This question is from my assignment in smooth manifolds and I was unable to solve this problem. I am asking it here as I think I will not be able to solve it by myself.
Question : Let $A\in M(n, \mathbb{R})$. Prove that:
(a) $det(e^A)= e^{ tr(A)}$.
Attempt: It is clear what $e^{tr(A)}$ means. $Det (e^A)= Det ( lim N\to \infty \sum_{k=0}^N \frac {A^k} {k!})$. But I am not able to simplify $ lim N\to \infty \sum_{k=0}^N \frac {A^k} {k!}$ and hence not able to prove what I am asked. Can you please help with that?
(b) The map $\psi : M(n,\mathbb{R}) \to GL(n,\mathbb{R}) $ defined by $\psi(B)= e^B$ is not surjective.
Attempt: The first way is to find an element of $GL(n,\mathbb{R})$ for which no element of $M(n,\mathbb{R})$ exists under $\psi$. But I was unable to find such an element of $GL(n,\mathbb{R})$.
I tried another method: Assuming $\psi$ is surjective and proving it is injective, I will get a contradiction as $GL(n,\mathbb{R})$ is not isomorphic to $M(n , \mathbb{R})$. But I am not able to prove that it is injective. Let $e^A =e^B$. It implies that $( lim N\to \infty \sum_{k=0}^N \frac {A^k} {k!} -\sum_{k=0}^{N} \frac {B^k} {k!})=0 $ which means that $( lim N\to \infty \sum_{k=0}^N \frac {(A-B) (...)} {k!})=0 $. But I am unable to see how can I prove that A=B.
Can you please help me with this?
Thanks!
If $Av=\lambda v$, then $A^kv=\lambda^kv$, so $(\sum (A^k/k!))v=\sum(\lambda^k/k!)=e^{\lambda}$.
Since $A$ has real entries, its trace, call it $S$, is real, so $e^S$ is positive, so $\det e^A$ is positive. But the general linear group has lots of matrices with negative determinant, so the map $A\mapsto e^A$ is not surjective.