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

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The discussion centers on finding an explicit formula for the n-th power of a specific 3x3 matrix. The matrix is expressed as a combination of a scalar multiple of the identity matrix and a nilpotent matrix. Using the Newton's binomial theorem, the n-th power of the matrix can be derived, resulting in a formula that incorporates powers of "a" and coefficients based on "b." The final expression includes terms for the identity matrix and the nilpotent matrix raised to various powers. This approach effectively provides the explicit form sought by the original poster.
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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.
 
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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.
 
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