# Excercise dealing with subgroups of $\text{GL}_n(K)$

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

$$U_n,T_n,D_n$$ 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.

• 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) – Enkidu Jan 28 '19 at 15:04
• 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 – RM777 Jan 28 '19 at 15:07
• 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). – Enkidu Jan 28 '19 at 15:08

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, $$(U_n)_{ij} = \frac{(T_n)_{ij}}{(D_n)_{ii}}$$.