Tangent space of $SL(2)$ at $A$ Consider $SL(2)=\{A \in \mathbb R^{2 \times 2}|\det A=1\}$.
I want to determine the tangent space of $SL(2)$ at $A \in SL(2)$. Let's call it $T_A$.
$A=\begin{pmatrix}a_1 && a_2 \\ a_3 && a_4\end{pmatrix}$.
$SL(2)=\{A \in \mathbb R^{2 \times 2}|F(A)=0\}$,
where $F:\mathbb R^{2 \times 2}\to \mathbb R$, $F(A)=\det(A)-1=a_1a_4-a_2a_3$.
$dF(A)=(a_4,-a_3,-a_2,a_1)$. So $SL(2)$ is a $3$-dimensional manifold in $\mathbb R^4$.
I want to use this to determine the tangent space described above.
 $dF(A)$ should be a normal vector of $T_A$ if I'm not mistaken. So $T_A=(\mathrm{span} [dF(A)])^\perp$?

That's one way to do this. That said, there are a couple possible improvements.
First one : $F$ is a function from $\mathbb{R}^{2 \times 2} \to \mathbb{R}$. The derivative a real function of multiple real variables is not a vector, but a covector (or a row-vector).
Here, you are, in order :


*

*working with the gradient (essentially, the transposee of $dF(A)$). Note that, by the way, you should have written $\nabla F (A)$ instead of $dF(A)$ to distinguish between derivative and gradient.

*taking the orthogonal of the image of what you got. 


That's correct, but if you start by seeing $DF(A)$ as a covector -- so a linear transformation from $\mathbb{R}^{2 \times 2} \to \mathbb{R}$ -- you just get $T_A = Ker (dF(A))$. There is no need for computation of transposee, image, orthogonal space.
Second one : compute directly with matrices. For $A \in SL(2)$, 
$$F(A+H) -F(A)= det (A+H) = det(A)det(I+A^{-1}H) = 1 + dF(I)(A^{-1}H) + O(\|H\|^2),$$
so that
$$dF (A) (H) =  dF(I)(A^{-1}H) = Tr(A^{-1}H).$$
This works well thanks to the algebraic propertis of the determinant, although the same trick extends to other similar computations (tangent space of $O(n)$, $Sp(2n)$...).
What you get works in any dimension, and is easier to manipulate than identities written in a basis of $\mathbb{R}^{2 \times 2}$.
