Is there an efficient systematic way to find relations between powers of Matrices?

Question

Hi I have a 3 by 3 matrix of the following form: $$ A=\begin{bmatrix} a & b & 0 \\ -b & a & c \\ 0 & -c & d \\ \end{bmatrix} $$ $$a,b,c,d\in\Re $$ I'm trying to explore special relations between powers of this matrix of the form: $$ A^n = \alpha A^m $$ or even for the simpler case where $$a=d=0$$ The motivation is to use one of the closed form special cases for the matrix exponential, I didn't find anything so far from exploring the literature, and I wonder if there is a systematic efficient way to go about it?

Thanks!

You need to focus on one question, otherwise it might get closed. The "simpler case" is much simpler. Also, are the coefficients real or complex? — Chrystomath, Jan 29 '21 at 08:33
you can form its characteristic polynomial and then you can easily form a recurrence relation for the powers of the matrix — , Jan 31 '21 at 09:32
Thanks for the answer, I didn't understand fully what characteristic polynomial are you referring to and how to use it further? I'll be happy if you can elaborate — Alex Finkelshtein, Jan 31 '21 at 09:36

score 2 · Answer 1 · answered Jan 31 '21 at 11:07

This answer is restricted to the case $a=d=0$. Then one can check that $$A^3=(b^2+c^2)A$$ So writing $\beta:=\sqrt{b^2+c^2}$, then \begin{align} e^A&=I+A+\frac{A^2}{2!}+\cdots+\frac{A^n}{n!}+\cdots\\ &=I+\left(1+\frac{\beta^2}{3!}+\frac{\beta^4}{5!}+\cdots\right)A+\left(\frac{1}{2!}+\frac{\beta^2}{4!}+\cdots\right)A^2\\ &=I+\frac{\sinh\beta}{\beta}A + \frac{\cosh\beta-1}{\beta^2}A^2\\ &=I+uA+vA^2\\ &=\begin{pmatrix}1-vb^2&ub&vbc\\-ub&1-\beta^2v&uc\\vbc&-uc&1-vc^2\end{pmatrix} \end{align}

This is very helpful. Thank you! – Alex Finkelshtein Jan 31 '21 at 15:59 — Alex Finkelshtein, Jan 31 '21 at 15:59

Is there an efficient systematic way to find relations between powers of Matrices?

1 Answers1