Given matrix $$A = \begin{bmatrix} 2 & 4 \\-2 & -4\end{bmatrix}$$ find the general formula of $A^n$. Hint: diagonalize matrix $A$.
What does it mean to find a general formula of $A^n$? I don't understand the exercise itself. I have a guess, that seems logically correct:
$A^2=\begin{bmatrix} 2 & 4 \\ -2 & -4 \\ \end{bmatrix}\begin{bmatrix} 2 & 4 \\ -2 & -4 \\ \end{bmatrix}=\begin{bmatrix} -4 & -8 \\ 4 & 8 \\ \end{bmatrix}=(-2)^1A$
$A^3=\begin{bmatrix} 2 & 4 \\ -2 & -4 \\ \end{bmatrix}\begin{bmatrix} 2 & 4 \\ -2 & -4 \\ \end{bmatrix}\begin{bmatrix} 2 & 4 \\ -2 & -4 \\ \end{bmatrix}=\begin{bmatrix} -8 & -16 \\ 8 & 16 \\ \end{bmatrix}=(-2)^2A$
Therefore: $A^n=(-2)^{n-1}A$
But I'm not sure whether is what's needed, and I didn't had to diagonalize the matrix using this method.
Let me explain the "diagonalize" part:
If we could write $A = PDP^{-1}$ for a diagonal matrix $D$, then it would follow that $A^n = PD^n P^{-1}$. The advantage is that $D^n$ is simple to calculate: if $D = \text{diag}(a_1, \cdots, a_k)$, then $D^n = \text{diag}(a_1^n, \cdots, a^n_k)$.