Quick question on a group definition

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The discussion centers on the definition of the group of unitriangular matrices, UT(n, ℝ), which consists of upper triangular matrices with unity on the diagonal. Participants clarify that the definition includes the condition that diagonal elements must equal one, while elements below the diagonal are zero. A user shares their progress on a coursework question, expressing confidence in their approach and noting the complexity of a subsequent question. They provide counterexamples to challenge a previously stated assertion about the group's center and conjecture a pattern for general elements of the group. The conversation highlights collaborative problem-solving and the application of mathematical concepts to coursework.
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In a piece 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?
 
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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.
 
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 :smile:
 
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.
 
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}<br /> 1 &amp; 5 &amp; 7 \\<br /> 0 &amp; 1 &amp; 9 \\<br /> 0 &amp; 0 &amp; 1<br /> \end{array} \right)

And:

B = \left( \begin{array}{ccc}<br /> 1 &amp; 3 &amp; 2 \\<br /> 0 &amp; 1 &amp; 3 \\<br /> 0 &amp; 0 &amp; 1<br /> \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 &amp; 0 &amp; a \\0 &amp; 1 &amp; 0 \\0 &amp; 0 &amp; 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.
 
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Indeed I've already worked this out, however I've not tested the trend for n > 3, I will later set up it in Mathematica to check for much larger matrices, but general life stuff to get on with now. I expect I will have determined and proven the centre by the end of the day and started on the 2nd problem :smile:
 
If it helps I've shown the conjecture is true for n = 4 and n = 5, I'm going to attempt to show it in general now.
 

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