I have to show that the derivative of 'the matrix exponential' $exp: \mathbb{C}^{n\times n}\mapsto\mathbb{C}^{n\times n}$ at the zero matrix $0$ is $id_{C^{n\times n}}$, i.e. $exp(0)=id$. The above map isn't given explicitly, so maybe someone can tell me what it does to a matrix $A$? I think the obvious guess would be $$A\mapsto e^A:=\sum_{k=0}^\infty \frac{A^k}{k!}$$, but I wouldn't know how to find the derivation of such a map since there is no variable.
So far I tried considering $$Ax\mapsto e^Ax:=\sum_{k=0}^\infty \frac{(Ax)^k}{k!}$$ instead, which is a map $\mathbb{C}\rightarrow\mathbb{C}$ I suppose.
As derivative I get $Ax\mapsto Ae^{Ax}$, which gives the identity for $x=0$, but $0$ for $A=0$.
Can somebody tell me what I got wrong, and how it is true that $exp(0)=id$?
$e^A$ is indeed $\sum_{n=0}^\infty \frac{A^k}{k!}$. To find the derivative at the zero matrix, let $B$ be an arbitrary $n \times n$ matrix which is unit in some norm, let $\epsilon>0$ be a small number, and consider
$$e^{\epsilon B}-e^0=\sum_{k=0}^\infty \frac{\epsilon^k B^k}{k!}-\sum_{k=0}^\infty \frac{0^k}{k!} = \sum_{k=1}^\infty \frac{\epsilon^k B^k}{k!}.$$
(Here by definition $A^0=I$ for any $n \times n$ matrix $A$, including the zero matrix.) Any idea what to do now? (Hint: what is the relative size of the terms?)