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I need to build a matrix $A\in Gl^+_n(\mathbb{R})$ ($det(A)>0$) that is not an exponent, i.e. there is no $B\in Mat_n(\mathbb{R})$ such that $A=exp(B)$.
Could you give me a hint to note some special property of an exponent?

Perhaps the first part duplicates some past questions but I also want to know if it is possible to prove that $\{exp(A): A\in Mat_n(\mathbb{R})\}$ is not $GL^+_n$ topologically (for instance, using that $G_A=\{exp(tA): t\in \mathbb{R}\}$ is a one dimensional subgroup in $GL_n$)?

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Consider $n=2$. If the eigenvalues of $B$ are $s$ and $t$, then the eigenvalues of $\exp(B)$ are $e^s$ and $e^t$. There are 2 possibilities: Either $s$ and $t$ are both real, or they are complex conjugates. In either cases, there are some constraints on $e^s$ and $e^t$. Find the constraints in the 2 cases, then find a matrix in $A \in GL_2^+(\mathbb{R})$ whose eigenvalues satisfy neither constraint, and therefore $A$ cannot be equal to any $\exp(B)$.

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