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Math Help - Product of atomic triangular matrices

  1. #1
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    Product of atomic triangular matrices

    Hello everybody,
    I am trying to prove the following: I am having a triangular n\times n matrix A (with a_{k,j} = 0 \forall k < j \land a_{j,j} = 1), I can create atomic triangular matrices A_{i} = A, a_{kj} = 0 \forall (k \neq j \land j \neq i) (in other words: For A_i I set all elements that are not on the main diagonal and that are not in column i to be zero.) Then I need to show that this is true:

    A = A_1 \cdot A_2 \cdot \dots \cdot A_n

    So far I have found: (what might even be the wrong attempt)

    \begin{aligned}{}^1\alpha_{ij} &= \sum_{k_{1}=1}^n {}^{1}a_{ik_{1}}  {{}^{2}a_{k_{1}j}}\\{}^2\alpha_{ij} &= \sum_{k_{1}=1}^n {}^{1}\alpha_{ik_{1}}  {{}^{3}a_{k_{1}j}} = \sum_{k_{1} = 1}^n \Big(\sum_{k_{2}=1}^n {}^{1}a_{ik_{2}} {{}^{2}a_{k_{2}k_{1}}}\Big) {{}^{3}a_{k_{1}j}} = \sum_{k_{1}=k_{2}=1}^n  {}^{1}a_{ik_{2}} {{}^{2}a_{k_{2}k_{1}}} {{}^{3}a_{k_{1}j}} \\ & \vdots\\ {}^{n-1}\alpha_{ij} &= \sum_{\{k_{i} = 1\}_{i \in \{1,\dots,n\}}}^n {}^1 a_{ik_{1}}{{}^{n}a_{k_{n-1}j}}\prod_{s=2}^{n-1} {}^s a_{k_{s-1}k_{s}}\end{aligned}

    where: {}^{s}\alpha_{ij} is an Element of the result after I performed the first s multiplications. and {}^{h}a_{ij} is an Element of the atomoc triangular matrix A_h.

    However, this is where I am stuck: I cannot figure out a way how to prove by the last equation that all the elements must be as in the original matrix.

    Thank you for ideas / suggestions / solutions!
    Best, Rafael
    Last edited by Opalg; May 28th 2011 at 12:21 AM. Reason: Attempted to clean up TeX
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  2. #2
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    Found a way by complete induction. Thank you!
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