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.

  • $\begingroup$ Have you tried the subgroup test? $\endgroup$ – Michael Burr Feb 17 '18 at 14:17

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.