walters
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
terenzreignz
  • terenzreignz
This matrix. I got you the first time :) \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]
walters
  • walters
A=|dw:1362839192398:dw|
terenzreignz
  • terenzreignz
I have a better handwriting!!! Just kidding. So, we're taking a cyclic group, generated by that matrix above, using matrix multiplication. Evaluate this...\[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]

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terenzreignz
  • terenzreignz
If matrices make you dizzy, (they make me dizzy) use this formula... \[\huge \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}p & q \\ r & s\end{matrix}\right]=\left[\begin{matrix}ap+br & aq+bs \\ cp+dr & cq+ds\end{matrix}\right]\]
experimentX
  • experimentX
looks like we could take out i's
walters
  • walters
|dw:1362839499204:dw|
terenzreignz
  • terenzreignz
And of course, I was doing it inefficiently :D Bloody matrices... Thanks @experimentX @walters As you 'heard', we have this convenient setup... \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=i\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]\]
walters
  • walters
before going too far wat if i use A and other matrix that in invertible
terenzreignz
  • terenzreignz
Now THAT's going too far :) We're only asked to figure out what the elements of the cyclic group generate by \[\huge\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]
walters
  • walters
are we not suppose to use those axioms of subgroups
terenzreignz
  • terenzreignz
You can do that, but why bother? Let G be a group. For any element g in 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 :)
terenzreignz
  • terenzreignz
Shall we proceed? :)
walters
  • walters
yes, it seems i have no idea "how to do it"
terenzreignz
  • terenzreignz
Okay, you can verify this, but \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\]
terenzreignz
  • terenzreignz
You know what? For convenience, we can let \[\huge A=\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]
experimentX
  • experimentX
isn't it obvious when you get negative of identity matrix?? when you get that matrix ... you should be very happy.
terenzreignz
  • terenzreignz
Yes. I'm very happy :D
experimentX
  • experimentX
I love identity matrix :DDD
terenzreignz
  • terenzreignz
So, @walters You're good so far with \[\huge A^2=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\]?
walters
  • walters
yes
walters
  • walters
we can proceed
terenzreignz
  • terenzreignz
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?
walters
  • walters
yes
terenzreignz
  • terenzreignz
So now, we compute A^4
terenzreignz
  • terenzreignz
\[\large A^4=A^2A^2=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\]
terenzreignz
  • terenzreignz
See where it has taken us?
terenzreignz
  • terenzreignz
So, A^4 it seems, is the identity matrix.
walters
  • walters
yes it is the identity
terenzreignz
  • terenzreignz
So... we started with A, and it turns out A^4 is the identity. So that must mean has four elements, right?
walters
  • walters
wow this is interesting
terenzreignz
  • terenzreignz
So... have you arrived at your conclusion?
walters
  • walters
no it is just that i have flash even if i don't know wat to do next
walters
  • walters
so it means there is no need to show those axioms of subgroup
terenzreignz
  • terenzreignz
Well, I'm sleepy, but I think we covered the gist of this question :) Signing off now :D ---------------------------------- Terence out

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