) for all v in R^n, and scalar c
  5. Oct 19, 2005 #4


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    Yes I knew the two statements that you wrote. I guess I may just be a little unclear on how to handle this problem as far as notation goes.

    I have some work and an answer. Does this look sound?

    L(a*p(x) + b*p(x)) = L(a*x*p(x)) + L(b*x*p(x)) = a*x*L(p(x)) + b*x*L(p(x))

    Therefore this is a linear transformation.

    I know this is a linear transform but I just want to make sure my proof is right.
  6. Oct 19, 2005 #5
    Well, let's try this one step at a time.

    i) If we take two polynomials p'(x) and p''(x) in P2, and multiply them each by x...what do we get? I think this is the main thing that your missing in your proof.
  7. Oct 19, 2005 #6
    i meant to type "you're"*
  8. Oct 19, 2005 #7
    actually...u know what?..let me start over. maybe im getting carried away. =\
  9. Oct 19, 2005 #8


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    So the x is used as a constant basically which is practically the original definition?
  10. Oct 19, 2005 #9
    I believe your method for proving it is on the right track. There are just a phew things wrong with it.

    Firstly, in "L(a*p(x) + b*p(x))", both polynomials should noted as the same. You should probably distinguish the two, as the def'n states that states for any u,v.

    secondly, we don't know this "L(a*p(x) + b*p(x)) = L(a*x*p(x)) + L(b*x*p(x))". What we know is that L(p(x)= x(p(x). So L(a*p(x) + b*p(x)) = x(a*p(x) + b*p'(x)).

    perhaps you can take it from there. Sorry about the frustration earlier. I really did lose track.
  11. Oct 19, 2005 #10
    also...phew=few. I'm not sure whats wrong with me.
  12. Oct 19, 2005 #11


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    Im curious if i can do it this way. It just came to me and I may be breaking a rule.

    Can I put:

    L(xp(x)) = xL(p(x)) = xp(x)

    And that satisfies my question?
  13. Oct 19, 2005 #12
    xL(p(x))=x(x(px)), by def'n of L(p(x)).
  14. Oct 19, 2005 #13


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    Ok I see what you mean. Maybe I will get this straight sometime. Im terrible at linear algebra. Calculus and differential equations are more my thing.
  15. Oct 19, 2005 #14


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    Ok I think I might have it this time. Can someone tell me if I am correct?

    L(a(p(x)) = x(ap(x)) = a(xp(x)) = aL(p(x))

    Does this look correct?
  16. Oct 19, 2005 #15


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    I also forgot to add this part to prove the second part.

    L(p(x) + p'(x)) = x(p(x) + p'(x)) = xp(x) + xp'(x) = L(p(x)) + L(p'(x))

    I think this should be right but would like to be corrected if wrong.
  17. Oct 20, 2005 #16


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    Homework Helper

    Both are right as far as I can see.
    Note that you can always replace the two statements that you check by just one, which is usually faster. For more complex transformation, it may be better to prove the two statements apart, but for easier ones: just show that "L of a lineair combination gives a lineair combination of L's"

    [tex]L\left( {ap\left( x \right) + bq\left( x \right)} \right) = x\left( {ap\left( x \right) + bq\left( x \right)} \right) = axp\left( x \right) + bxq\left( x \right) = aL\left( {p\left( x \right)} \right) + bL\left( {q\left( x \right)} \right)[/tex]
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