Power of a matrix

Question

I'm following a YouTube linear algebra course. (https://www.youtube.com/watch?v=PFDu9oVAE-g&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=14)
In part 9 there's the following question:

I don't know what the formula is. What I figured out is that there is a link with the fibonacci sequence.

I also tried to convert A to eigenbasis. I get this:

\begin{bmatrix}\frac{{-\sqrt5 + 1}}{2}&0\\0&\frac{{\sqrt5 + 1}}{2}\\\end{bmatrix}

How do I go back to normal basis and what is the formula?

This is what I have now:

Answer 1

$\phi = \frac {1 + \sqrt 5}{2}\\ \phi' = \frac {1 - \sqrt 5}{2} = -\phi^{-1}$

You have these two eigenvectors, and they have the following propety:

$Av_1 = \phi v_1\\ Av_2 = \phi' v_2$

$A[v_1,v_2] = [v_1,v_2]\begin{bmatrix} \phi&\\&\phi'\end{bmatrix}$

lets say $P = \frac 12[v_1,v_2] = \begin{bmatrix} 1&1\\\phi&\phi'\end{bmatrix}$ and $D = \begin{bmatrix} \phi&\\&\phi'\end{bmatrix}$

$A P = PD\\ A = P D P^{-1}\\ P^{-1}= \frac {1}{\phi-phi'}\begin{bmatrix}-\phi'& 1\\\phi & -1\end{bmatrix}$

Now what happens when we multiply A by itself?

$A^2 = P D P^{-1}P D P^{-1} = P D^2 P^{-1}\\ A^n = P D^n P^{-1}\\ D^n=\begin{bmatrix} \phi^n&\\&\phi'^n\end{bmatrix}$

The Fibbonacci sequence

$\begin{bmatrix} F_n\\F_{n+1}\end{bmatrix} =$$ A \begin{bmatrix} F_n\\f_{n+1}\end{bmatrix}\\ A^n \begin{bmatrix} 0\\1\end{bmatrix}\\ P D^n P^{-1}\begin{bmatrix} 0\\1\end{bmatrix}\\ \frac{1}{\phi-\phi'}P D^n \begin{bmatrix} 1\\-1\end{bmatrix}\\ \frac{1}{\phi-\phi'}P \begin{bmatrix} \phi^n\\-\phi'^n\end{bmatrix}\\ \frac{1}{\phi-\phi'}\begin{bmatrix} \phi^n-\phi'^n\\\phi^{n+1}-\phi'^{n+1}\end{bmatrix}\\$

$F_n = \frac {\phi^n -\phi'^n}{\phi -\phi'}$

Answer 2

Hint:

Compute the first powers of a generic diagonal matrix: $$\begin{bmatrix} a&0\\ 0&b\end{bmatrix}$$ Then use the fact that the diagonal form $D$ of the matrix in the bassis of eigenvectors and $A$ are related this way: $$D=P^{-1}AP,\enspace\text{whence}\quad A=P^{-1}DP,\enspace\text{so that}\quad A^n=P^{-1}D^nP .$$