I have shown the easy case, when A is diagonalizable.
But I am stuck on the case when A is not.
So, I put A in its Jordan canonical form. Then say A = $SJS^{-1}$.
Then $e^A = Se^JS^{-1}$,
where $e^J$ is an upper triangular matrix with the Jordan blocks exponentiated.
Now for each Jordan block, I have that
$$e^{J_i} = e^{\lambda_i I + N}$$
where N is elementary nilpotent. But since scalar matrices commute with all matrices, it commutes with N.
Then
$$e^{J_i} = e^{\lambda_i I + N} = e^{\lambda_i}e^{N}$$
where $e^{\lambda_i}$ is a diagonal matrix with non-zero diagonal, hence it is invertible / has non-zero determinant. What can I do with the $e^N$ factor? I know that it has a finite expansion, since it is nilpotent:
$$e^N = I + N + … + \frac {N^k}{k!}$$
I'm not sure where to go from here…
Thanks,
Best Answer
The simpler way to shows that $e^A$ is invertible is to note that, for commuting matrices $A,B$ we have, from the definition of the exponential, $e^{A+B}=e^Ae^B$. So, since $A$ and $-A$ commute, we have: $$ e^Ae^{-A}=e^{A-A}=e^O=I $$
Also, using Jacobi's formula (see here) we can find: $$ \det(e^A)=e^{Tr\;(A)} $$ that confirms the invertibility of $e^A$.