Say, I have the following recurrence relation with $\hat{A}, ~\hat{B}$ known square matrices of complex numbers.
$$\begin{cases}\hat{W}^{(0)}=1 \\ \hat{W}^{(1)}=-\hat{A} \\ \hat{W}^{(n+2)}=-\hat{A}\hat{W}^{(n+1)}-\hat{B}\hat{W}^{(n)} \end{cases}$$
The corresponding recurrence for numbers has a simple closed form solution, which could be found by solving a quadratic:
$$x^2+ax+b=0$$
Then the general solution can be expressed as $C_1 x_1^n+C_2 x_2^n$.
However, I believe for matrices (which may not commute) the solution is not as simple. Could you point me in the right direction?
While the question is general enough, if you are curious about my motivation, I have been trying to invent a new perturbation method for time dependent Schroedinger equation, and obtaining an explicit solution for this recurrence would make my life easier.
Googling "matrix linear recurrence" has produced very little in the way of general results and nothing for second order recurrences...
(Partial answer, too long for a comment.) An explicit form can be found in case there exists an invertible matrix $\,C\,$ that satisfies the quadratic $\,C^2-CA+B=0\,$ and commutes with $\,B\,$.
Multiplying the recurrence by $\,C\,$ gives:
$$ C\,W_{n+2}=-\underbrace{\,CA\,}_{C^2+B}W_{n+1}-CB\,W_{n} \;\; \iff \;\; C\,(W_{n+2}+ C\, W_{n+1}) = -B\,(W_{n+1}+C\,W_{n}) $$
It follows that:
$$ W_{n+2}+ C\, W_{n+1} = -C^{-1}B\,(W_{n+1}+C\,W_{n}) = \ldots = \left(-C^{-1}B\right)^{n+1} \,(W_1 + C\,W_0) $$
This reduces the problem to a first order linear recurrence of the form $\,Z_{n+1}=U\,Z_n+V^n\,W\,$ where $\,U=-C\,$ and $\,V = -C^{-1}B\,$, which telescopes to:
$$ Z_{n+1}=U^{n+1}\,Z_0 + \left(V^n + V^{n-1} \, U+ \ldots + V\,U^{n-1}+ U^n\right)\,W $$
In the case where $\,V-U\,$ is invertible, and given that $\,U,V\,$ commute, this further simplifies to:
$$ Z_{n+1} = U^{n+1}\,Z_0+ \left(V-U\right)^{-1}\,\left(V^{n+1}-U^{n+1}\right)\,W $$