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Math Help - Change of Basis wrt Linear Transformation

  1. #1
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    Change of Basis wrt Linear Transformation

    Hi, I have one part of a two-part question related to a linear transformation.

    ------------

    (a) Let a linear transformation T: R2 --> R3 be given by T[1,3]=[2,3,4] and T[-2,5]=[1,0,2]. Find the matrix of T relative to the standard basis in each space.

    I actually got this part. You just have to find how T operates on [x,y] by using the fact that T is linear.

    From there, it was easy to show that T[1,0] = [7/11,15/11,14/11] and that T[0,1] = [5/11,6/11,10/11].

    So, the matrix of T relative to the standard bases is [ T[1,0] ; T[0,1] ].

    ------------

    (b) Find a basis B of R2 and a basis C of R3 such that T relative to B,C =

    [1 0]
    [0 1]
    [0 0]


    I have a notion of what to do, but don't know how to set it up. The image under this matrix would be {[x,y,z]: z=0}, as opposed to a different planar subspace of R3 as in part (a). Do I use a change of basis procedure here? And how would I set it up?



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    Last edited by joeyjoejoe; May 11th 2008 at 07:59 PM.
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  2. #2
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    Quote Originally Posted by joeyjoejoe View Post
    Let a linear transformation T: R2 --> R3 be given by T[1,3]=[2,3,4] and T[-2,5]=[1,0,2].

    ------------

    (b) Find a basis B of R2 and a basis C of R3 such that T relative to B,C =

    [1 0]
    [0 1]
    [0 0]
    I would take B to consist of the vectors [1,3] and [-2,5], and C to consist of [2,3,4], [1,0,2] and a third vector [a,b,c] chosen so as to be linearly independent of the other two vectors in C.
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  3. #3
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    You know, that was actually one of my first thoughts about this problem. So what you are saying is:

    T[x,y] = T(a[1,3] + b[-2,5]) = aT[1,3] + bT[-2,5] = a[1,3,0] + b[-2,5,0].

    Then it doesn't matter that T[x,y] = [a-2b,3a+5b,0], because anything with a non-zero third coordinate would be out of the original image as found in part (a) under this basis. That makes sense. New basis.

    I got it, thanks.
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