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Math Help - linear transformation standard matrix

  1. #1
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    linear transformation standard matrix

    I have a linear transformation satisfying F(1,0,0) = (0,1), F(0,2,1) = (1,1) and F(0,1,−2) = (−2,0) is the standard matrix {[0,1,-2],[1,1,0]} ? or if not how do i find the stardard matrix correctly?

    Thanks.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: linear transformation standard matrix

    yes your matrix is correct. Since the three vectors you gave,  \vec{v_1}, \vec{v_2}, \vec{v_3} from  \mathbb{R}^3 are linearly independent. They form a basis for  \mathbb{R}^3.

    so any vector in v \in \mathbb{R}^3 can be written as  v = c_1 * \vec{v_1} + c_2 * \vec{v_2} + c_3 * \vec{v_3} .
    so  L(v) = L(c_1 * \vec{v_1} + c_2 * \vec{v_2} + c_3 * \vec{v_3}) = c_1*L(\vec{v_1}) + c_2*L(\vec{v_2}) + c_3*L(\vec{v_3}) = \begin{bmatrix} L(v_1) & L(v_2) & L(v_3) \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}

    errr. I think they want the linear transformation written interm of the standard basis for \mathbb{R}^3
    So since  \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & -2 \end{bmatrix} is the change of basis matrix from the standard to the basis with ur given vectors.
    so  \begin{bmatrix} 0 & 1 & -2 \\ 1 & 1 & 0 \end{bmatrix} * \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & -2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 5 \\ 1 & 2 & 1 \end{bmatrix} is the standard matrix for your linear transformation.
    Last edited by jakncoke; January 28th 2013 at 01:02 AM.
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  3. #3
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    Re: linear transformation standard matrix

    no, that is the matrix in the basis B' = {(1,0,0),(0,2,1),(0,1,-2)}.

    as jakncoke points out, you need to employ a change of basis similarity transform, to obtain the "standard matrix". if the change of basis matrix is P (from B' to B, the standard basis B = {(1,0,0),(0,1,0),(0,0,1)})

    the matrix you want is:

    [F]_B = [F]_{B'}P^{-1}.

    the matrix you have is [F]B'. P is given by jakncoke above.

    you COULD compute the inverse using the adjugate, but i always found that to be tedious and time-consuming.

    we know that (1,0,0)--->[1,0,0]B', so we just need to see how to express (0,1,0) and (0,0,1) as linear combinations of (0,2,1) and (0,1,-2).

    a little fooling around gives:

    (0,1,0) = (2/5)(0,2,1) + (1/5)(0,1,-2)

    (0,0,1) = (1/5)(0,2,1) + (-2/5)(0,1,-2) so:

    P^{-1} = \begin{bmatrix}1&0&0\\0&\frac{2}{5}&\frac{1}{5}\\0  &\frac{1}{5}&\frac{-2}{5} \end{bmatrix}

    this gives:

    [F]_B = \begin{bmatrix}0&0&1\\1&\frac{2}{5}&\frac{1}{5} \end{bmatrix}

    as the correct matrix for F in the standard basis. we can verify this like so:

    F(1,0,0) = (0,1) since this is the first column of F.
    F(0,2,1) = (0+0+1,0+4/5+1/5) = (1,1)
    F(0,1,-2) = (0+0-2,0+2/5-2/5) = (-2,0).
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  4. #4
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    Re: linear transformation standard matrix

    ok i used the linear combination method to get the standard matrix equal to {[0,0,1],[1,2/5,1/5]} could anybody confirm this? Thanks
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  5. #5
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    Re: linear transformation standard matrix

    Am i also correct in saying the range of F is =span{[0,1],[1,1/5]} and the kernel of F is ={0}?
    Thanks
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  6. #6
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    Re: linear transformation standard matrix

    since the range has dimension 2 (and our matrix has RANK 2) the range is all of R2. F is onto.

    by the rank-nullity theorem, this means the dimension of the kernel is 1, so there HAS to be more in the kernel than just {0}.
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  7. #7
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    Re: linear transformation standard matrix

    Yes, i see my mistake. Thanks
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