Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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
 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
 one year ago

This Question is Closed

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

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

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

terenzreignzBest ResponseYou've already chosen the best response.2
If matrices make you dizzy, (they make me dizzy) use this formula... \[\huge \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}p & q \\ r & s\end{matrix}\right]=\left[\begin{matrix}ap+br & aq+bs \\ cp+dr & cq+ds\end{matrix}\right]\]
 one year ago

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

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

waltersBest ResponseYou've already chosen the best response.0
before going too far wat if i use A and other matrix that in invertible
 one year ago

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

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

terenzreignzBest ResponseYou've already chosen the best response.2
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 :)
 one year ago

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

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

terenzreignzBest ResponseYou've already chosen the best response.2
Okay, you can verify this, but \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\]
 one year ago

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

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

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

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

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

terenzreignzBest ResponseYou've already chosen the best response.2
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?
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

terenzreignzBest ResponseYou've already chosen the best response.2
Well, I'm sleepy, but I think we covered the gist of this question :) Signing off now :D  Terence out
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.