Say we can find a matrix \(A \in M_n(\mathbb{R})\) such that \(\exp(A) = B\) for \(B \in \operatorname{SL}_n(\mathbb{R})\). Does \(A\) also have to be in \(\operatorname{SL}_n(\mathbb{R})\)? Does \(A\) have to satisfy \(\operatorname{tr}(A) = 0\) (i.e. \(A \in \mathfrak{sl}_n(\mathbb{R})\))?
Solution
For every real matrix \(A\), \[ \det(e^A)=e^{\operatorname{tr}A}. \] If \(e^A=B\in\SL_n(\RR)\), then \[ 1=\det B=e^{\operatorname{tr}A}. \] Because \(\operatorname{tr}A\in\RR\), this implies \[ \operatorname{tr}A=0. \] Hence \[ A\in\mathfrak{sl}_n(\RR). \]
However, \(A\) need not lie in the group \(\SL_n(\RR)\): that would require \(\det A=1\). The zero matrix already gives a counterexample, \[ e^0=I_n\in\SL_n(\RR), \qquad \det0=0. \] Thus \[ \boxed{e^A\in\SL_n(\RR)\implies A\in\mathfrak{sl}_n(\RR), \quad\text{but not necessarily }A\in\SL_n(\RR).} \]