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Recall that any unit lower triangular matrix $L\in\mathbb{R}^{n\times n}$ can be written in factored form as \begin{equation} L = M_1 M_2\ldots M_{n-1} \end{equation} where $M_i = I + l_i e_i^{T}$ is an elementary unit lower triangular matrix (column form). Given a simple elementary unit lower triangular matrix (element form) that differs from the identity in one off-diagonal element in the strict lower triangular part, i.e. $$E_{ij} = I + \lambda_{ij}e_i e_j^{T}$$ where $i\neq j$.

i.) Write a column form elementary matrix $M_i$ in terms of element form elementary matrices. Does the order of the $E_{ji}$ matter in this product?

ii.) Show how it follows that the factorization of $L$ is easily expressed in terms of element form elementary matrices.

iii.) Show that the expression from part (ii) can be rearranged to form $L = R_2\ldots R_n$ where $R_i = I + e_i r_i^{T}$ is an elementary unit lower triangular matrix in row form.

i.) From the answer provided we let $\lambda_{ij}$ denote the $j$th entry of the vector $l_i$. We note that $$M_i = I + l_i e_{i}^{T} = \prod_{j=1}^{i-1}(I + \lambda_{ij}e_je_i^{T}) = \prod_{j=1}^{i-1}E_{ji}$$

ii.) So we have $$M_i = \prod_{j=1}^{i-1}E_{ji}$$ and $$L = M_1\ldots M_{n-1} = \prod_{j=1}^{i-1}E_{ji}$$ where do we go from here?

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Let $\lambda_{ij}$ denote the $j$th entry of the vector $l_i$. We note that $$ (I + l_i e_i^T) = \prod_{j=1}^{i-1} (I + \lambda_{ij} e_je_i^T) $$ Note that these $E_{ji}$ commute with $[e_je_i^T][e_{j'}e_i^T] = 0$ so long as neither $j$ nor $j'$ is equal to $i$.

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  • $\begingroup$ Could you expand on your answer I am not sure I completely understand $\endgroup$ – Wolfy May 22 '16 at 10:59
  • $\begingroup$ I'm not sure what you're expecting me to add. Can you be more specific about what you're not getting? $\endgroup$ – Omnomnomnom May 22 '16 at 12:36
  • $\begingroup$ Never mind I actually understand it now. This answer you provided is for (i) correct? If so do you know how I would proceed with (ii)? $\endgroup$ – Wolfy May 22 '16 at 12:41
  • $\begingroup$ I re-edited my question to include your solution and the set up for (ii) $\endgroup$ – Wolfy May 22 '16 at 12:49
  • $\begingroup$ To get an answer for (ii), note that we can just put the factorizations for each $M_i$ next to each other. Note, however, that our factorization will have one matrix per entry above the diagonal. $\endgroup$ – Omnomnomnom May 22 '16 at 12:56

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