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## jango_IN_DTOWN one year ago Please help

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1. jango_IN_DTOWN

2. jango_IN_DTOWN

I am having trouble with the notations

3. ganeshie8

Hey!

4. jango_IN_DTOWN

hiii @ganeshie8

5. ganeshie8

are you familiar with special linear group ?

6. jango_IN_DTOWN

yes

7. ganeshie8

$$SL_3(\mathbb{Z})$$ is the special linear group of 3x3 matrices with "integer" entries

8. jango_IN_DTOWN

and the determinant value will be 1

9. ganeshie8

yes, also we're considering only the matrices whose element at first row first column is 1

10. ganeshie8

as a start, consider any two such matrices, just to get a feel..

11. jango_IN_DTOWN

ok

12. ganeshie8

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

|dw:1444570260285:dw|

14. jango_IN_DTOWN

yes the determinant value is 1

15. ganeshie8

more importantly the element at first row first column is 1

16. jango_IN_DTOWN

i guess closure property is not satisfied

17. ganeshie8

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

thanks we are done with the first one..

19. ganeshie8

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

yes but i dont know the notation what does it mean

21. ganeshie8

$$\mathbb{R}^{2*2}$$ means, all 2x2 matrices with real entries

22. ganeshie8

$A\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix}$

23. ganeshie8

$\begin{bmatrix}a&b\\c& d\end{bmatrix}\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix}$

24. ganeshie8

to get a feel, try finding few matrices that satisfy above relation

25. jango_IN_DTOWN

oh now it make sense.

26. jango_IN_DTOWN

|dw:1444571091740:dw|

27. ganeshie8

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

yes... the second column doesn't contribute anything

29. ganeshie8

so, now that we know what things belong to the set, we can test if those elements form a group

30. ganeshie8

before even starting to test anything, whats ur gut feeling, do they form a group ?

31. ganeshie8

My hunch is that they should form a group because I see immediately that the identity element (I) belong to the set

32. ganeshie8

so, lets try and prove that the set is indeed a group under multiplication

33. jango_IN_DTOWN

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

right, what about inverses ? does every element in the set has an inverse ?

35. ganeshie8

consider below matrix : $\begin{bmatrix}1&0\\0& 0\end{bmatrix}$ does it belong to the set ?

36. jango_IN_DTOWN

yes

37. ganeshie8

what can you say about its inverse

38. jango_IN_DTOWN

its singular matrix and hence not invertible

39. jango_IN_DTOWN

so it will not form a group?

40. jango_IN_DTOWN

|dw:1444572164470:dw|

41. ganeshie8

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

so infinitely many elements dont have inverse...

43. jango_IN_DTOWN

thanks

44. jango_IN_DTOWN

and the last one we have orthogonal matrices?

45. ganeshie8

Yes, 2x2 orthogonal matrices quick review http://mathworld.wolfram.com/OrthogonalGroup.html

46. jango_IN_DTOWN

wow

47. ganeshie8

By definition, $$A$$ is called orthogonal matrix iff $$AA^T = I$$. so clearly, the inverse of $$A$$ is $$A^T$$

48. ganeshie8

so all the elements in the set trivially have inverses

49. jango_IN_DTOWN

yes

50. ganeshie8

we have seen earlier in part ii that closure is satisfied, so...

51. jango_IN_DTOWN

so it will form a group

52. ganeshie8

I feel so, let me think a bit more..

53. jango_IN_DTOWN

|dw:1444572740427:dw|

54. ganeshie8

we have identity element in the set we have closure, and we have inverses so we must have a group !

55. ganeshie8

ofcourse associativity is always there because matrix multiplication is associative

56. jango_IN_DTOWN

|dw:1444572917065:dw|

57. jango_IN_DTOWN

closure will fail if we can contruct bd not equal to +1 or -1

58. jango_IN_DTOWN

oh b and b will be always 1or -1

59. jango_IN_DTOWN

since the determinant is always +1 or -1

60. jango_IN_DTOWN

thanks you helped me a lot, I really had problems with the notations.

61. ganeshie8

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

np, do tag me in all your future group theory questions, i wanted to review these too...

63. jango_IN_DTOWN

Yes sure I will.

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