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I am having trouble with the notations
Hey!

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are you familiar with special linear group ?
yes
\(SL_3(\mathbb{Z})\) is the special linear group of 3x3 matrices with "integer" entries
and the determinant value will be 1
yes, also we're considering only the matrices whose element at first row first column is 1
as a start, consider any two such matrices, just to get a feel..
ok
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 ?
|dw:1444570260285:dw|
yes the determinant value is 1
more importantly the element at first row first column is 1
i guess closure property is not satisfied
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.
thanks we are done with the first one..
for part ii, you need to think of (1, 0) as column vector, not a row vector as it was shown in the problem
yes but i dont know the notation what does it mean
\(\mathbb{R}^{2*2}\) means, all 2x2 matrices with real entries
\[A\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \]
\[\begin{bmatrix}a&b\\c& d\end{bmatrix}\begin{bmatrix} 1\\0\end{bmatrix} =\begin{bmatrix} 1\\0\end{bmatrix} \]
to get a feel, try finding few matrices that satisfy above relation
oh now it make sense.
|dw:1444571091740:dw|
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 \(a,b\in\mathbb{R}\)
yes... the second column doesn't contribute anything
so, now that we know what things belong to the set, we can test if those elements form a group
before even starting to test anything, whats ur gut feeling, do they form a group ?
My hunch is that they should form a group because I see immediately that the identity element (I) belong to the set
so, lets try and prove that the set is indeed a group under multiplication
yes and matrix multiplication is associative and closure will be satisfied since the product will give us members of R in the last column
right, what about inverses ? does every element in the set has an inverse ?
consider below matrix : \[\begin{bmatrix}1&0\\0& 0\end{bmatrix} \] does it belong to the set ?
yes
what can you say about its inverse
its singular matrix and hence not invertible
so it will not form a group?
|dw:1444572164470:dw|
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
so infinitely many elements dont have inverse...
thanks
and the last one we have orthogonal matrices?
Yes, 2x2 orthogonal matrices quick review http://mathworld.wolfram.com/OrthogonalGroup.html
wow
By definition, \(A\) is called orthogonal matrix iff \(AA^T = I\). so clearly, the inverse of \(A\) is \(A^T\)
so all the elements in the set trivially have inverses
yes
we have seen earlier in part ii that closure is satisfied, so...
so it will form a group
I feel so, let me think a bit more..
|dw:1444572740427:dw|
we have identity element in the set we have closure, and we have inverses so we must have a group !
ofcourse associativity is always there because matrix multiplication is associative
|dw:1444572917065:dw|
closure will fail if we can contruct bd not equal to +1 or -1
oh b and b will be always 1or -1
since the determinant is always +1 or -1
thanks you helped me a lot, I really had problems with the notations.
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
np, do tag me in all your future group theory questions, i wanted to review these too...
Yes sure I will.

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