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Math Help - Invertible matrix is in same image as linear transformation

  1. #1
    Senior Member sfspitfire23's Avatar
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    Invertible matrix is in same image as linear transformation

    Let A be a 4x3 matrix and let B be a 4x4 matrix. Then, BA is a 4x3 matrix. Let L be the linear
    transformation from R4 to R3 defined by L(x) = xA, and let M be the
    linear transformation from R4 to R3 defined by M(x) = xBA. Now, assume that B is an invertible matrix so that C is a
    4x4 matrix and CB=BC=Identity matrix.

    Show that the image of L is the same as the image of M.

    Attempt-

    Let y be in the image of L so that y = xA.

    Now, to get to y = xBA, I have to assume that x has an inverse, invert both sides, multiply each side by B, then C, and then move the x vector back over? I'm stuck.

    Thanks
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  2. #2
    GJA
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    Re: Invertible matrix is in same image as linear transformation

    Hi, sfspitfire23.

    I'll outline what came to my mind here; hopefully it will get things going in the right direction.

    I think your idea of trying to express elements as images is the right one, we just need to apply it in a slightly different way. Since B is invertible, we know it is a matrix of full rank, i.e. rank(B)=4; this comes from the rank-nullity theorem (because an invertible matrix has only 0 in its kernel, i.e. its nullity is 0). Since B has full rank, for any x\in \mathbb{R}^{1\times 4}, there is a unique y\in \mathbb{R}^{1\times n} such that x=yB. Then xA=yBA and the images are the same.

    Does this help? Let me know if anything is unclear or in error.

    Good luck!
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  3. #3
    Senior Member sfspitfire23's Avatar
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    Re: Invertible matrix is in same image as linear transformation

    Hmm, we havent studied rank yet, so I'm not sure that is needed.
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  4. #4
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    Re: Invertible matrix is in same image as linear transformation

    suppose y is in Im(M).

    then y = xBA = (xB)A = L(xB), so y is in Im(L). thus Im(M) is contained in Im(L).

    on the other hand, suppose y is in Im(L), so y = xA. let z = xB-1 (we can do this since B is invertible).

    then M(z) = zBA = (xB-1)BA = x(B-1B)A = x(IA) = xA = y, so y is in Im(M). thus Im(L) = Im(M)

    (the two sets contain each other).
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