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Find all 2x2 matrices $ A \in SL(2,\mathbb {R}) $such that $det(A) =1$ and $A\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix} $. Do these matrices form a group?

My thoughts: I have considered the matrices \begin{pmatrix} 1 & 0\\ x & 1 \end{pmatrix} and \begin{pmatrix} 1 & x\\ 0 & 1 \end{pmatrix} I have realised that the matrices will not satisfy the condition if $x \neq 0$. These two have the identity element, and are invertible. Another matrix of interest was \begin{pmatrix} 0 & -1\\ 1 & 2 \end{pmatrix}. My problem is, how does one compactly answer this question, and I believe that the group is finite.

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  • $\begingroup$ Have you tried the subgroup test? $\endgroup$ – Michael Burr Feb 17 '18 at 14:17
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A nonempty subset $S$ of a group $G$ forms a subgroup if for all $A,B\in S$, $AB^{-1}\in S$. In this case, $S$ is nonempty because the identity is in $S$. You then assume that $A$ and $B$ are two matrices with the given property and consider (the $\frac{1}{\sqrt{2}}$ doesn't matter in this problem since it's a scaling factor). $$ AB^{-1}\begin{pmatrix}1\\-1\end{pmatrix}. $$ Since $$ B\begin{pmatrix}1\\-1\end{pmatrix}=\begin{pmatrix}1\\-1\end{pmatrix}, $$ $$ \begin{pmatrix}1\\-1\end{pmatrix}=B^{-1}\begin{pmatrix}1\\-1\end{pmatrix}. $$ Moreover, since $$ A\begin{pmatrix}1\\-1\end{pmatrix}=\begin{pmatrix}1\\-1\end{pmatrix}, $$ the product has the correct property and the set $S$ forms a group. I'm leaving the details at this point for you to fill in.

If you want to describe the group, write a general group element as $$ \begin{pmatrix}a&b\\c&d\end{pmatrix} $$ and observe that the following three equalities must hold: \begin{align*} a-b&=1\\ c-d&=-1\\ ad-bc&=1 \end{align*} You can write $a=1+b$ and $c=-1+d$. By substituting these into the determinant equation, you can solve for $c$ in terms of $a$. This gives a one-parameter family of solutions for different $a$'s.

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  • $\begingroup$ Thank you Micheal. Let me work on this , and come back to you. $\endgroup$ – TICHA Feb 17 '18 at 14:32
  • $\begingroup$ Thank you Micheal. I now have the matrix \begin{pmatrix} 1-c & -c\\c & 1+c \end{pmatrix} This matrix works like a bomb $ \forall c \in \mathbb{R}$ $\endgroup$ – TICHA Feb 17 '18 at 16:10

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