DSphery's question at Yahoo Answers (matrix powering question)

[Fernando Revilla]

Here is the question:

Consider you have a 3x3 matrix like this:
a b 0
0 a b
0 0 a
The question is the explicit formule for the n-th power. It's simple to see that the "a"-s will convert to "a^n" and I've also worked out a formula for the rest of the elements, but those are implicit forms. Can anyone help me with an explicit form of it for the n-th power? Thank you in advance!

Here is a link to the question:

Matrix powering question, with specific matrix.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello DSphery,

The simplest way in this case: we can express
$$A=\begin{bmatrix}{a}&{b}&{0}\\{0}&{a}&{b}\\{0}&{0}&{a}\end{bmatrix}=aI+bN, \mbox{ where } N=\begin{bmatrix}{0}&{1}&{0}\\{0}&{0}&{1}\\{0}&{0}&{0}\end{bmatrix}$$
The matrix $N$ is nilpotent that is,

$$N^2=\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{bmatrix},N^3=0,N^4=0,\ldots$$
As $(aI)(bN)=(bN)(aI)$ we can use the Newton's binomial theorem:
$$A^n=(aI+bN)^n=\displaystyle\binom{n}{0}(aI)^n+ \displaystyle\binom{n}{1}(aI)^{n-1}(bN)+\binom{n}{2}(aI)^{n-2}(bN)^2$$
Equivalently:
$$A^n=a^n\begin{bmatrix}{1}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0}&{1}\end{bmatrix}+na^{n-1}b\begin{bmatrix}{0}&{1}&{0}\\{0}&{0}&{1}\\{0}&{0}&{0}\end{bmatrix}+\dfrac{n(n\color{red}-1)}{2}a^{n-2}b^2\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{bmatrix}$$
Now, we can conclude.