# Powering a sum of two easily powered matrices

I am currently studying matrices and in order to understand them better I want to know why I can't do certain things in my calculations. This question is just about that.

The task is to calculate $A^n$ if

$$A=\begin{bmatrix} a & b \\ 0 & c \\ \end{bmatrix}$$

I started of by calculating smaller powers, $A^2$, $A^3$, but I did not recognize the pattern at first. I tried an alternative approach, writing the matrix in a form of a sum of two matrices that will be easier do power.

$A=\begin{bmatrix} a & 0 \\ 0 & c \\ \end{bmatrix}$ $+\begin{bmatrix} 0 & b \\ 0 & 0 \\ \end{bmatrix}$

Let's denote these matrices as $C=\begin{bmatrix} a & 0 \\ 0 & c \\ \end{bmatrix}$ and $D=\begin{bmatrix} 0 & b \\ 0 & 0 \\ \end{bmatrix}$

When we apply Binomial Theorem, we get:

$$A^n = (C+D)^n=\binom{n}{0}C^n + \binom{n}{1}C^{n-1}D + \binom{n}{2}C^{n-2}D^2 \dots + \binom{n}{n-1}CD^{n-1} + \binom{n}{n}D^n$$

I tested powering both $C$ and $D$ for smaller powers to see if there is a pattern. As it turns out:

$C^n = \begin{bmatrix} a^n & 0 \\ 0 & c^n \\ \end{bmatrix}$ and
$D^n = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$

Every matrix multiplied by zero-matrix $O$ is equal to zero, which leaves us with:

$$A^n = C^n$$

which is not the correct solution to the problem.

What interests me is: which step did I do wrong and why ? I am aware that it would have been easier to recognize the pattern before turning to Binomial Theorem, but I want to know why is this particular method of solving wrong.

When applying the binomial theorem in this way you are assuming that the two matrices commute. The usual proof of that theorem for real numbers freely interchanges $x$ and $y.$