The matrix exponential is smooth Let $ \exp : M(n, \mathbb{C}) \rightarrow GL(n, \mathbb{C}) $ be the matrix exponential defined by $$ \exp(X) = \sum_{k=0}^{\infty} \frac{X^k}{k!} $$
Is this map smooth as a map from $ \mathbb{C}^{n^2} \rightarrow \mathbb{C}^{n^2} $ ?
My attempt: I show that the derivative at $ 0 $, $ D\exp(0) $ is the identity linear transformation on $ \mathbb{C}^{n^2} $, thus the derivative is nonsingular. This is because we have $$ \frac{|| \exp(H) - \exp(0) - H ||}{||H||} = \frac{||\sum_{k=2}^{\infty} \frac{H^k}{k!}||}{||H||} \le \sum_{k=1}^{\infty} \frac{||H||^{k}}{(k+1)!} $$ and the limit of the last expression as $ ||H|| \rightarrow 0 $ is clearly $ 0 $. 
My problems begin at the following: (1) I can calculate the derivative only at scalar matrices in $ M(n, \mathbb{C}) $ (I need commutativity for the identity $ \exp(A+B)=\exp(A)\exp(B) $ to hold) (2) I cannot apply the inverse function theorem yet, because I have established differentiability at only a point.
How would one get around these difficulties? I know that the inverse function theorem holds for analytic functions too, but I would like to avoid it, and in any case, I would need to show that the derivative is nonsingular everywhere. I cannot see a coordinate free approach.
 A: It is indeed a smooth function. A direct inductive argument can be given which is analogous (but a bit more tricky) to the argument that a complex function defined by a power series is smooth in the domain of convergence. In fact, it is actually easier to prove a more general statement and then apply it to deduce that $\exp$ is smooth.
Let $(A, \cdot, \| \cdot \|)$ be a finite dimensional complex Banach algebra. The reason we want to generalize our discussion to an arbitrary Banach algebra and not work only with $A = M_n(\mathbb{C})$ is that it makes the inductive argument easier. Show first that the power maps $p_k \colon A \rightarrow A$ given by $p_k(X) = X^k$ are continuously differentiable with differential given by
$$ dp_k|_{X}(Y) = X^{k-1}Y + X^{k-2}YX + \dots + XYX^{k-2} + YX^{k-1}.$$
The differential has this "strange" form because you don't know if $X$ and $Y$ commute. If they do, the formula above reduces to the usual formula $dp_k|_{X}(Y) = k X^{k-1} Y$.
Then show the following lemma:
Lemma: Let $(c_k)_{k=0}^{\infty}$ be a sequence of complex numbers such that $\sum_{k=0}^{\infty} c_k z^k$ converges on $B_{\mathbb{C}}(0,r)$. Define $f \colon B_A(0,r) \rightarrow A$ by $f(X) = \sum_{k=0}^{\infty} c_k X^k$. Then $f$ is well-defined and continuously differentiable. The differential of $f$ is given by
$$ df|_{X}(Y) = \sum_{k=0}^{\infty} c_k dp_k|_{X}(Y). $$
Finally, you can use the lemma inductively to deduce that $f$ is actually smooth and not just $C^1$. For full details, see Chapter 3 of the book "Structure and Geometry of Lie Groups" by Joachim Hilgert and Karl-Hermann Neeb.
A: Define a vector field $V$ on $GL(n,\mathbb{C}\times M(n,\mathbb C))$ by $$V_{(g,X)}=(gX,0),$$where we used the canonical identification $T_{(g,X)}(GL(n,\mathbb{C}\times M(n,\mathbb C))\cong M(n,\mathbb C)\oplus M(n,\mathbb C).$ Its flow $\theta$ is given by$$\theta(t,(g,X))=(g\exp(tX),X),$$as is easily checked. Thus $\theta (1,(I_n,X))=(\exp(X),X),$ and  the smoothness of $\exp$ follows from the smoothness of $\theta$. (An essentially same argument applies to the exponential map on Lie groups.)
