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Math Help - Transpose theorem proof

  1. #1
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    Transpose theorem proof

    (A*B)^t = B^t * A^t

    So, my problem is, how to proof this is true?

    What I could do:
    Lets Say A axb and B bxc, so, in this case, AB axc, and obviously AB^T cxa. The only way of getting AB^T cxa is multiplying B^t cxb for A^t bxa, which would be false if we used A^t bxa * B^t cxa, since a =/= c. Even though this makes sense, is this a valid proof? And is there a cleaner one?

    PS: A and B are matrices.
    Last edited by Capes; March 23rd 2011 at 01:35 PM. Reason: forgot to add something
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  2. #2
    Senior Member roninpro's Avatar
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    To be honest, I have no idea what you are talking about, in the statement and your "proof". What do you mean by (ab)^t=a^t b^t? Are a,b supposed to be matrices? If so, your formula is false, in general. You really want (ab)^t=b^ta^t. Also, what are Aaxb and Bbxc?
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  3. #3
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    Ahh, sorry about that, I meant (A*B)^t = B^T * A^T, yes, where both of them are matrices. Aaxb means a general matrix A with a rows and b columns.
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  4. #4
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    Let \mathbf{r}_i be the ith row of A.
    Let \mathbf{c}_j be the jth column of B.

    So the d_{ij} i,jth entry D of the product AB=D

    \mathbf{d}_{ij}=\mathbf{r}_i\cdot \mathbf{c}_j

    (AB)^T=D^T Now lets compare with

    B^TA^T

    Now this exchanges the rows and columns of A and B

    So

    \mathbf{d}_{ji}=\mathbf{c}_j\cdot \mathbf{r}_i

    Now what can we conclude?
    Last edited by TheEmptySet; March 23rd 2011 at 03:33 PM. Reason: used the same letter for two different things.
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  5. #5
    Senior Member roninpro's Avatar
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    Quote Originally Posted by Capes View Post
    Ahh, sorry about that, I meant (A*B)^t = B^T * A^T, yes, where both of them are matrices. Aaxb means a general matrix A with a rows and b columns.
    No problem. You might want to try something like A_{a\times b} (code: A_{a\times b}) next time.
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