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Math Help - Change of Basis of Linear Transformation

  1. #1
    Senior Member sfspitfire23's Avatar
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    Change of Basis of Linear Transformation

    Hi all,

    Let B be a basis for R^2. Let L be the linear transformation represented in standard coordinates by the matrix R. Then, to represent L with respect to the standard coordinates, we take A^(-1)RA where A is a column matrix of the B vectors.

    Now, what if R was expressed in terms of the basis B and we want to express L with respect to the standard coordinates? How would we go this way?

    Thanks
    Last edited by sfspitfire23; November 2nd 2012 at 08:33 PM.
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  2. #2
    MHF Contributor
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    Re: Change of Basis of Linear Transformation

    Hey sfspitfire23.

    The easiest way to look at this is to go from first principles.

    Now in one basis we have some vector v in Basis B, and in another basis we have v' in Basis B'. Let x be the true vector that is invariant to both bases: then this implies

    Bx = v and
    B'x = v'. If B and B' are both basis then B and B' are invertible giving

    x = B^(-1)v = B'^(-1)v'. Writing these equations to get v in terms of v' (and vice versa) gives:

    B*B'^(-1)v' = v and
    B'*B^(-1)v = v'

    Now a linear transformation acting on R^2 is simply Ax = b taking x and mapping it to b, and the above can relate Av to Av' in a straight-forward manner.
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  3. #3
    MHF Contributor

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    Re: Change of Basis of Linear Transformation

    suppose B and B' are two bases for a vector space V, and L:V-->V is a linear transformation.

    if [L]B = R, and the change-of-basis matrix from B' to B is A (whose columns consist of the basis vectors of B' expressed in B-coordinates), then:

    [L]B' = A-1RA, as you correctly deduced.

    it therefore stands to reason that:

    R = A[L]B'A-1 (use matrix multiplication).

    here is how it works:

    A-1([v]B) = [v]B' (since A changes B'-vectors to B-vectors, A-1 "changes them back" to B'-vectors).

    [L]B'([v]B') = [Lv]B' (the matrix for L in the basis B' takes v in B'-coordinates to Lv in B'-coordinates)

    A([Lv]B') = [Lv]B (A turns Lv in B'-coordinates to Lv in B-coordinates).

    so A[L]B'A-1([v]B') = A([L]B'(A-1([v]B))) = A([L]B'([v]B')) = A([Lv]B') = [Lv]B, that is:

    A[L]B'A-1 is a linear transformation which takes [v]B to [Lv]B for every v in V, and so is the matrix for L in the basis B, that is: R.

    the algebra "looks prettier" if we call the matrix for L in the basis B', S.

    then S = A-1RA so:

    AS = AA-1RA = IRA = RA

    ASA-1 = RAA-1 = RI = R
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