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Math Help - Linear transformations and matrices

  1. #1
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    Linear transformations and matrices

    Hi

    I got a question regarding the matrix of linear transformation.

    If a linear transformation T: R^3 -> R^2 implies that T(1,2-1) = (1,4) ; T(-1,2,1) = (5,9) and T(2,2,1) = (2,1).

    Is the most commen the method of determining the matrix of linear transformation???

    M * T = n => T = M^-1 * n.

    Where M is matrix the set equations in R^3, n being the vector coordinants in R^2, and T the matrix of linear transformation.

    /Fred
    Last edited by Mathman24; April 26th 2005 at 01:00 AM.
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  2. #2
    hpe
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    Quote Originally Posted by Mathman24
    Is the most commen the method of determining the matrix of linear transformation???

    M * T = n => T = M^-1 * n.
    Not quite. Write M for the 3x3 matrix whose columns are the vectors in R^3, i.e. (1,2,-1)' etc, and write m for the 2x3 matrix whose columns are the corresponding vectors in R^2, i.e. (1,4)' etc. Here (1,2,-1)' is the transpose of the row vector (1,2,-1) and so on..

    The conditions are that T times first column of M equals first column of m, and so on, or in matrix notation T*M = m.

    Then T = m*M^(-1).

    Your approach would work if matrices operate on vectors by multiplication from the right. But convention has it that in this operation the matrix is on the left and the vector is on the right.

    Hope this helps.
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  3. #3
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    Hello and thanks for Your answer

    Let me ask in another way. Then one is presented with a linear transformation T:R^3 -> R^2 which implies that T(1,2,-1) = (0,0) ; T(1,2,1) = (1,4) and T(2,2,1) = (9,4) and one has to find the matrix of linear transformation:

    This can be view as a system of linear eqautions:

    T_(1,1) + 2 T_(1,2) - T_(1,3) = 0
    T_(2,1) + 2 T_(2,2) - T_(2,3) = 0
    T_(1,1) + 2 T_(1,2) + T_(1,3) = 1
    T_(2,1) + 2 T_(2,2) + T_(2,3) = 4
    2T_(1,1) + 2 T_(1,2) + T_(1,3) = 9
    2T_(1,1) + 2 T_(1,2) + T_(1,3) = 4


    Which then can we viewed as a 6x6 matrix.

    Quote Originally Posted by hpe
    Not quite. Write M for the 3x3 matrix whose columns are the vectors in R^3, i.e. (1,2,-1)' etc, and write m for the 2x3 matrix whose columns are the corresponding vectors in R^2, i.e. (1,4)' etc. Here (1,2,-1)' is the transpose of the row vector (1,2,-1) and so on..

    The conditions are that T times first column of M equals first column of m, and so on, or in matrix notation T*M = m.

    Then T = m*M^(-1).

    Your approach would work if matrices operate on vectors by multiplication from the right. But convention has it that in this operation the matrix is on the left and the vector is on the right.

    Hope this helps.
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  4. #4
    hpe
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    Quote Originally Posted by Mathman24
    This can be view as a system of linear equations:

    ...

    Which then can we viewed as a 6x6 matrix.
    Right, that's a possible way to write the problem, six equations for six unknowns. Note that the 6x6 matrix for this system has a particular structure - only(!) 18 coefficients are non-zero, the zero coefficients appear in a pattern, and the non-zero coefficients each appear twice. If you write down the matrix for the six equations, you will notice this.

    That's because the system comes from a "matrix times matrix equals matrix" equation. The technical term for systems that arise in this fashion is Kronecker Product or matrix direct product.
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  5. #5
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    I am having trouble remembering just quite how to do this, but here goes, let
    B (beta) ={(1,2-1), (-1,2,1), (2,2,1)} be your basis for R^3, G (gamma) = {(1,0),(0,1)}, be the basis for R^2,
    T(1,2-1) = (1,4) = 1(1,0) + 4(0,1)
    T(-1,2,1) = (5,9) = 5(1,0) + 9(0,1)
    T(2,2,1) = (2,1) = 2(1,0) +1(0,1)

    Reading off the columns and writing as rows we get the matrix for the transformation.
    Then T(x,y,z) = [1 5 2 ; 4 9 1]from B to G * [(x,y,z) relative to B]
    (note: semicolon means next row)
    T(x,y,z) relative to B = (x+5y+2z,4x+9y+z),
    Now, (1,2,-1) = 1(1,2,-1) + 0(-1,2,1) + 0(2,2,1) So (1,2,-1) relative to B is (1,0,0) by just taking the coefficients.
    And T(1,0,0) = (1 + 5(0) +2(0), 4(1)+9(0)+(0)) = (1,4), you can see it works provided you rewrite the original vectors relative to B. I believe that if you want to have the transformation go from standard basis to standard basis, you will first have to create another transformation G that will transform vectors in the standard basis to vectors relative to B.
    We need the matrix representation A from S = {(1,0,0),(0,1,0),(0,0,1)} to B.
    But obviously, all we can find is the representation from B to S, from the method similar to above,
    G(1,2,-1) = 1(1,0,0) + 2 (0,1,0) + -1 (0,0,1)
    G(-1,2,1) = -1(1,0,0) + 2 (0,1,0) + 1(0,0,1)
    G(2,2,1) = 2(1,0,0) + 2 (0,1,0) + 1(0,0,1)

    G(x,y,z) = [1,-1,2;2,2,2;-1,1,1] from B to S [x,y,z] relative to B (notice the pattern, just read off the columns in the B vectors when going to the standard basis)
    G(x,y,z) = (x-y+2z,2x+2y+2z,-x+y+z)
    (1,2,-1) relative to B is simply (1,0,0) and G(1,0,0) = (1,2,-1) so haha it works. Now take the inverse of this matrix, you will get a transformation from S to B. Use the calculator to do this, the result is
    [0,1/4,-1/2;-1/3,1/4,1/6;1/3,0,1/3] from S to B. Now, we want
    T(G(x,y,z) so we can multiply their matrix representations,
    [1 5 2 ; 4 9 1]*[0,1/4,-1/2;-1/3,1/4,1/6;1/3,0,1/3] =
    [-1,3/2,1;-2 2/3 , 3 1/4, -1/6]

    This is the matrix you desire. Let T:R^3->R^2 be given by
    T(x,y,z) = [-1,3/2,1;-2 2/3 , 3 1/4, -1/6][x;y;z]

    Checking with the calculator, I get
    T(1,2,-1) = (1,4)
    T(-1,2,1) = (5,9)
    T(2,2,1) = (2,1)
    Last edited by beepnoodle; April 30th 2005 at 05:30 PM.
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