Expressing a $SL_2(\mathbb{R})$ matrix as product of.... If $\begin{bmatrix} a&b \\ c&d \end{bmatrix}$ is some matrix in $SL_2(\mathbb{R})$, then how can we express it as a product of matrices of the following type:
$$\begin{bmatrix} s&0 \\ 0&s^{-1} \end{bmatrix}, \begin{bmatrix} 1&t \\ 0&1 \end{bmatrix}, \begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix} \text{where $s,t \neq 0$}$$ ?
I can't seem to find anything that works, and my linear algebra is a bit rusty. Thanks for any help.
 A: One way to do this would be to simply see what happens to your reference matrix if you start applying the matrices you have in your toolbox. By taking inverses appropriately (do you allow inverses in your product?), it is sufficient to write $\begin{pmatrix} s & 0 \\ 0 & s^{-1} \end{pmatrix}$ as a product of $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and the other two matrices, so let's try to do that.
So, we want to turn some of the entries of $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ into zeroes. Multiplying by $\begin{pmatrix} 1 & t_1 \\ 0 & 1 \end{pmatrix}$ from the right, we get $\begin{pmatrix} a & b+t_1a \\ c & d+t_1c \end{pmatrix}$, so choose $t_1$ so that $d + t_1c = 0$ (this is of course hard if $c = 0$ so save that case). So now we have something of the form $\begin{pmatrix} a & x \\ c & 0 \end{pmatrix}$. Apply the third matrix to this to obtain $\begin{pmatrix} -x & a \\ 0 & c \end{pmatrix}$. Now we just want to get rid of the $a$, so, as before, find a $t_2$ such that by applying $\begin{pmatrix} 1 & t_2 \\ 0 & 1 \end{pmatrix}$ form the right you're left with $\begin{pmatrix} -x & 0 \\ 0 & c \end{pmatrix}$. By letting $s = -x$, you see that this matrix is of the desired form if $-x = c^{-1}$; this, on the other hand, is where the assumption that $ad-bc = 1$ comes into play.
All we've done here is taking products of matrices so now you can work your way back.
