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$U_n=$$\left( \begin{array}{rrrr} 1 & * & \cdots & * \\ 0 & \ddots & * & \vdots \\ \vdots & 0 & \ddots & * \\ 0 & \cdots & 0 & 1 \\ \end{array}\right) $$T_n=$$\left( \begin{array}{rrrr} * & * & \cdots & * \\ 0 & \ddots & * & \vdots \\ \vdots & 0 & \ddots & * \\ 0 & \cdots & 0 & * \\ \end{array}\right) $$D_n=$ $\left( \begin{array}{rrrr} * & 0 & \cdots & 0 \\ 0 & \ddots & 0 & \vdots \\ \vdots & 0 & \ddots & 0 \\ 0 & \cdots & 0 & * \\ \end{array}\right) $

$U_n,T_n,D_n$ are all sets which describe Matrices of the forms above. Furthermore they are all subsets of $\text{GL}_n(K)$. I have to Show that a Matrix $A$ of the set $T_n$ can be describes as a product of a Matrix $U_A$ from the set $U_n$ with another Matrix $D_A$ from the set $D_n$.

Is there a way to prove it by induction?

I have looked at the Problem for $n=2$

$\begin{pmatrix} *_{11} & *_{12}\\ 0 & *_{22} \\ \end{pmatrix} $ =$\begin{pmatrix} 1 & *_{12}*_{22}^{-1}\\ 0 & 1 \\ \end{pmatrix} $$\begin{pmatrix} *_{11} & 0\\ 0 & *_{22} \\ \end{pmatrix} $

I know that I have to set the trace-elements of $D_A$ equal to the trace Elements of $A$. And from my Observation my guess for the Elements which are not within the trace of $U_A$ (i.e. they are not $1$) is that they can be calculated inductively by the element in $A$ which is in the same Position as the element we want to calculate and the inverse of the element which is below the element we want to calculate.

How can I formalize this thought, and how can I then prove this property inductively. Would appreciate an Approach which does not make use of bloxmatrices.

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  • $\begingroup$ do you know quotients of groups by subsets? I would use that, since as soon as you can proof that $D_n$ is a normal subgroup, you could just devide it out and be done, (respectively, you could do that either way, just a little less nice) $\endgroup$ – Enkidu Jan 28 '19 at 15:04
  • $\begingroup$ I have heard something About normal Groups we have defined them as subgroups of a Group where the left coset is Always equal to the Right coset. They create a Quotient set $\endgroup$ – RM777 Jan 28 '19 at 15:07
  • $\begingroup$ yep, then just use that $D_n \subset T_n$, kill it, and proof that you can identify this with $U_n$ (and as you actually do not need a groupstructure on the quotient, you can ignore the reauirements on the cosets). $\endgroup$ – Enkidu Jan 28 '19 at 15:08
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Just set $D_n := \text{diag}(T_n)$; that is, $D_n$ is the diagonal matrix consisting of the diagonal elements of $T_n$. Then, since multiplying on the right scales the rows of a matrix, set the elements of $U_n$ to be the elements of $T_n$ with each row divided by the corresponding diagonal. In other words, $(U_n)_{ij} = \frac{(T_n)_{ij}}{(D_n)_{ii}}$.

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