How to construct the matrix Lie group from the corresponding Lie algebra, using the exponential map? Let $\frak{g}$ be the (real) Lie algebra:
$${\frak g}=\{A =M_n(\mathbb{R}) \mid \text{tr}(A)=0\}$$
with Lie bracket $[x,y]=xy-yx$ for $x,y\in \frak g$
Than how do I construct the corresponding Lie group by using the exponential map?
 A: Let's collect a few facts:
(i) For any matrix $A$, $e^A$ is well defined.
(ii) We have $\det e^A = e^{\mathrm{Tr} A}$. Hence exponential map takes Lie algebra of all matrices $\mathfrak{gl}(n)$ into group of nonsingular matrices $\mathrm{GL}(n)$. Furthermore $\mathrm{Tr}A=0$ if and only if $\det e^A =1$.
From that you can see that your Lie algebra, known as $\mathfrak{sl}(n)$ is Lie algebra of $\mathrm{SL}(n)$ - group of matrices of unit determinant.
Edit:
As Dietrich Burde correctly pointed out, group you are looking for is not uniquely determined. Given two groups with isomorphic Lie algebras you can only infer that they are isomorphic in some neighbourhood of identity. Well known examples of this phenomenon are: $SU(2)$, $SO(3)$, $O(3)$. They all have the same Lie algebra. However $O(3)$ is distinct from $SO(3)$ as it contains two disjoint connected components (both diffeomorphic to $SO(3)$). $SU(2)$ on the other hand is connected and even simply-connected (which $SO(3)$ is not). In fact it is double cover of $SO(3)$. Even simpler example is circle group $SO(2) \cong U(1) \cong S^1$ which is locally isomorphic to $\mathbb R$ with addition. In fact $\mathbb R$ is simply connected cover of circle group. So without some extra conditions you can't quite pin down what group is that.
