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anonymous
 3 years ago
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
anonymous
 3 years ago
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

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terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2This matrix. I got you the first time :) \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A=dw:1362839192398:dw

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2I have a better handwriting!!! Just kidding. So, we're taking a cyclic group, generated by that matrix above, using matrix multiplication. Evaluate this...\[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2If 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
 3 years ago
Best ResponseYou've already chosen the best response.0looks like we could take out i's

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362839499204:dw

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2And of course, I was doing it inefficiently :D Bloody matrices... Thanks @experimentX @walters As you 'heard', we have this convenient setup... \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=i\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0before going too far wat if i use A and other matrix that in invertible

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Now THAT's going too far :) We're only asked to figure out what the elements of the cyclic group generate by \[\huge\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are we not suppose to use those axioms of subgroups

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2You 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 :)

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Shall we proceed? :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, it seems i have no idea "how to do it"

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Okay, 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
 3 years ago
Best ResponseYou've already chosen the best response.2You know what? For convenience, we can let \[\huge A=\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0isn't it obvious when you get negative of identity matrix?? when you get that matrix ... you should be very happy.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Yes. I'm very happy :D

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0I love identity matrix :DDD

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2So, @walters You're good so far with \[\huge A^2=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\]?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2So, 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?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2So now, we compute A^4

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2\[\large A^4=A^2A^2=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\]

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2See where it has taken us?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2So, A^4 it seems, is the identity matrix.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes it is the identity

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2So... we started with A, and it turns out A^4 is the identity. So that must mean <A> has four elements, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wow this is interesting

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2So... have you arrived at your conclusion?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no it is just that i have flash even if i don't know wat to do next

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Well, the 4 elements are just \[\huge <A>=\left\{I_2 \ , \ A \ , \ A^2 \ , \ A^3 \right\}\] Where \[\huge A=\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so it means there is no need to show those axioms of subgroup

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2No need at all :) It has already been assumed (correctly) that <A> is a subgroup.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.2Well, I'm sleepy, but I think we covered the gist of this question :) Signing off now :D  Terence out
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