Bounding $\|A^n-B^n\|_F$ by $\|A-B\|_F$

Question

Given $$ \epsilon = \|A-B\|_F, $$ it is clear that $$ \epsilon^n > \|(A-B)^n\|_F, $$ which follows from submultiplicativity.

I wonder if something related can be said about $\|A^n-B^n\|_F$?

Answer 1

\begin{align}A^n-B^n&=A^n-A^{n-1}B+A^{n-1}B-A^{n-2}B^2+\cdots-B^n\\ &=A^{n-1}(A-B)+A^{n-2}(A-B)B+\cdots+(A-B)B^{n-1} \end{align} $$\therefore\ \|A^n-B^n\|\le(\|A^{n-1}\|+\|A^{n-2}\|\|B\|+\cdots+\|B^{n-1}\|)\|A-B\|\le\frac{\|A\|^n-\|B\|^n}{\|A\|-\|B\|}\|A-B\|$$

If the intent of the question is whether $\|A^n-B^n\|$ is small compared to $\|A-B\|$ then the answer is no. Take $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, $B=\begin{pmatrix}1&0\\1&1\end{pmatrix}$, then $$\|A-B\|_F=2,\qquad\|A^n-B^n\|_F=2n^2$$