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Math Help - Transformation in Euclidean vector space.

  1. #1
    fkf
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    Transformation in Euclidean vector space.

    Consider the Euclidean vector space  R^2 with the basis  B = {(2 \quad 0),(1 \quad 1)}
    and the linear transformation img.top {vertical-align:15%;} T:x \rightarrow Ax" alt=" T:x \rightarrow Ax" />, where

    A = [1 2 ; 0 1]

    Find the B-matrix of T, that is find [T]_B

    I thought that I was just going to find the transform matrix from B to A, but that doesn't seem right?
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    Re: Transformation in Euclidean vector space.

    what you need to do is a two-step process:

    step one: find the "change of basis matrix" P. to be explicit, the P we are going to use is the one that changes B-coordinates to standard coordinates.

    since [1,0]B = 1(2,0) + 0(1,1) = (2,0), we have:

    P\begin{bmatrix}1\\0 \end{bmatrix} = \begin{bmatrix}2\\0 \end{bmatrix}

    which makes it clear P is of the form:

    P = \begin{bmatrix}2&\ast \\ 0&\ast \end{bmatrix}

    similarly, since [0,1]B = (1,1), we have that:

    P = \begin{bmatrix}2&1 \\ 0&1 \end{bmatrix}

    so the overall procedure is this:

    B-coordinate input ---> change to standard coordinates --->apply standard T (matrix A) ---> change back to B-coordinates.

    to change back to B--coordinates, we need to use P-1, which is:

    P^{-1} = \frac{1}{(2^*1 - 1^*0)}\begin{bmatrix}1&-1\\0&2 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}&-\frac{1}{2}\\0&1 \end{bmatrix}

    step 2: find the desired matrix for T, which is:

    [T]_B = P^{-1}AP = \begin{bmatrix} \frac{1}{2}&-\frac{1}{2}\\0&1 \end{bmatrix} \begin{bmatrix}1&2\\0&1 \end{bmatrix} \begin{bmatrix}2&1 \\ 0&1 \end{bmatrix}

     = \begin{bmatrix}1&1 \\ 0&1 \end{bmatrix}.

    let's verify that this actually works:

    we know that T(x,y) = (x+2y,y).

    writing (x,y) in B-coordinates, we get: (x,y) = x(1,0) + y(0,1) = x[1/2,0]B + y[-1/2,1]B = [(x-y)/2,y]B.

    applying our [T]B to [(x-y)/2,y]B (to get [T(x,y)]B), we obtain:

    [T]B([(x-y)/2,y]B) = [(x-y)/2+y,y]B = [(x+y)/2,y]B.

    changing this back to standard coordinates, we have:

    [(x+y)/2,y]B = ((x+y)/2)[1,0]B + y[0,1]B = ((x+y)/2)(2,0) + y(1,1) = (x+y,0) + (y,y) = (x+2y,y) = T(x,y).
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