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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.

User43029
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  • Why not take the $\log$ (not necessarily unique) of the upper triangular part? All eigenvalues are non-zero. – copper.hat Sep 06 '13 at 15:20
  • Do I have how to prove that the trace of the log of that upper triangular matrix is zero? – User43029 Sep 06 '13 at 17:25
  • @copper.hat ??? – User43029 Sep 06 '13 at 17:57
  • Actually, I'm not sure what you are looking for exactly. Any member of this group is non-singular and can be written as $A=e^{\log A}$. Do you need a real log? – copper.hat Sep 06 '13 at 18:47
  • @Shalop: Try $e^{\begin{bmatrix} i\pi & -1 \ 0 & i \pi \end{bmatrix}}$. – copper.hat Oct 13 '18 at 17:53
  • @copper.hat What I meant was that $\begin{bmatrix} -1 & 1 \ 0 & -1 \end{bmatrix}$ cannot be written as the exponential of any matrix with real entries. For sure it is true that the exponential is surjective from $M_n \Bbb C$ to $GL_n \Bbb C$, but this fails quite badly if you replace $\Bbb C$ with $\Bbb R$ (although the OP did not specify, I assume he/she meant this for $\Bbb R$). – shalop Oct 14 '18 at 01:02
  • @Shalop: I understand, but the OP never responded to my inquiry on my 3rd comment. It is not clear what context is needed, for proof purposes a complex $\log$ might suffice? – copper.hat Oct 14 '18 at 01:10
  • Sorry, I hadn't seen the sentence "Do you need a real log?" So, my complaints had no basis in the first place. – shalop Oct 15 '18 at 01:30

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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 $a,b$.

Max
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