1. anonymous

2. anonymous

I am having trouble with the notations

3. ganeshie8

Hey!

4. anonymous

hiii @ganeshie8

5. ganeshie8

are you familiar with special linear group ?

6. anonymous

yes

7. ganeshie8

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

8. anonymous

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. anonymous

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. anonymous

|dw:1444570260285:dw|

14. anonymous

yes the determinant value is 1

15. ganeshie8

more importantly the element at first row first column is 1

16. anonymous

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. anonymous

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. anonymous

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. anonymous

oh now it make sense.

26. anonymous

|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. anonymous

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. anonymous

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. anonymous

yes

37. ganeshie8

what can you say about its inverse

38. anonymous

its singular matrix and hence not invertible

39. anonymous

so it will not form a group?

40. anonymous

|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. anonymous

so infinitely many elements dont have inverse...

43. anonymous

thanks

44. anonymous

and the last one we have orthogonal matrices?

45. ganeshie8

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

46. anonymous

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. anonymous

yes

50. ganeshie8

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

51. anonymous

so it will form a group

52. ganeshie8

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

53. anonymous

|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. anonymous

|dw:1444572917065:dw|

57. anonymous

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

58. anonymous

oh b and b will be always 1or -1

59. anonymous

since the determinant is always +1 or -1

60. anonymous

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. anonymous

Yes sure I will.