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Math Help - Linear Algebra Proof Regarding Linear Transformations

  1. #1
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    Linear Algebra Proof Regarding Linear Transformations

    Given the following vectors in R^2:
    a(sub 1) = (1; -1), a(sub 2) = (2; -1), a(sub 3) = (-3;2),
    b(sub 1) = (1;0), b(sub 2) = (0;1), b(sub 3) = (1;1),
    determine whether there is a linear transformation T from R^2 into R^2 such that Ta(sub i) = b(sub i) for i = 1, 2 and 3.

    Basically, I don't quite understand the process of figuring out linear transformations, so I'd like at least a few helpful pointers in figuring this out. Thank you.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Linear Algebra Proof Regarding Linear Transformations

    so assume we have a linear transformation. L which takes a_i \to b_i
     A = \begin{bmatrix} 1 & 2 & -3 \\ -1 & -1 & 2 \end{bmatrix}
     B = \begin{bmatrix} 1 & 0 & 1 \\  0 & 1 & 1 \end{bmatrix}
    For the matrix A, you can write the 3rd column as the linear combination of the first and second columns.
     A = \begin{bmatrix} 1 & 2  \\ -1 & -1  \end{bmatrix} \begin{bmatrix} -1 \\ -1 \end{bmatrix} = \begin{bmatrix} -3 \\ 2 \end{bmatrix}
    since a linear transformation has the property L( c_1 v_1 + c_2 v_2)=  c_1 L(v_1) + c_2 L(v_2) then
    if the first column of A went to the first column of B, and the second column of A went to the second column of B then
    L(-1 * \begin{bmatrix} 1  \\ -1   \end{bmatrix} + -1 * \begin{bmatrix} 2  \\ -1   \end{bmatrix})  = -1*L(\begin{bmatrix} 1  \\ -1   \end{bmatrix}) + -1 * L(\begin{bmatrix} 2  \\ -1   \end{bmatrix}) = -1 * \begin{bmatrix} 1  \\ 0   \end{bmatrix} + -1 * \begin{bmatrix} 0  \\ 1   \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}


    since this is not equal to the 3rd column of B, there is no Linear transformation which takes the first column of A to the first column of B, second ....
    Thanks from zachoon
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