I wanted to confirm my intuition about a problem that I got wrong relating to an application of the power method to recursion relations. The question is as follows:
For the context of the question, the matrix is $$A = \left[ \begin{matrix} a & -1 \\ 1 & 0 \end{matrix} \right]$$ so that $$\left[ \begin{matrix} x_{n+1} \\ x_n \end{matrix} \right] = A = \left[ \begin{matrix} a & -1 \\ 1 & 0 \end{matrix} \right] \left[ \begin{matrix} x_n \\ x_{n-1} \end{matrix} \right]$$
What I wanted to check was the following: 
The eigenvalues and eigenvectors in terms of $a$ are as follows: $$\lambda_\pm = \frac{a \pm \sqrt{a^2 - 4}}{2}$$
and $$v_\pm = \left[ \begin{matrix} \lambda_\pm \\\ 1\end{matrix} \right]$$
So my question is, this relationship of having complex eigenvalues and the solution to the recursion relation as $n \rightarrow \infty$ only holds in the case where $A$ is a $(2 \times 2)$ matrix right ? Also, why is it the case that for any $a^2 < 4$, then it follows that $A$ will have a bounded solution to the recursion relation ? I can see why such as case holds for $\lambda_\pm = 1\pm i$ but I don't exactly see why this holds for $\lambda_\pm = 1 \pm 2i$ for instance.
Would someone mind explaining this please ?
Thank you
As far as I can tell, this relationship between complex eigenvalues and solutions is specific to this problem; it is tied to the quadratic formula, which in turn is tied to the form of this particular matrix $A$.
As for why $a^2 < 4$ implies boundedness, note that (as the solution says) when $a^2<4$, $\lambda_-$ and $\lambda_+$ are a conjugate pair, and in particular, $|\lambda_-| = |\lambda_+|$. We also know that $\det(A) = a \cdot 0 - (-1) \cdot 1 = 1$, and so $\lambda_- \cdot \lambda_+ = 1$. Hence, we must have have $|\lambda_-| = 1 = |\lambda_+|$.