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Math Help - Proof of Transpose Property of Matrix

  1. #1
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    Proof of Transpose Property of Matrix

    prove (AB)^T = (B^T)*(A^T)
    using summation notation

    i know how to prove it without summation notation but with summation notation i'm not getting anywhere
    if someone can prove it and then explain the summation steps involved that would be of great help
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  2. #2
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    Quote Originally Posted by razorfever View Post
    prove (AB)^T = (B^T)*(A^T)
    using summation notation

    i know how to prove it without summation notation but with summation notation i'm not getting anywhere
    if someone can prove it and then explain the summation steps involved that would be of great help
    (AB)^T_{i,j}=(AB)_{j,i}=\sum_k A_{j,k}B_{k,i}=\sum_k B_{k,i}A_{j,k}=\sum_k B^T_{i,k}A^T_{k,j}=(B^TA^T)_{i,j}

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  3. #3
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    If transpose was defined to you as [A^T]_{ij} = [A]_{ji}, then "summation sign" method as explained in the prev. post is most likely the only thing you could do.

    Ignore whatever's below, I replied in a haste.
    ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

    However if you are doing linear algebra, and you are familiar with the notion of a linear functional [x,y](Halmos' notation), then one could define A', adjoint of linear transformation A as [Ax,y] = [x,A'y].If [A]_{\cal{X}} is the matrix of linear transformation A w.r.t the basis \cal{X}, then its easy to verify that the matrix of linear transformation A' is [A]^T_{\cal{X}}.

    Now with all this setup, its extremely easy to prove (AB)^T = B^TA^T. Note that it suffices to prove (AB)' = B'A'.

    But \forall x,y : [x,(AB)'y] = [ABx,y] = [A(Bx),y] = [Bx, A'y] = [x, B'A'y]

    Thus (AB)' = B'A'
    Last edited by Isomorphism; January 19th 2009 at 08:07 AM. Reason: Thanks Constatine11
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  4. #4
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    Quote Originally Posted by Isomorphism View Post
    If transpose was defined to you as [A^T]_{ij} = [A]_{ji}, then "summation sign" method as explained in the prev. post is most likely the only thing you could do.


    However if you are doing linear algebra, and you are familiar with the notion of a linear functional [x,y](Halmos' notation), then one could define A', adjoint of linear transformation A as [Ax,y] = [x,A'y].If [A]_{\cal{X}} is the matrix of linear transformation A w.r.t the basis \cal{X}, then its easy to verify that the matrix of linear transformation A' is [A]^T_{\cal{X}}.

    Now with all this setup, its extremely easy to prove (AB)^T = B^TA^T. Note that it suffices to prove (AB)' = B'A'.

    But \forall x,y : [x,(AB)'y] = [ABx,y] = [A(Bx),y] = [Bx, A'y] = [x, B'A'y]

    Thus (AB)' = B'A'
    The question explicity asks for a demonstration using summation notation.

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