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Jacobi's formula says that:

$$\det e^{X}=e^{\operatorname{Tr}(X)}$$

So for any matrix $A$, I could try to find a matrix $X$ (the equivalent to a group generator) such that $A=e^{X}$ holds. But if $\det A$ is negative or zero, then $\det A=\det e^{X}=e^{\operatorname{Tr}(X)}$ would not be true, because $e^{\operatorname{Tr}(X)}$ is always bigger than zero.

Where is the failure in my reasoning?

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up vote 4 down vote accepted

There's no failure, unless you count assuming that such a matrix $X$ exists for any $A$. You've in fact given a proof that if $\det{A}\leq0$, there is no $X$ (with real trace) such that $e^X=A$.

If you allow $\operatorname{Tr}(X)$ to be complex (so it is possible for $e^{\operatorname{Tr}(X)}$ to be negative), then there does exist $X$ with $e^X=A$ for any invertible $A$, but not for general $A$. This $X$ is not generally unique.

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¿So does this mean that for any non-special linear group, it's generators must be complex? – jinawee Apr 22 '13 at 17:45
No group of complex matrices is generated as a group by real matrices, as real numbers are closed under the required operation for matrix multiplication. As the exponential of a real matrix is also real, it isn't generated by exponentials of real elements of the Lie algebra either. If you restrict to working with groups of real matrices, I'm not sure what happens exactly. – Matthew Pressland Apr 22 '13 at 19:08

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