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walters
Let G be the group of all invertible 2 x 2 matrics over the C of complex numbers with matrix multiplication a)Find the subgroup of G generated by
This matrix. I got you the first time :) \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]
A=|dw:1362839192398:dw|
I have a better handwriting!!! Just kidding. So, we're taking a cyclic group, generated by that matrix above, using matrix multiplication. Evaluate this...\[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]
If matrices make you dizzy, (they make me dizzy) use this formula... \[\huge \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}p & q \\ r & s\end{matrix}\right]=\left[\begin{matrix}ap+br & aq+bs \\ cp+dr & cq+ds\end{matrix}\right]\]
looks like we could take out i's
And of course, I was doing it inefficiently :D Bloody matrices... Thanks @experimentX @walters As you 'heard', we have this convenient setup... \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=i\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]\]
before going too far wat if i use A and other matrix that in invertible
Now THAT's going too far :) We're only asked to figure out what the elements of the cyclic group generate by \[\huge\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]
are we not suppose to use those axioms of subgroups
You can do that, but why bother? Let G be a group. For any element g in G, <g> (the set generated by repeatedly operating g on itself) is always a subgroup of G. In other words, the cyclic group generated by an element of a group is a subgroup :)
Shall we proceed? :)
yes, it seems i have no idea "how to do it"
Okay, you can verify this, but \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\]
You know what? For convenience, we can let \[\huge A=\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]
isn't it obvious when you get negative of identity matrix?? when you get that matrix ... you should be very happy.
Yes. I'm very happy :D
I love identity matrix :DDD
So, @walters You're good so far with \[\huge A^2=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\]?
So, now, let's work on A^3... We get... \[\huge A^3 = A^2A=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=\left[\begin{matrix}0 & -i \\ -i & 0\end{matrix}\right]\] Following this so far?
So now, we compute A^4
\[\large A^4=A^2A^2=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\]
See where it has taken us?
So, A^4 it seems, is the identity matrix.
So... we started with A, and it turns out A^4 is the identity. So that must mean <A> has four elements, right?
wow this is interesting
So... have you arrived at your conclusion?
no it is just that i have flash even if i don't know wat to do next
Well, the 4 elements are just \[\huge <A>=\left\{I_2 \ , \ A \ , \ A^2 \ , \ A^3 \right\}\] Where \[\huge A=\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]
so it means there is no need to show those axioms of subgroup
No need at all :) It has already been assumed (correctly) that <A> is a subgroup.
Well, I'm sleepy, but I think we covered the gist of this question :) Signing off now :D ---------------------------------- Terence out