I want to find a formula for $a_n$, where:
$a_0 = a,$
$a_1=b,$
$a_{n+2}=6a_{n+1}-9a_n$
By looking at $a_2, a_3, a_4$... I did manage to see some formula, but I don't think this is the right way.
Instead, in an earlier question, I'm asked to find the Jordan form of
$ A=\begin{pmatrix} 6 & -9 \\ 1 & 0 \end{pmatrix}$
I found it to be
$G=\begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$
Also, $G^n=3^{n-1}\begin{pmatrix} 3 & n \\ 0 & 3 \end{pmatrix}$
Now, I noticed that if $v=\begin{pmatrix} a_{n+1}\\ a_n \end{pmatrix}$, then $Av=\begin{pmatrix} 6a_{n+1}-9a_n\\ a_{n+1} \end{pmatrix}$ and for our sequence,
we need $Av=\begin{pmatrix} 6a_{n+1}-9a_n\\ a_{n+1} \end{pmatrix}=\begin{pmatrix} a_{n+2}\\ a_{n+1} \end{pmatrix}$
But from here on I'm not sure how to proceed.