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Math Help - need help with proof

  1. #1
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    need help with proof

    A= [aij] n x n lower triangular matrix
    B= [bij] n x n lower triangular matrix
    C= [cij] = AB

    Proof cij= 0 for i < j or lower diagonal

    cij= [ ai1b1j + ai2b2j + ...+ai(j-1)b(j-1)j ] + [ aijbjj +....+ainbnj ]


    I am a little confuse of what they are trying to show within the 2 brackets.
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  2. #2
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    Quote Originally Posted by justin016 View Post
    A= [aij] n x n lower triangular matrix
    B= [bij] n x n lower triangular matrix
    C= [cij] = AB

    Proof cij= 0 for i < j or lower diagonal

    cij= [ ai1b1j + ai2b2j + ...+ai(j-1)b(j-1)j ] + [ aijbjj +....+ainbnj ]


    I am a little confuse of what they are trying to show within the 2 brackets.
    Let A = (a_{ij}) be a lower triangular n\times n matrix, this means a_{ij} = 0 for 1\leq i<j \leq n.
    Let B = (b_{ij}) be a lower triangular n\times n matrix, this means b_{ij} = 0 for 1\leq i < j \leq n.
    Define, C = AB = (c_{ij}), then c_{ij} = \Sigma_k a_{ik}b_{kj}.
    Say that i<j then a_{ik}b_{kj}=0 for all k=1,2,...,n so c_{ij}=0*.

    *)If k<j then b_{kj}=0 otherwise i<j\leq k and then a_{ik}=0.
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  3. #3
    Senior Member Twig's Avatar
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    nice reply Hacker, thx!

    btw, is this thought process correct:
    If we multiply a matrix A by a matrix B, assuse dimensions are so that
    multiplication is defined, then AB=C, and the columns of C
    will be a linear combination of the columns of A, using the the elements in
    column j, j=1,....,n as weights ?
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  4. #4
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    thanx hacker for the reply

    if k is the column of a, how can k be greater than itself? in the statement

    i < j < k?
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  5. #5
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    Quote Originally Posted by justin016 View Post
    thanx hacker for the reply

    if k is the column of a, how can k be greater than itself? in the statement

    i < j < k?
    I consider two cases: (i) k<j or (ii) j\leq k.
    In case (i) since k<j it means b_{kj} = 0.
    In case (ii) since j\leq k and i<j it implies i<k and so a_{ik}=0.
    Therefore, a_{ik}b_{kj}=0 for k=1,2,...,n thus c_{ij}=\Sigma_k a_{ik}b_{kj} = 0 for i<j.
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