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Matrices of linear transformations

  1. Sep 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Let T: P2 - P2 be the linear operator defined by
    T(a0 + a1x + a2x2) = a0 + a1(x - 1) + a2(x - 1)2

    (a) Find the matrix for T with respect to the standard basis B = {1, x, x2}.

    2. Relevant equations

    [T]B[x]B = [T(x)]B

    3. The attempt at a solution

    T(1) = a0 + a1(1 - 1) + a2(1 - 1)2
    = a0

    Okay...I don't know if I'm going in the right direction here?...and if I am, I'm not sure how to obtain vectors from the proceeding expressions for T(x) and T(x2)??

    Any help would be appreciated.

    Thanks.

    Derryck.
     
  2. jcsd
  3. Sep 1, 2011 #2

    tiny-tim

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    Hi Derryck! :smile:
    nooo :redface:

    suppose it was T(a0 + a1x + a2x2) = a0 + a1(x + 1) + a2(x + 1)2

    then your method would give you T(1) = a0 + a1(1 + 1) + a2(1 + 1)2
    = a0 + 2a1 + 4a2

    but the correct way is to start by saying 1 = 1 + 0x + 0x2, so T(1) = … :wink:

    carry on from there :smile:
     
  4. Sep 1, 2011 #3
    Wo...?

    But then is this the same as saying (in vector form):

    1 1
    0 = 0
    0 0 ?

    I'm not sure what you mean by 1 = 1 + 0x + 0x2, because what I see from this, I would think that it would be more correct to say:

    1 = a0 + a1x + a2x2, then to solve...? Flip, I'm so bad at this I know...but please, just humor me a little...it will help a great deal...

    Cheers

    Derryck
     
  5. Sep 1, 2011 #4

    tiny-tim

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    Yes, and the solution is a0 = 1, a1 = a2 = 0, isn't it? :wink:

    In other words: in the basis {1,x,x2}, 1 has components (1,0,0).

    ok, now what are the components of x ?

    and what are the components of x2 ? :smile:
     
  6. Sep 1, 2011 #5
    Ok, the components of x would be found by solving:

    x = a0 + a1(x - 1) + a2(x - 1)2

    So I would then say

    a1(x - 1) = x
    Therefore a1 = x / (x - 1)... Am I going along the right lines? I'm not too sure what to do from here.

    Thanks for the help so far though really!

    Cheers

    Derryck
     
  7. Sep 1, 2011 #6

    tiny-tim

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    No!!

    In the basis {1,x,x2}, what is x?

    x = ?1 + ?x + ?x2
     
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