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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!

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    Might the polar decomposition of $x$ help? From what you have already done, you now only have to deal with the orthogonal part. – 2017-02-19
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    @HaraldHanche-Olsen I am trying to follow your suggestion, but I don't think any orthogonal matrix can be realized as the image under $\exp$ of an antisymmetric matrix. – 2017-02-19
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    @MadsFriis Each *special* orthogonal matrix can be realized as the image under $\exp$ of an antisymmetric matrix, which should be sufficient in this case. – 2017-02-19
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    @ReinhardMeier Yes, I noticed that myself, and I too believe it is sufficient. Do you have an idea on how to prove this? – 2017-02-19
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    Not an idea of my own. I have found a proof in www.cis.upenn.edu/~cis610/geombchap14.pdf where the surjectivity of the $\exp$ is stated in theorem 14.2.2. – 2017-02-19

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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.

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    Considering the last sentence: what do you mean by "spectral decomposition of orthogonal matrix"? – 2017-02-19