Show that $ \exp \left(\mathfrak{sl}(2,R)\right)$ is the set of all matrices with positive trace $\geq -2$ Using the fact that every matrix in $SL(2,\mathbb{R})$ is conjugate in $SL(2,\mathbb{R})$ to one of the following matrices:
$$
\left(\begin{array}{rr}
a & 0\\
0 & \frac{1}{a}
\end{array}\right), \quad 
\left(\begin{array}{rr}
1 & t\\
0 & 1
\end{array}\right), \quad 
\left(\begin{array}{rr}
-1 & t\\
0 & -1
\end{array}\right) \quad \mbox{y} \quad
\left(\begin{array}{rr}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{array}\right),  
$$
Show that the image of the exponential map
$$\exp\: : \: \left\{ \left(\begin{array}{rr}
x & y \\
z & -x
\end{array}\right) \: : \: x,y,z \in \mathbb{R}\right\} \longrightarrow SL(2,\mathbb{R})$$
is
\begin{equation}\left\{M\in SL(2,\mathbb{R}) \: : \: tr(M)>-2\right\} \cup \left\{-I\right\}. \qquad  (*)\end{equation}
Remark: I have tried many things, for example, we know that
$$\exp(PJP^{-1}) =P\exp(J)P^{-1}$$
We also know that the trace is invariant under cujugación, this is
$$tr\left(\exp(PJP^{-1}) \right)=tr\left(\exp(J)\right).$$
Therefore, if $M \in \exp \left(SL(2,R)\right)$, then there is $A\in SL(2,R)$ such that $M=\exp(A)$, then, as $A$ is conjugate to one of above matrices, then 
$$tr(M)= \left\{
\begin{array}{}
tr\left(\exp\left(\left(\begin{array}{rr}
a & 0\\
0 & \frac{1}{a}
\end{array}\right) \right) \right) & \mbox{if } A \mbox{ is conjugate to } \left(\begin{array}{rr}
a & 0\\
0 & \frac{1}{a}
\end{array}\right)\\
tr\left(\exp\left( \left(\begin{array}{rr}
1 & t\\
0 & 1
\end{array}\right)\right) \right) & \mbox{if } A \mbox{ is conjugate to } \left(\begin{array}{rr}
1 & t\\
0 & 1
\end{array}\right)\\
tr\left(\exp\left(\left(\begin{array}{rr}
-1 & t\\
0 & -1
\end{array}\right) \right) \right) & \mbox{if } A \mbox{ is conjugate to } \left(\begin{array}{rr}
-1 & t\\
0 & -1
\end{array}\right)\\
tr\left(\exp\left( \left(\begin{array}{rr}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{array}\right) \right) \right) & \mbox{if } A \mbox{ is conjugate to } \left(\begin{array}{rr}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{array}\right)\\
\end{array}
\right.$$
From the latter, the problem that I have is to show that all matris in $ \exp \left(SL(2,R)\right)$ is in the set $(*)$. The biggest problem I have when I want to show that all matris in set $(*)$ is in $ \exp \left(SL(2,R)\right)$. 
 A: Notation: We will use the following notations. 
$$ r(\theta) = \left ( \begin{matrix} 0 & \theta \\ -\theta & 0 \end{matrix} \right ) $$
$$ n(x) = \left ( \begin{matrix} 0 & x \\ 0 & 0 \end{matrix} \right ) $$
$$ a(t) = \left ( \begin{matrix} t & 0 \\ 0 & -t \end{matrix} \right ) $$
where $\theta$, $x$, $t$ are real numbers.
We will use  Iwasawa decomposition . It says that any matrix in $SL(2,\mathbb{R})$ is of the form $KAN$ where $K=$ a rotation matrix, $A$ is a diagonal matrix (with positive entries) and $N$ is a unipotent matrix. 
We will show that we can find trace zero real matrices $(k,a,n)$ such that $e^{k} = K$, $e^{a} = A$ etc.
For any diagonal matrix $A$ we can take $a$ = diagonal matrix with entries $ln(a)$ and $ -ln(a)$ on the diagonal.
For any unipotent matrix $N(x)$ with $x$ on the off-diagonal matrix we can take $n(x)$. 
Finally if $K(\theta)$ is the rotation matrix, which rotates the plane by $\theta$ then consider the matrix $r(\theta)$ above. The $n$-th product $r(\theta)^{n}$ will only have diagonal or off-diagonal elements depending on whether $n$ is odd or even. It is easy to see that $e^{r(\theta)} = K(\theta)$ when you substitute power-series expressions for $\cos(\theta)$ and $\sin(\theta)$.  
The estimates on the eigenvalues follow easily. 
