Finding the norm of the differential of product matrix function

Assume that $f_p:M_n(R) \to M_n(R)$ is a function from the set of all $n$ by $n$ matrices over the reals with the subordinate matrix norm $ ||.|| $, defined as $f_p(A) = A^p$ where $p \in N$. I proved that this function is differentiable and its differential $df_p(A)(H) = A^{p-1} H + A^{p-2} HA + ... + HA^{p-1}$. However I can't figure out the norm of $d_pf$. I only found that $||d_pf(A)|| \leq p||A||^{p-1}$ by using the definition of the norm of the differential which is not necessarily the same as $p||A^{p-1}||$. I would appreciate if anyone could provide a hint as to how I should find the norm of the differential. Thanks.

Solution 1:

I don't think there is a nice formula. The norm depends heavily on the relation between $A$ and $p$.

For instance, if $A=I$, then $df_p(A)H=pH$, and so $\|df_p(I)\|=p=p\|A\|^{p-1}$.

Now if $A=E_{12}=\begin{bmatrix} 0&1\\0&0\end{bmatrix}$ and $p=2$, then $ df_p(E_{12})I=2E_{12},$ so $\|df_p(E_{12})\|=2\|E_{12}\|^{p-1}$.

But if $A=E_{12}$ and $p=3$, now $df_p(E_{12})H=\begin{bmatrix}0& H_{21}\\0&0\end{bmatrix}$, so $\|df_3(E_{12})\|=1$.

And for $p\geq4$ we have $df_p(E_{12})=0$, and so $\|df_p(E_{12})\|=0$.

In summary, $$ \|df_p(E_{12})\|=\begin{cases} 2,&\ p=2\\ 1,&\ p=3\\ 0,&\ p\geq4\end{cases} $$ This is neither $p\|A\|^{p-1}$ nor $p\|A^{p-1}\|$.