# Showing $f\left(A\right)=e^{A^{2}}$ is differentiable.

Let $f\left(A\right)=e^{A^{2}}$ where $A$ is an $n\times n$ matrix. Show that $f$ is differentiable and compute its derivative.

I know this is kind of a basic question, but I am not sure how to solve it. By definition, we want to show that there exists a matrix $B$ such that $$\lim_{H\to0}\frac{\left|f\left(A+H\right)-f\left(A\right)-BH\right|}{\left|H\right|}=0.$$ Is this right? But since we're working with matrices I am not sure what to do. Any guidence would be appreciated.

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No. You want to show that there is a linear mapping $\phi\colon M_n(\mathbb R) \to M_n(\mathbb R)$ such that $$\lim_{H \to 0} \frac{|f(A+H) - f(A) - \phi(H)|}{|H|} = 0$$ –  martini May 9 '12 at 19:40
What are you allowed to use? This is immediate via functional calculus. –  Theo May 9 '12 at 19:50
@Theo Just real analysis. –  Galois May 9 '12 at 19:57
@Martini makes sense. How do we find this $\phi$. –  Galois May 9 '12 at 19:57
Well the only problem is the exponential. You can use the chain rule to come to a conclusion once you show that the exponential is a differentiable map. The way to go is to explicitely write down the power series for $\mathrm{exp}(A+H)$, develop it not forgetting the fact that $A$ and $H$ do not commute in general, and single out the constant term ($\mathrm{exp}(A)$), the first order terms in $H$ and show that what remains is of the oreder of $|H|^2$. You also need to show that the linear term in $H$ does indeed converge (...) and defines a linear map (...). –  Olivier Bégassat May 9 '12 at 20:09
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$$Df(A):H\mapsto\sum_{n=1}^{+\infty}\frac1{n!}\sum_{i=1}^{2n}A^{i-1}HA^{2n-i}$$