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walters

  • 2 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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  1. terenzreignz
    • 2 years ago
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    This matrix. I got you the first time :) \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]

  2. walters
    • 2 years ago
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    A=|dw:1362839192398:dw|

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

  4. terenzreignz
    • 2 years ago
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    If matrices make you dizzy, (they make me dizzy) use this formula... \[\huge \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}p & q \\ r & s\end{matrix}\right]=\left[\begin{matrix}ap+br & aq+bs \\ cp+dr & cq+ds\end{matrix}\right]\]

  5. experimentX
    • 2 years ago
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    looks like we could take out i's

  6. walters
    • 2 years ago
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    |dw:1362839499204:dw|

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

  8. walters
    • 2 years ago
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    before going too far wat if i use A and other matrix that in invertible

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

  10. walters
    • 2 years ago
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    are we not suppose to use those axioms of subgroups

  11. terenzreignz
    • 2 years ago
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    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 :)

  12. terenzreignz
    • 2 years ago
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    Shall we proceed? :)

  13. walters
    • 2 years ago
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    yes, it seems i have no idea "how to do it"

  14. terenzreignz
    • 2 years ago
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    Okay, you can verify this, but \[\huge \left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\]

  15. terenzreignz
    • 2 years ago
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    You know what? For convenience, we can let \[\huge A=\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]

  16. experimentX
    • 2 years ago
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    isn't it obvious when you get negative of identity matrix?? when you get that matrix ... you should be very happy.

  17. terenzreignz
    • 2 years ago
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    Yes. I'm very happy :D

  18. experimentX
    • 2 years ago
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    I love identity matrix :DDD

  19. terenzreignz
    • 2 years ago
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    So, @walters You're good so far with \[\huge A^2=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\]?

  20. walters
    • 2 years ago
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    yes

  21. walters
    • 2 years ago
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    we can proceed

  22. terenzreignz
    • 2 years ago
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    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?

  23. walters
    • 2 years ago
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    yes

  24. terenzreignz
    • 2 years ago
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    So now, we compute A^4

  25. terenzreignz
    • 2 years ago
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    \[\large A^4=A^2A^2=\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]\left[\begin{matrix}-1 & 0 \\ 0 & -1\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\]

  26. terenzreignz
    • 2 years ago
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    See where it has taken us?

  27. terenzreignz
    • 2 years ago
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    So, A^4 it seems, is the identity matrix.

  28. walters
    • 2 years ago
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    yes it is the identity

  29. terenzreignz
    • 2 years ago
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    So... we started with A, and it turns out A^4 is the identity. So that must mean <A> has four elements, right?

  30. walters
    • 2 years ago
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    wow this is interesting

  31. terenzreignz
    • 2 years ago
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    So... have you arrived at your conclusion?

  32. walters
    • 2 years ago
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    no it is just that i have flash even if i don't know wat to do next

  33. terenzreignz
    • 2 years ago
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    Well, the 4 elements are just \[\huge <A>=\left\{I_2 \ , \ A \ , \ A^2 \ , \ A^3 \right\}\] Where \[\huge A=\left[\begin{matrix}0 & i \\ i & 0\end{matrix}\right]\]

  34. walters
    • 2 years ago
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    so it means there is no need to show those axioms of subgroup

  35. terenzreignz
    • 2 years ago
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    No need at all :) It has already been assumed (correctly) that <A> is a subgroup.

  36. terenzreignz
    • 2 years ago
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    Well, I'm sleepy, but I think we covered the gist of this question :) Signing off now :D ---------------------------------- Terence out

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