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Thread: 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 $\displaystyle A = (a_{ij})$ be a lower triangular $\displaystyle n\times n$ matrix, this means $\displaystyle a_{ij} = 0$ for $\displaystyle 1\leq i<j \leq n$.
    Let $\displaystyle B = (b_{ij})$ be a lower triangular $\displaystyle n\times n$ matrix, this means $\displaystyle b_{ij} = 0$ for $\displaystyle 1\leq i < j \leq n$.
    Define, $\displaystyle C = AB = (c_{ij})$, then $\displaystyle c_{ij} = \Sigma_k a_{ik}b_{kj}$.
    Say that $\displaystyle i<j$ then $\displaystyle a_{ik}b_{kj}=0$ for all $\displaystyle k=1,2,...,n$ so $\displaystyle c_{ij}=0$*.

    *)If $\displaystyle k<j$ then $\displaystyle b_{kj}=0$ otherwise $\displaystyle i<j\leq k$ and then $\displaystyle a_{ik}=0$.
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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)$\displaystyle k<j$ or (ii)$\displaystyle j\leq k$.
    In case (i) since $\displaystyle k<j$ it means $\displaystyle b_{kj} = 0$.
    In case (ii) since $\displaystyle j\leq k$ and $\displaystyle i<j$ it implies $\displaystyle i<k$ and so $\displaystyle a_{ik}=0$.
    Therefore, $\displaystyle a_{ik}b_{kj}=0$ for $\displaystyle k=1,2,...,n$ thus $\displaystyle c_{ij}=\Sigma_k a_{ik}b_{kj} = 0$ for $\displaystyle i<j$.
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