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I'm trying to calculate the powers of a 2 by 2 matrix with entries in $\mathbb{Z} \left[ t,t^{-1} \right]$.

The matrix is \begin{bmatrix} 0 & 1 \\ 1 & t \end{bmatrix}

I tought of writing my matrix in the following form $$ \begin{bmatrix} 0 & 1 \\ 1 & t \end{bmatrix} = t \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$ Now we get $$ \begin{bmatrix} 0 & 1 \\ 1 & t \end{bmatrix} ^n = \sum_{k=0}^n \binom{n}{k} t^{n-k} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}^{n-k} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}^{k} $$

The only issue is that if $\mathcal{M}_n(\mathbb{Z} \left[ t,t^{-1} \right])$ is not isomorphic to $\mathcal{M}_n(\mathbb{Z}) \left[ t,t^{-1} \right] $ then my logic will be wrong.

I also thought of diagonalizing the matrix, but I found no algorithme to get it done.

Thanks in advance for your help.

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    $\begingroup$ The issue is not a problem, because your matrices have finitely many entries. $\endgroup$
    – S. Carnahan
    May 2, 2017 at 21:18
  • $\begingroup$ So it is correct to reason the way i did ? $\endgroup$ May 2, 2017 at 21:39

2 Answers 2

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This is not much different from Robert Israel's answer, but here goes:

Let $A(t)=\begin{pmatrix}0&1\\1&t\end{pmatrix}$ and write $A^n(t)=\begin{pmatrix}c_n&b_n\\b_n&a_n\end{pmatrix}$. Then, $$\begin{pmatrix}c_{n+1}&b_{n+1}\\ b_{n+1}&a_{n+1}\end{pmatrix}=A^{n+1}=\begin{pmatrix}c_n&b_n\\b_n&a_n\end{pmatrix}\begin{pmatrix}0&1\\1&t\end{pmatrix} =\begin{pmatrix}b_n&c_n+tb_n\\a_n&b_n+ta_n\end{pmatrix}$$ implies $c_{n+1}=b_n, b_{n+1}=a_n, a_{n+1}=b_n+ta_n$. That means, $A^n=\begin{pmatrix}a_{n-2}&a_{n-1}\\a_{n-1}&a_n\end{pmatrix}$ and $a_n=ta_{n-1}+a_{n-2}$ with initial conditions $a_{-1}:=0$ and $a_0=1$. This recurrence can be utilized to prove the following explicit formula $$a_n(t)=\sum_{k=0}^{\lfloor\frac{n}2\rfloor}\binom{n-k}kt^{n-2k}.$$

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Your matrix (call it $A(t)$) has characteristic polynomial $\lambda^2 - t \lambda - 1$, so it satisfies $A(t)^2 - t A(t) - I = 0$ and thus $A(t)^{n+2} = t A(t)^{n+1} + A(t)^n$. For $n \ge 2$ I get

$$ A^n = \pmatrix{i^{n-2} U_{n-2}(-it/2) & i^{n-1} U_{n-1}(-it/2)\cr i^{n-1} U_{n-1}(-it/2) & i^n U_{n}(-it/2)\cr} $$

where $U_n$ are the Chebyshev polynomials of the second kind.

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