(1) Sequence $ \{a_n\} $ is defined as $ a_{0} = a_1 = a_2 = 1 $ and
$ a_n = 7a_{n-2} + 6a_{n-3}, for. \ n \geq 3$.
Find a general formula for $ a_n $
I have only solved recurrence relations for n >= 2 so far in my class, I am not sure as to how I should extend it to n >= 3?
Edit: P =
(2) $ C = \begin{pmatrix} 2 &2 & 0 \\ -1 & -1 & 0 \\ 2 & 2 & 3 \end{pmatrix}\ $. Diagonalise the matrix in 3 ways so that the diagonal matrices $ D_1, D_2, D_3 $ are different. I found it's eigenvalues to be
$ \lambda_1 = 0, $ corresponding vector is $ (-1,1,0) $.
$ \lambda_2 = 1, $ corresponding vector is $ (-2,1,1) $
$ \lambda_3 = 3, $ corresponding vector is $ (0,0,1) $
Is it correct to say that $ D_1 = \begin{pmatrix} 3 &0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\ $. $ D_2 = \begin{pmatrix} 1 &0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \end{pmatrix}\ $ $ D_1 = \begin{pmatrix} 3 &0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\ $?
Hint for (1): We have $$ \begin{pmatrix} a_n \\ a_{n-1} \\ a_{n-2} \end{pmatrix} = \begin{pmatrix} 0 &7 & 6 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} a_{n-1} \\ a_{n-2} \\ a_{n-3} \end{pmatrix} $$ whose characteristic polynomial is $x^3 - 7 x - 6$ (no surprises here).
Finally, $x^3 - 7 x - 6=(x + 1) (x + 2) (x - 3)$.