Show that the Matrix mapping f defined as $f(A) = A^p$ is differentiable I've been asked to show that for a function $f:M_{n\times n}(\mathbb R)\longrightarrow M_{n\times n}(\mathbb R) $
that defined as $f(A) = A^p$ for all $2\leq p\in \mathbb N$
is differentiable.
I've tried solving this using induction and showing that the partial derivatives are continues and then to conclude that f is differentiable.
On the case of p = 2 is pretty simple to show that, but for the case of multiplying p matrices is more complex and I feel like i'm not on the right way.
Any suggestions or alternative ways of solving it?
Edit: I found a different approach for this case, in addition to the solution mentioned below.
I can define the function $h:M_{n\times n}(\mathbb R)\longrightarrow M_{n\times n}(\mathbb R)\times M_{n\times n}(\mathbb R)$ which is defined by $h(X) = (X,X^{p-1})$
and then define the bilnear map $g:M_{n\times n}(\mathbb R)\times M_{n\times n}(\mathbb R) \longrightarrow M_{n\times n}(\mathbb R)$
and now we get that $f=g\circ h$.
now $g$ is a bilnear map thus it's differentiable, and now by induction on $p$ we can solve the problem using the following formula:
Derivative of the bilnear map $\beta:\mathbb R^m\times\mathbb R^n\longrightarrow\mathbb R^d$ is $D\beta(x_0,y_0)(h,k)=\beta (x_0,k)+\beta(h,y_0)$
And the chain rule.
 A: You can try to show directly that $f(A+H)=f(A) + f'(A)(H) + o(\|H\|)$ for some linear map $f'(A)$. Indeed
\begin{align*}
f(A+H)=(A&+H)^p \\
=A^p &+ A^{p-1}H+A^{p-2}HA+\dots+HA^{p-1}\\
&+\text{terms of order }H^2\text{ or higher}.
\end{align*}
But using $\|AB\|\leq\|A\|\|B\|$ we see that the higher order terms are $O(\|H\|^2)$ and hence $o(\|H\|)$. It follows that the derivative $f'(A)$ is given by the linear map
$$H\mapsto A^{p-1}H+A^{p-2}HA+\dots HA^{p-1}.$$
A: *

*First prove that matrix multiplication $M: \mathbb R^{2n^2} \Rightarrow \mathbb R^{n^2}$ (where $M(A,B)=AB$) is differentiable. This is straightforward since each element of the product is a bilinear combination of original elements.


*Then prove that squaring is differentiable: if $D: R^{n^2}\Rightarrow R^{2n^2}$ maps each matrix $A$ to the pair $(A,A)$, then squaring can be defined as $M\circ D: A \Rightarrow (A,A) \Rightarrow A^2$


*Now we can prove by induction: if $P_{n-1}:A\Rightarrow A^{n-1}$ is differentiable then $P_n = M(A,P_{n-1}(A))$ is a composition of two differentiable maps.
