# Thread: Product of atomic triangular matrices

1. ## 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

2. Found a way by complete induction. Thank you!