Let $x\in SL(n,\mathbb{R})$, where $$ SL(n,\mathbb{R}) = \{X\in GL(n,\mathbb{R}) \mid \det X = 1\} $$ Show that there exists $X_a$ an asymmetric and $X_s$ a symmetric matrix with $\text{tr}(X_a) = \text{tr}(X_s) = 0$ such that $x = \exp(X_a)\exp(X_s)$.
So far I have considered $x^Tx$, and since this matrix is a real symmetric, positive definite matrix, I have been able to deduce $x^Tx = \exp(X)$ for a symmetric, traceless matrix. I do not see how to proceed from here though :(
Any help is appreciated!
A great comment of @Harald_Hanche-Olsen is seems to be proper solution. You already showed, that in case of a symmetric matrice the problem is done. So consider $A$ to be a orthogonal matrix. It is a well known fact, that any orthogonal matrix is $exp(T)$, where $T = -T^{T}$. Last fact can be proved by using spectral decomposition of orthogonal matrix.