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jango_IN_DTOWN

  • one year ago

Please help

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  1. jango_IN_DTOWN
    • one year ago
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  2. jango_IN_DTOWN
    • one year ago
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    I am having trouble with the notations

  3. ganeshie8
    • one year ago
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    Hey!

  4. jango_IN_DTOWN
    • one year ago
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    hiii @ganeshie8

  5. ganeshie8
    • one year ago
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    are you familiar with special linear group ?

  6. jango_IN_DTOWN
    • one year ago
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    yes

  7. ganeshie8
    • one year ago
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    \(SL_3(\mathbb{Z})\) is the special linear group of 3x3 matrices with "integer" entries

  8. jango_IN_DTOWN
    • one year ago
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    and the determinant value will be 1

  9. ganeshie8
    • one year ago
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    yes, also we're considering only the matrices whose element at first row first column is 1

  10. ganeshie8
    • one year ago
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    as a start, consider any two such matrices, just to get a feel..

  11. jango_IN_DTOWN
    • one year ago
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    ok

  12. ganeshie8
    • one year ago
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    try below two matrices : \(\begin{bmatrix} 1&2&1\\0&1&3\\0&0&1 \end{bmatrix}\) \(\begin{bmatrix} 1&1&2\\3&0&1\\1&0&0 \end{bmatrix}\) do they belong to the given set in part i ?

  13. jango_IN_DTOWN
    • one year ago
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    |dw:1444570260285:dw|

  14. jango_IN_DTOWN
    • one year ago
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    yes the determinant value is 1

  15. ganeshie8
    • one year ago
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    more importantly the element at first row first column is 1

  16. jango_IN_DTOWN
    • one year ago
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    i guess closure property is not satisfied

  17. ganeshie8
    • one year ago
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    Exactly! \(A=\begin{bmatrix} 1&2&1\\0&1&3\\0&0&1 \end{bmatrix}\) \(B=\begin{bmatrix} 1&1&2\\3&0&1\\1&0&0 \end{bmatrix}\) \(BA\) belongs to the given set, but \(AB\) does not.

  18. jango_IN_DTOWN
    • one year ago
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    thanks we are done with the first one..

  19. ganeshie8
    • one year ago
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    for part ii, you need to think of (1, 0) as column vector, not a row vector as it was shown in the problem

  20. jango_IN_DTOWN
    • one year ago
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    yes but i dont know the notation what does it mean

  21. ganeshie8
    • one year ago
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    \(\mathbb{R}^{2*2}\) means, all 2x2 matrices with real entries

  22. ganeshie8
    • one year ago
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    \[A\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \]

  23. ganeshie8
    • one year ago
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    \[\begin{bmatrix}a&b\\c& d\end{bmatrix}\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \]

  24. ganeshie8
    • one year ago
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    to get a feel, try finding few matrices that satisfy above relation

  25. jango_IN_DTOWN
    • one year ago
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    oh now it make sense.

  26. jango_IN_DTOWN
    • one year ago
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    |dw:1444571091740:dw|

  27. ganeshie8
    • one year ago
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    Good, as you can see the first column of the matrix has to be fixed at \(\begin{array}{} 1\\ 0\end{array}\) and the second column could be anything : \[\begin{bmatrix}1&b\\0& d\end{bmatrix}\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \] above relation works for all \(a,b\in\mathbb{R}\)

  28. jango_IN_DTOWN
    • one year ago
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    yes... the second column doesn't contribute anything

  29. ganeshie8
    • one year ago
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    so, now that we know what things belong to the set, we can test if those elements form a group

  30. ganeshie8
    • one year ago
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    before even starting to test anything, whats ur gut feeling, do they form a group ?

  31. ganeshie8
    • one year ago
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    My hunch is that they should form a group because I see immediately that the identity element (I) belong to the set

  32. ganeshie8
    • one year ago
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    so, lets try and prove that the set is indeed a group under multiplication

  33. jango_IN_DTOWN
    • one year ago
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    yes and matrix multiplication is associative and closure will be satisfied since the product will give us members of R in the last column

  34. ganeshie8
    • one year ago
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    right, what about inverses ? does every element in the set has an inverse ?

  35. ganeshie8
    • one year ago
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    consider below matrix : \[\begin{bmatrix}1&0\\0& 0\end{bmatrix} \] does it belong to the set ?

  36. jango_IN_DTOWN
    • one year ago
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    yes

  37. ganeshie8
    • one year ago
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    what can you say about its inverse

  38. jango_IN_DTOWN
    • one year ago
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    its singular matrix and hence not invertible

  39. jango_IN_DTOWN
    • one year ago
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    so it will not form a group?

  40. jango_IN_DTOWN
    • one year ago
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    |dw:1444572164470:dw|

  41. ganeshie8
    • one year ago
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    Yes, every element in a group must have an inverse. Since few matrices in the set don't have inverses, the given set doesn't form a group under multiplication

  42. jango_IN_DTOWN
    • one year ago
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    so infinitely many elements dont have inverse...

  43. jango_IN_DTOWN
    • one year ago
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    thanks

  44. jango_IN_DTOWN
    • one year ago
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    and the last one we have orthogonal matrices?

  45. ganeshie8
    • one year ago
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    Yes, 2x2 orthogonal matrices quick review http://mathworld.wolfram.com/OrthogonalGroup.html

  46. jango_IN_DTOWN
    • one year ago
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    wow

  47. ganeshie8
    • one year ago
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    By definition, \(A\) is called orthogonal matrix iff \(AA^T = I\). so clearly, the inverse of \(A\) is \(A^T\)

  48. ganeshie8
    • one year ago
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    so all the elements in the set trivially have inverses

  49. jango_IN_DTOWN
    • one year ago
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    yes

  50. ganeshie8
    • one year ago
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    we have seen earlier in part ii that closure is satisfied, so...

  51. jango_IN_DTOWN
    • one year ago
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    so it will form a group

  52. ganeshie8
    • one year ago
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    I feel so, let me think a bit more..

  53. jango_IN_DTOWN
    • one year ago
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    |dw:1444572740427:dw|

  54. ganeshie8
    • one year ago
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    we have identity element in the set we have closure, and we have inverses so we must have a group !

  55. ganeshie8
    • one year ago
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    ofcourse associativity is always there because matrix multiplication is associative

  56. jango_IN_DTOWN
    • one year ago
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    |dw:1444572917065:dw|

  57. jango_IN_DTOWN
    • one year ago
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    closure will fail if we can contruct bd not equal to +1 or -1

  58. jango_IN_DTOWN
    • one year ago
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    oh b and b will be always 1or -1

  59. jango_IN_DTOWN
    • one year ago
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    since the determinant is always +1 or -1

  60. jango_IN_DTOWN
    • one year ago
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    thanks you helped me a lot, I really had problems with the notations.

  61. ganeshie8
    • one year ago
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    good observation, yeah we only have two matrices in the group I think http://www.wolframalpha.com/input/?i=solve+%7B%7B1%2Ca%7D%2C%7B0%2Cb%7D%7D*%7B%7B1%2C0%7D%2C%7Ba%2Cb%7D%7D+%3D+%7B%7B1%2C0%7D%2C%7B0%2C1%7D%7D

  62. ganeshie8
    • one year ago
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    np, do tag me in all your future group theory questions, i wanted to review these too...

  63. jango_IN_DTOWN
    • one year ago
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    Yes sure I will.

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