Q:

Newton's binomial for matrices that don't commute?

I'll give a bit of background info as to why I'm asking. I need to find the directional derivative of $f(A)=A^m$ where $m>0$ and $A$ is an $n$ by $n$ matrix with real entries.
I want to do this via the definition:
$$Df(A)H=\lim_{t \to 0} \frac{f(A+tH)-f(A)}{t} =\lim_{t \to 0} \frac{(A+tH)^m-A^m}{t} $$
It is clear to see that if I expand $(A+tH)^m$ then I get $A^m+t\alpha_1+t^2\alpha_2+...+t^mH^m$ where $\alpha_i$ is some combination of $A$ and $H$ that I don't know how to calculate.
So overall, in the original formula if I expand it, I'll get $$\lim_{t \to 0} \frac{t\alpha_1+t^2\alpha_2+...+t^mH^m}{t}=\alpha_1$$
If $AH=HA$ then I could use newton's binomial to find $\alpha_i$, but sadly we can't assume that holds true here. So how would I find $\alpha_1$? how would I expand $(A+tH)^m$

A:

Hint: just focus on the terms linear in $t$ as these are the only ones which survive the limit. The product is non-commutative, but $A$ commutes with itself. For $n=3$ we have
\begin{align} (A+tH)^3 &= (A+tH)(A^2+t(AH+HA)+t^2H^2) \\
&= A^3+t(A^2H+AHA)+t^2AH^2+tHA^2+t^2(HAH+H^2A)+t^3H^3 \\
&= A^3+t(A^2H+AHA+HA^2)+ \cdots
\end{align}
The terms with just one $H$ are the ones which actually give the derivative. The others will be shown to vanish in the quotient of the directional derivative (a similar comment applies to the Frechet quotient). So, continuing to higher orders, the pattern is easy enough to see: for $n=4$ the linear terms are:
$$ t(A^3H+A^2HA+AHA^2+HA^3)$$
notice there are $4$ terms now. If $HA=AH$ then it would squash to $t4A^3H$ which is reminiscent of $\frac{d}{dx}x^4=4x^3$. I think the pattern starts to be clear. Perhaps this helps.

