series ${a_n}$ is define by $a_1=1$ , $a_2=5$ , $a_3=1$, $a_{n+3}=a_{n+2}+4a_{n+1}-4a_n$, ($n \geq 1$).
$\begin{pmatrix}a_{n+3} \\ a_{n+2} \\ a_{n+1} \\ \end{pmatrix}=B\begin{pmatrix}a_{n+2} \\ a_{n+1} \\ a_{n} \\ \end{pmatrix}$
find $B$
find general term of $a_n$ series
I found matrix b is $B=\begin{bmatrix}1&4&-4\\1&0&0\\0&1&0\end{bmatrix}$
but how can I find $a_n$?
I can compute that $\begin{pmatrix}a_{4} \\ a_{3} \\ a_{2} \\ \end{pmatrix}=B\begin{pmatrix}a_{3} \\ a_{2} \\ a_{1} \\ \end{pmatrix}$
$\begin{pmatrix}a_{4} \\ a_{3} \\ a_{2} \\ \end{pmatrix}=B\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix}$
$\begin{pmatrix}a_{5} \\ a_{4} \\ a_{3} \\ \end{pmatrix}=B^2\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix}$
i see there is relation about this.
$a_{n+1}=B^na_n$
suppose D is diagonal matrix that similar to B
$a_{n+1}=TD^nT^- a_n$
which mean i need to find diagonal matrix and its inver to find $a_n$? is my assumption wrong?
The formula is $$ \begin{pmatrix} a_{n+2} \\ a_{n+1} \\ a_n \end{pmatrix}=B^{n-1}\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix} $$
EDIT : PROOF
By induction on $n \in \mathbb{N}^{*}$.
Let $n=1$, then we have $$ \begin{pmatrix} a_{3} \\ a_{2} \\ a_1 \end{pmatrix}=B^{0}\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix} $$ which is true.
Then suppose it is true until a certain index $n$. Then $$ \begin{pmatrix} a_{n+3} \\ a_{n+2} \\ a_{n+1} \end{pmatrix}=B\begin{pmatrix} a_{n+2} \\ a_{n+1} \\ a_{n} \end{pmatrix} $$ By hypothesis $$\begin{pmatrix} a_{n+3} \\ a_{n+2} \\ a_{n+1} \end{pmatrix}=BB^{n-1}\begin{pmatrix} 1 \\ 5 \\ 1 \end{pmatrix}=B^{n}\begin{pmatrix} 1 \\ 5 \\ 1 \end{pmatrix}$$
Then the formula has been proved. Now, suppose you diagonalize $B$. Then you have $$ \begin{pmatrix} a_{n+2} \\ a_{n+1} \\ a_n \end{pmatrix}=PD^{n-1}P^{-1}\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix} $$
You know that $1$ is an eigenvalue because $$ B\begin{pmatrix}1 \\ 1 \\ 1 \\ \end{pmatrix}=\begin{pmatrix}1 \\ 1 \\ 1 \\ \end{pmatrix} $$ Then you know that $$Tr(B)=1+\lambda_1+\lambda_2=1 \ \text{ and } \ \text{det}B=\lambda_1\lambda_2=-4 $$ Hence it comes $\lambda_1=2$ and $\lambda_2=-2$.
Now you have the three eigan value, and one eigenspace so you can deduce the two other eigenspace. You will find $$D=\begin{pmatrix} 1 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 &-2\end{pmatrix}$$ And $$B-2I_3=\begin{pmatrix} -1 & 4 & -4\\ 1 & -2 & 0 \\ 0 & 1 &-2\end{pmatrix}$$ Then it appears thart your eigenvector is $(4,2,1)$
$$B+2I_3=\begin{pmatrix} 3 & 4 & -4\\ 1 & 2 & 0 \\ 0 & 1 &2\end{pmatrix}$$ Then it appears thart your eigenvector is $(4,-2,1)$.
Then you have $$ P=\begin{pmatrix} 1 & 4 & 4\\ 1 & 2 & -2 \\ 1 & 1 &1\end{pmatrix} $$ So finally computing $PDP^{-1}\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix}$ gives you $a_n$.