How to calculate the natural powers of a 2x2 matrix? Given $A \in \mathit{M_2}(\mathbb{C})$ and its eigenvalues $\lambda_1, \lambda_2 \in \mathbb{C}$  there exists $B, C \in \mathit{M_2}(\mathbb{C})$,
so that for any $n \in \mathbb{N^*}$. ($\mathbb{N^* = \mathbb{N} - \{0\}}$)
$$
A^n = \left\{ \begin{array}{ll}
\lambda_1^nB + \lambda_2^nC, & \text{if} \ \lambda_1 \ne \lambda_2, \\ 
\lambda_1^nB + n\lambda_2^{n-1}C, \ & \text{if} \ \lambda_1 = \lambda_2.
\end{array} \right.
$$
This is a theorem that was proven with the help of the Cayley-Hamilton theorem but I don't understand the proof.    
Could someone provide an explanation or a link to some articles/papers with an explication on how to calculate the natural powers of a 2x2 matrix and the logic behind it?
 A: As others have stated, the proof which you explained in a previous post is needlessly convoluted.  Here is an alternative proof.
First, consider the case where $\lambda_1 \neq \lambda_2$.  $A$ must have two linearly independent eigenvectors, and so $A$ must be diagonalizable.  That is, there exist an invertible $2 \times 2$ matrix $P$ such that $A = PDP^{-1}$, where
$$
D = \pmatrix{\lambda_1 & 0\\0 & \lambda_2}.
$$
In particular, the columns of $A$ are the eigenvectors associated with $\lambda_1,\lambda_2$ (in that order).  With that established, we note that
$$
D^n = \pmatrix{\lambda_1^n & 0\\0 & \lambda_2^n}.
$$
This allows us to calculated $A^n$ since we have
$$
A^n = \overbrace{(PDP^{-1}) \cdots (PDP^{-1})}^{n \text{ times }} = \\
P \cdot \overbrace{(DP^{-1}P) \cdots (DP^{-1}P)}^{(n-1) \text{ times }} \cdot DP^{-1} = \\
P \cdot D^{n-1} \cdot DP^{-1} =\\
PD^n P^{-1}.
$$
Now, let $v_1,v_2$ denote the columns of $P$, and let $w_1^T, w_2^T$ denote the rows of $P^{-1}$.  Now, let $E_1,E_2$ be the matrices
$$
E_1 = \pmatrix{1&0\\0&0}, \quad E_2 = \pmatrix{0&0\\0&1}.
$$
Using the properties of matrix multiplication we see that
$$
A^n = PD^n P^{-1} = P(\lambda_1^n E_1 + \lambda_2^n E_2) P^{-1} = \\
P(\lambda_1^n E_1) P^{-1} + P(\lambda_2^n E_2) P^{-1} =\\
\lambda_1^n (PE_1P^{-1}) + \lambda_2^n (PE_2P^{-1}).
$$
That is, your statement holds with $B = PE_1P^{-1}$ and $C = PE_2P^{-1}$.

The proof in the second case is similar.  If $\lambda_1 = \lambda_2$, call this common eigenvalue $\lambda$. If $A = \lambda I $ then the statement holds trivially.  If $A \neq \lambda I$, then we can show that there exists a $2 \times 2$ matrix $J$ such that $A = PJP^{-1}$, where
$$
J = \pmatrix{\lambda&1\\ 0 & \lambda} = \lambda I + N.
$$
Just like last time, it is easy to calculate the $n$th power of $J$.  Since $\lambda I $ and $N$ commute, we can expand using the binomial theorem.  Since $N^2 = 0$, this simplifies to 
$$
J^n = (\lambda I)^n + n\,(\lambda I)^{n-1}(N)^1 = \lambda^n I + n \lambda^{n-1}N.
$$
Just as before, we have $A^n = PJ^nP^{-1}$. It follows that
$$
A^n = PJ^nP^{-1} = 
P(\lambda^n I + n \lambda^{n-1}N)P^{-1} = \\
P(\lambda^n I)P^{-1} + P(n \lambda^{n-1} N)P^{-1} = \\
\lambda^n I + n \lambda^{n-1} (PNP^{-1}).
$$
The conclusion follows.
