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Math Help - Matrix Representation of Linear Transformation

  1. #1
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    Matrix Representation of Linear Transformation

    When gamma = {1}, I found the answer to be:

    [T] = [1 0 0 1]

    However, how do I calculate the answer if gamma = {2}?

    Is it just [2 0 0 2]?

    Thanks in advance for your help.
    Attached Thumbnails Attached Thumbnails Matrix Representation of Linear Transformation-la.jpg  
    Last edited by divinelogos; October 1st 2011 at 05:31 PM.
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  2. #2
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    Re: Matrix Representation of Linear Transformation

    what is α? there are many possible bases for M2,2(F). assuming it is the "standard basis" Eij, then this should be [1/2 0 0 1/2] (see 2nd post below).
    Last edited by Deveno; October 1st 2011 at 07:39 PM.
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  3. #3
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    Re: Matrix Representation of Linear Transformation

    My apologies, α is actually {(1,0;1,0),(0,1;0,0),(0,1;1,0),(0,0;0,1)}, but I believe it still comes out to [2,0,0,2].

    Could you confirm this is correct?

    In general, I am having trouble with this type of problem when a non-standard basis is given for W. Here is another example:


    T: p2(R)->R and T(f(x))=f(2)

    The given bases are: B={x^2,x+1,1-x} and y(gamma)={2}

    So, I need to calculate the matrix representation of T with respect to B and y.

    I can easily calculate the four images:


    T(x^2)= 4
    T(x+1)= 3
    T(1-x)= -1


    So, if we had the standard basis ({1}) for W then the answer would just be:

    [4 3 -1]


    However, since the basis is {2}, does this mean the answer is?:

    [8 6 -2]


    I feel uncomfortable "just multiplying by 2", because I'm not really sure why I'm doing that.


    Could you explain a little?


    Thanks!
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  4. #4
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    Re: Matrix Representation of Linear Transformation

    you know, using a 1-dimensional vector space is really confusing for just this reason. in the basis {2}, the number 2 has "coordinates" (1) (since 2 = (2)1).

    think of it this way: when we make {2} our basis for R, we are now calling 2 our unit "length" (we're stretching the scale of our axis).

    you can verify you're doing it right by checking for some easy values of p(x). let's use p(x) = x + 1. in our basis for P2, this has coordinates (0,1,0).

    now we know that T(x+1) = 2+1 = 3. in the basis {2}, 3 has coordinates (3/2) (3 is 1-1/2 "twos").

    so let's verify the matrix multiplication comes out right: T(0,1,0) should be (3/2).

    well (8,6,-2).(0,1,0) = 6, so...oh dear. it appears that the get the "right" answer for [T], the matrix wrt the bases {x^2, x+1, 1-x}, {2},

    should be [2 3/2, -1/2] (and thus the answer to your first question should be [1/2 0 0 1/2]). let's try another, more random polynomial.

    let's let g(x) = 3 - 2x + 4x^2. now T(g(x)) = g(2) = 3 - 4 + 16 = 15, and in the basis {2} 15 is (15/2).

    now the icky part, we have to find g(x) in terms of our basis for P2:

    3 - 2x + 4x^2 = a(x^2) + b(x+1) + c(1-x) = ax^2 + (b-c)x + (b+c).

    so a = 4, b-c = -2, b+c = 3.

    using c = 3-b, and substituting in b-c = -2, we get: -2 = b-c = b - (3-b) = 2b - 3, so 1 = 2b, so b =1/2.

    and thus c = 5/2, so 4x^2 + (1/2)(x+1) + (5/2)(1 - x) = 4x^2 + x/2 + 1/2 + 5/2 - 5x/2 = (1/2 + 5/2) - (5/2 - 1/2)x + 4x^2

    = 3 - 2x + 4x^2 = g(x). so g(x) has β-coordinates (4,1/2,5/2). now to multiply by T:

    (2, 3/2,-1/2).(4,1/2,5/2) = 8 + 3/4 - 5/4 = 8 - 1/2 = 15/2, the correct answer.
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