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

walters
 one year ago
Best ResponseYou've already chosen the best response.0A=dw:1362839192398:dw

terenzreignz
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0looks like we could take out i's

terenzreignz
 one year 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... \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=i\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]\]

walters
 one year 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
 one year 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 \[\huge\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]

walters
 one year ago
Best ResponseYou've already chosen the best response.0are we not suppose to use those axioms of subgroups

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

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

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

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

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

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

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

walters
 one year ago
Best ResponseYou've already chosen the best response.0yes it is the identity

terenzreignz
 one year 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?

walters
 one year ago
Best ResponseYou've already chosen the best response.0wow this is interesting

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

walters
 one year 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
 one year 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 \[\huge A=\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]

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

terenzreignz
 one year 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
 one year 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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