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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.

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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 $y.$

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The binomial formula is only true for elements that commute, which is not the case of your two matrices.

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