Suppose we wish to solve a differnece equation by using linear algebra, just like presented in Strang's Linear Algebra book.
Now let's say we have the difference equation: $h[n]+\frac{2}{3}h[n-1] +\frac{1}{9}h[n-2] = 0$.
Using the method we have the system: $\begin{bmatrix} h[n] \\ h[n-1]\end{bmatrix} = \begin{bmatrix} -2/3 &-1/9 \\ 1 & 0\end{bmatrix} \cdot \begin{bmatrix} h[n-1] \\ h[n-2]\end{bmatrix}$. where we are given that h[0]=1, $h[1]=-\frac{2}{3}$.
The above system can be written as $ u_k = A u_{k-1}$ and solving recursively we have $u_k = A^{k-1}u_1.$
We can write $A^{k-1}$ as : $A^{k-1} = S\cdot Λ ^{k-1}\cdot S^{-1}$ where S is the matrix containing the eigenvectors of A and Λ is the matrix containing the eigenvalues of A, raised to the power k-1.
The problem is that S is not ivertible as there is a repeated eigenvector. In that case ,the above form of $A^{k-1}$ is not valid.
However there is still a solution to the above difference equation.
My question is how do we get a solution of difference equations where the eigenvalue is repeated as in this case?
Any source that contains a derivation to the solution of the above difference equation is preffered. Thanks.