I'm trying to prove that every matrix in $Sl(2,\mathbb{R})$ can be written as a product of two exponential matrix. First I noted that every matrix in $Sl(2,\mathbb{R})$ can be written as a product of a orthogonal matrix and a upper triangular matrix, so a orthogonal matrix can be written as exponential of some matrix, but my problem is with the upper triangular matrix. If the diagonal elements were the same my problem will be done, but they are different, I'm stucked here, any help will be welcome.
Thank you.
You already know you can write an element $A \in \rm{SL}(2,\Bbb{R})$ as $A=KR$ for $K$ orthogonal and $R$ upper triangular, so $$ R = \left \{ \begin{pmatrix} a & b \\ 0 & d \end{pmatrix} \mid a,d \ne 0\right\}. $$ If $a$ or $d$ is negative you can left multiply by a diagonal matrix $D$ with $\pm 1$ on the diagonal and rewrite $A = (KD)(DR)$ so that $DR$ has positive entries on the diagonal. Since $A$ and $DR$ have positive determinant, $KD$ has determinant $1$ and is an element of $\rm{SO}(2)$, so it is the exponential of a skew-symmetric matrix. Then $DR$ has determinant $1$ and its diagonal entries are multiplicative inverses. It remains to show that the matrix $$ DR = \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} $$ with $a>0$ has a logarithm. But you can check that the matrix exponential of $$ \begin{pmatrix} x & y \\ 0 & -x \end{pmatrix} $$ is $$ \begin{pmatrix} e^x & \frac{e^x - e^{-x}}{2 x} y \\ 0 & e^{-x} \end{pmatrix} $$ and you can solve for $x,y$ in terms of $a,b$.