1
$\begingroup$

Let $T_n$ be a sequence of $2 \times 2$ matrices defined by the difference equation $$ T_{n+1} = B T_{n}+AT_{n-1} $$ with initial conditions $T_0 =1, T_1 = B$ where $B$ and $ A$ are also $2 \times 2$ matrices which don't commute (the only thing I know about $A$ and $B$ is that the determinant of $A$ is $1$). I want to solve to be able to give a closed form expression for $T_n$ (actually, I would be happy to get a closed form expression for $\det(T_n)$, not $T_n$ itself). I've tried solving this expression using a matrix valued $Z$ transform like one would when $T_n$ is just a sequence of numbers and not matrices.

In particular, define $Z$ transform as $$ \Gamma(z) = \sum_{n=0}^{\infty}z^{-n}T_n \equiv \mathcal{Z}[T_n] $$ Using several properties of the $Z$ transform ($\mathcal{Z}[T_{n-1}] =z^{-1}\Gamma(z)$ and $\mathcal{Z}[T_{n+1}] = z \Gamma(z)-z T_0$ ) we arrive at the $Z$ transform of the sequence $T_n$ $$ \Gamma(z) = \dfrac{z^2}{z^2-zB-A} $$ The matrix $T_n$ is then give by $$ T_n = \dfrac{1}{2\pi i}\oint_\Omega \dfrac{z^{n+1}}{z^2-zB-A}dz $$ where $\Omega$ is a closed curve which circles all the points for which $\det(z^2-zB-A) = 0$. This is a fourth order polynomial in $z$ so let's assume that it has 4 distinct roots in the complex plane. I would like to use the residue theorem $$ T_n = \sum_{i=1}^4 \dfrac{z_i^{n+1}}{2z_i-B} $$ where $z_i$ is a root of the polynomial $\det(z^2-zB-A)$. But I'm almost sure this is wrong since, for example, I don't think that $T_0 = 1$. Does anyone have a solution to this problem? Thanks!

$\endgroup$
0
$\begingroup$

In your last equality, $T_n$ is a polynomial in $B$, that is false in general.

A closed form fo $T_n$ seems to me hopeless. It would work if we had (for example) a relation between $A,B$, as $(AB)^2=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.