solving linear homogeneous recurrence relations of second-order $$\begin{equation}
  x_k=\begin{cases}
    1, & \text{for } k =0.\\
   7, & \text{for } k =1. \\
    -16x_{k-2}+8x_{k-1}
  \end{cases}
\end{equation}$$
I have to  find the commun for $x_k$ using  ,
$\textbf{characteristic polynomial  and Companion matrix and eigenvalue}$ 
till i was able to  resolve alle  those recurrence relations in which  the polynome is  equal  ${X^2}-a_1X-a_0$ in which the  root of this polynome are  $b_1$ and $b_2$ and $b_1 \neq b_2$
but in my case  i have only one root and i don t know how  to proceed 
.
$\textbf{my Try}$
$$x_{k+2}=-16x_k+8k_{k+1}$$
then i define a matrix $$f_k=\begin{bmatrix}
-16X_{k} \\
8X_{k+1}
\end{bmatrix}$$ 
so that  $$f_{k+1}=\begin{bmatrix}
-16X_{k+1} \\
8X_{k+2}
\end{bmatrix}=\begin{bmatrix}
0&-2 \\
8&8
\end{bmatrix}*\begin{bmatrix}
-16X_{k} \\
8X_{k+1}
\end{bmatrix}=\begin{bmatrix}
0&-2 \\
8&8
\end{bmatrix}*f_k$$
then let  $$ A =\begin{bmatrix}
0&-2 \\
8&8 \\
\end{bmatrix}$$
$$f_{k+1}= Af_k$$
in this Step i  would look for the characteristic polynomial of $A$
which is $$P_A(x) =x^2-8x+16$$
and then i will be able to  create the Capanion Matrix .
$$C(p_A(x))= \begin{bmatrix}
0 &-16 \\
1 &8
\end{bmatrix}$$
then is $$C^T(p_A(x))= \begin{bmatrix}
0 &1 \\
-16 &8
\end{bmatrix}$$
 and i have  a theorem in my  Skript that said 
   $$x_k=((C^T(p_A(x)))^k*\begin{bmatrix}
1\\
7
\end{bmatrix})_1$$
and 
$$(C^T(p_A(x)))^k⁼ P\begin{bmatrix}
a_1^k& & & & & &0 \\ 
& & &\ddots \\
0 & & & & & & a_n^k
\end{bmatrix}P^{-1}$$
where $a_i$ are the eigenvalue of $C^T(p_A(x))$
and $P = (s_1 ,\dots ,s_n)$ and $s_i \in \operatorname{Eig}_{a_i}(C^T(p_A(x)))$
but  in my case the  polynome $p_A(x) = x^2-8x+16 =(x-4)^2$ which has only 4 as root  and  $Eig_4(C^T(p_A(x)))= \langle  \begin{bmatrix}
1 \\
4
\end{bmatrix}\rangle $
and  in this  case i am not able to  create P which must be inverted matrix ...... help in this Step please .
 A: If you really like to follow this approach, you can redefine the problem:
Solve the recursion relation
$$
x_k = 8 x_{k-1} - (16 -\epsilon^2) x_{k-2}
$$
with $x_0=1$, $x_1=7$ and $\epsilon>0$. 
The characteristic polynomial is
$$
P(x) = x^2 - 8x + 16 - \epsilon^2 = (x - 4 +\epsilon)(x - 4 -\epsilon)
$$
with two distinct roots. 
The various matrices, eigenvalues, and eigenvectors can all be computed and once you reach the desired result take the limit $\epsilon \downarrow 0$.
The result will of course be the same as was mentioned before.
A: Unfortunately, since your characteristic polynomial has repeated roots, it isn't necessarily diagonalizable, and since you've found that the eigenspace associated to $4$ has dimension $1,$ we're out of luck on that approach. Instead, we will have to use generalized eigenvectors to get the job done.
What we should do now is find a solution $\vec w$ to $$\Bigl(C^T\bigl(p_A(x)\bigr)-4I_2\Bigr)\vec w=\begin{bmatrix}1\\4\end{bmatrix}.$$ These solutions will not be unique, but regardless of which one we pick, it will be linearly independent from $\begin{bmatrix}1\\4\end{bmatrix}.$ Letting $\begin{bmatrix}1\\4\end{bmatrix}$ and our chosen $\vec w$ be the columns of $P,$ we will have that $$C^T\bigl(p_A(x)\bigr)=P\begin{bmatrix}4 & 1\\0 & 4\end{bmatrix}P^{-1},$$ from which we can derive a more appropriate form for the powers of $C^T\bigl(p_A(x)\bigr),$ and take it from there.
Let me know if you have trouble with any of these steps, or if you just want to bounce your work off of someone to make sure you're on the right track.
