# Quick question on a group definition

Homework Helper

## Main Question or Discussion Point

In a peice of coursework I need to work out some stuff about $UT \left( n, \mathbb{R} \right)$.
I forget what the definition of this, is it: given $M \in UT \left( n, \mathbb{R} \right)$ and represented by $M = \left(a_{i, j} \right)$ Where $a_{i, j}$ is a typical element of M then if i > j $a_{i, j} = 0$ else $a_{i, j} \in \mathbb{R}$.
Or was there the added condition that if i = j then $a_{i, j} = 1$?

Last edited:

## Answers and Replies

Related Linear and Abstract Algebra News on Phys.org
Group Definition

This group is called the group of unitriangular matrices, so called because it contains (upper) triangular matrices with the property that the diagonal elements are all unity. As such, the added condition is needed.
If you attended Dr Stöhr's lectures you should know this.

Homework Helper
Diophantus said:
This group is called the group of unitriangular matrices, so called because it contains (upper) triangular matrices with the property that the diagonal elements are all unity. As such, the added condition is needed.
If you attended Dr Stöhr's lectures you should know this.
Haha thanks, I would also know this if Ruby wasn't being weird and gave me chance to photocopy her notes.

Anyway, I've made quite a bit of progress on that question, I think I have a better method of tackling it than any that we discussed

Oh right, I'm quite sure that my approach will work adequately so I'm sticking with that. I have a feeling that the second question will be trickier though but I'm not going to get stuck into it for a few days. I'll probably be in touch when I've had the chance to work on it.

Homework Helper
kk, well, I just wanted to say some thing. I've just done a bit of work on the 1st question and you have the centre of the group wrong, I have a counter example.

Try something like:

$$A = \left( \begin{array}{ccc} 1 & 5 & 7 \\ 0 & 1 & 9 \\ 0 & 0 & 1 \end{array} \right)$$

And:

$$B = \left( \begin{array}{ccc} 1 & 3 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array} \right)$$

Particularly check the element of the 1st row, 3rd column for AB and BA

Alas, you are correct. That will teach me to use silly notation for the elements of my matrices(a, a', a'', a'''...). Closer examination leads to the conclusion that the general element of the group when n=3 is:

$$A = \left( \begin{array}{ccc}1 & 0 & a \\0 & 1 & 0 \\0 & 0 & 1\end{array} \right)$$

and I conjecture that for larger n, the trend will continue i.e the general element is simply I with the top right zero replaced with a variable free to take any real value.

This should not, however, affect my chosen method of showing this to be true for all n.

In fact it should simplify it.

Last edited: