Working thorugh the exercises in Thorpe's book "Elementary Topics in Differential Geometry" I've got stuck on the last step of this one:
Prove that tangent space of $SL_2(\mathbb{R})$ at $I=\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$ is the set of all matrices of trace zero.
Following the hints I get that if
$$\alpha(t)=\begin{pmatrix}x_1(t) & x_2(t) \\ x_3(t) & x_4(t) \end{pmatrix}$$
is a parametrized curve in $SL_2(\mathbb{R})$ with $\alpha(t_0)=\begin{pmatrix}x_1(t_0) & x_2(t_0) \\ x_3(t_0) & x_4(t_0) \end{pmatrix}=\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$, then $(dx_1/dt)(t_0)+(dx_4/dt)(t_0)=0$.
This is true because since $\alpha(t)\in SL_2(\mathbb{R})$, it follows that $x_1(t)x_4(t)-x_2(t)x_3(t)=1$. Derivating respect to $t$ and evaluating at $t_0$:
$$\frac{dx_1(t_0)}{dt}x_4(t_0)+x_1(t_0)\frac{dx_4(t_0)}{dt}=\frac{dx_1(t_0)}{dt}\cdot 1+1 \cdot\frac{dx_4(t_0)}{dt}=0$$
Now for the last step it says: "Then use a dimensional argument" I'm not sure how to proceed.
Can someone help me complete the proof? Of course, any comment if I'm wrong with the argument above is welcomed.
Note that
$$SL_2(\mathbb R) = \{ A : \det A = 1\}$$
is a three dimensional manifold. Thus it's tangent space is also three dimensional. You've shown that
$$T = \{ X : \text{tr} X = 0\}$$
lies in the tangent space of $SL_2(\mathbb R)$ at $I$. Since $T$ is also three dimensional, it must be that $T = T_I SL_2(\mathbb R)$.