# Powers of $2 \times 2$ matrices expressed in linear form

I recently reopened an old high school math textbook and came upon the matrices unit. Some of the questions were those rewrite-in-linear-form problems: given, say, $M^2 = 2M - I$, express in linear form ($aM+bI$) the matrices $M^3$ and $M^4$. The method of solution would be to express $M^n$ as $M^{n-1}M$ and continue expanding each consecutive set of powers until you end up substituting in the original $M^2 = 2M-I$, for answers of $M^3 = 3M-2I$ and $M^4 = 4M-3I$. Further, induction proof problems could also be posed, such as proving that $M^n = nM - (n-1)I$ for all $n \in \mathbb{Z}^+$. (Technically, it could also be for $n \in \mathbb{Z}$ if you consider $M^{-n} \equiv (M^{-1})^n \equiv (M^n)^{-1}$, which is relevant.)

I decided to take this a step further, and see if I could take powers of $M$ where $M^2 = aM + bI$. After some experimentation, one can define two related sequences $(a_n)$ and $(b_n)$ which describe linear form coefficients of $M^n$. Trivially, $M^1 = M = 1M + 0I$ and $M^0 = MM^{-1} = I = 0M + 1I$, which defines $a_0$, $a_1$, $b_0$, and $b_1$ as 0, 1, 1 and 0, respectively.

From this, and by the properties of expanding powers through $M^2 = a_2M + b_2I$, \begin{align} &\text{Basic} \\ a_0 &= b_1 = 0 \\ a_1 &= b_0 = 1 \\ \\ &\text{Following} \\ a_{n+1} &= a_2 a_n + b_n \\ b_{n+1} &= b_2 a_n \\ \\ &\text{Preceding} \\ a_{n-1} &= \frac{1}{b_2}b_n \\ b_{n-1} &= a_n - \frac{a_2}{b_2} b_n \\ \end{align} With a little playing around, one finds that $$a_n = a_2 a_{n-1} + b_2 a_{n-2}$$ which looks a little like the Fibonacci sequence, and since $b_n$ is already defined in terms of $a_n$, the crux of the problem lies in a non-recurring equation for $a_n$, given $a_2$ and $b_2$ as parameters.

Is there a formula for $a_n$, and subsequently, $b_n$? Am I on the right track with this method or is there a more elegant exploitability in these sequences?

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The general statement is a little messy, but in the special case that the polynomial $t^2 = at + b$ has distinct roots $r_1, r_2$, then $$a_n = \frac{r_1^n - r_2^n}{r_1 - r_2}$$
and $b_n$ has a similar expression. The idea is to look at the vector space of sequences $c_n$ satisfying $$c_{n+2} = a c_{n+1} + b c_n.$$
This vector space has dimension $2$ because such a sequence is uniquely determined by $c_0$ and $c_1$, but on the other hand $c_n = r_1^n$ and $c_n = r_2^n$ both clearly lie in it. It follows that these two sequences in fact form a basis of this vector space, so any solution can be expressed as a linear combination of them.