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anonymous
 one year ago
Please help
anonymous
 one year ago
Please help

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am having trouble with the notations

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2are you familiar with special linear group ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(SL_3(\mathbb{Z})\) is the special linear group of 3x3 matrices with "integer" entries

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and the determinant value will be 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2yes, also we're considering only the matrices whose element at first row first column is 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2as a start, consider any two such matrices, just to get a feel..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2try 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 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444570260285:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes the determinant value is 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2more importantly the element at first row first column is 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i guess closure property is not satisfied

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Exactly! \(A=\begin{bmatrix} 1&2&1\\0&1&3\\0&0&1 \end{bmatrix}\) \(B=\begin{bmatrix} 1&1&2\\3&0&1\\1&0&0 \end{bmatrix}\) \(BA\) belongs to the given set, but \(AB\) does not.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks we are done with the first one..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2for part ii, you need to think of (1, 0) as column vector, not a row vector as it was shown in the problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but i dont know the notation what does it mean

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(\mathbb{R}^{2*2}\) means, all 2x2 matrices with real entries

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[A\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\begin{bmatrix}a&b\\c& d\end{bmatrix}\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2to get a feel, try finding few matrices that satisfy above relation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh now it make sense.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444571091740:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Good, 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 \(a,b\in\mathbb{R}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes... the second column doesn't contribute anything

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so, now that we know what things belong to the set, we can test if those elements form a group

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2before even starting to test anything, whats ur gut feeling, do they form a group ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2My hunch is that they should form a group because I see immediately that the identity element (I) belong to the set

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so, lets try and prove that the set is indeed a group under multiplication

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes and matrix multiplication is associative and closure will be satisfied since the product will give us members of R in the last column

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2right, what about inverses ? does every element in the set has an inverse ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2consider below matrix : \[\begin{bmatrix}1&0\\0& 0\end{bmatrix} \] does it belong to the set ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2what can you say about its inverse

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its singular matrix and hence not invertible

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it will not form a group?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444572164470:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so infinitely many elements dont have inverse...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and the last one we have orthogonal matrices?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes, 2x2 orthogonal matrices quick review http://mathworld.wolfram.com/OrthogonalGroup.html

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2By definition, \(A\) is called orthogonal matrix iff \(AA^T = I\). so clearly, the inverse of \(A\) is \(A^T\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so all the elements in the set trivially have inverses

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2we have seen earlier in part ii that closure is satisfied, so...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it will form a group

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I feel so, let me think a bit more..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444572740427:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2we have identity element in the set we have closure, and we have inverses so we must have a group !

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2ofcourse associativity is always there because matrix multiplication is associative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444572917065:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0closure will fail if we can contruct bd not equal to +1 or 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh b and b will be always 1or 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since the determinant is always +1 or 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks you helped me a lot, I really had problems with the notations.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2good 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2np, do tag me in all your future group theory questions, i wanted to review these too...
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