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I am trying to find

$$\begin{bmatrix}t&1-t\\1-t&t\\\end{bmatrix}^n$$ in $\mathbb{Z}_n[t^{\pm 1}]/(t-1)^2$, where $n$ is a positive integer. My guess is that the result is the identity matrix but am not sure about a proof. Thanks for any help or hints!

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    try to find both eigenvectors (eigenvalues is 1 and $2*t-1$) and calculate images for it – 2017-01-24
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    My first thought is $$ \left[ \begin{array}{cc} t & 1-t\\ 1-t & t\\ \end{array} \right] = \left[ \begin{array}{cc} 0 & 1-t\\ 1-t & 0\\ \end{array} \right] + t \left[ \begin{array}{cc} 1 & 0\\ 0 & 1\\ \end{array} \right]. $$ Because the last matrix is a scalar multiple of the identity, the two matrices on the right-hand side commute. Therefore, Newton's binomial formula will work as it does for two commuting matrives, $A, B$: $$ (A + B)^n = \sum_{k=0}(^{n}_{k}) A^{n-k} B^{k}. $$ – 2017-01-24
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    sorry, you don't need calculate both eigenvectors (1, 1) for 1 and (1, -1) for 2*x - 1 You only need to prove what $(2t-1)^n = p(t)(t-1)^2 +1 $ in $\mathbb{Z}_n$ – 2017-01-24
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    Use the idea given by @avs but with slightly simpler matrices, taking into account that it is important to use at an early stage the fact that for example $t(t-1)=t-1$ ; for example for converting the initial matrix into $\left[ \begin{array}{cc} t & t(1-t)\\ t(1-t) & t\\ \end{array} \right]=t\left[ \begin{array}{cc} 1 & 1-t\\ 1-t & 1\\ \end{array} \right]=t(A+tB)$ with $A$ not depending on $t$ with $A$ and $B$ having simple powers... – 2017-01-24

3 Answers 3

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A variant of avs's strategy in comments: Write $$ \begin{bmatrix} t & 1-t \\ 1-t & t \end{bmatrix} = I_{2\times 2} + (t-1)\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} $$ and use the binomial theorem (which applies because the $I$ commutes with everything) to raise this to the $n$th power.

All but two of the resulting terms contain two or more factors of $t-1$ and therefore vanish. One the the remaining terms has a factor of $\binom n1$ which equals $0$ modulo $n$, so it vanishes too. The last one is $\binom 11 I^n = I$.

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    Thank you for this proof! It is crystal clear and neat! – 2017-01-24
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If $$M = \pmatrix{t & 1-t\cr 1-t & t}$$

show by induction that

$$ M^j = \pmatrix{j t - (j-1) & j - j t\cr j - jt & jt - (j-1)\cr} $$

and then take $j=n$.

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    Thanks for this answer! It is more than what I needed but it implies the result! – 2017-01-24
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$\newcommand{\bbx}[1]{\,\bbox[10px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\mbox{Note that}\quad \pars{\begin{array}{cc} \ds{t} & \ds{1 - t} \\ \ds{1 - t} & \ds{t} \end{array}}^{n} = t^{n}\pars{\begin{array}{cc} \ds{1} & \ds{\Lambda} \\ \ds{\Lambda} & \ds{1} \end{array}}^{n} = t^{n}\pars{\sigma_{0} + \Lambda\sigma_{x}}^{n} \label{1}\tag{1} \\[2mm] &\mbox{where}\quad\Lambda \equiv {1 - t \over t}\label{2}\tag{2} \end{align}

$\ds{\sigma_{0}}$ is the $\ds{2 \times 2\ identity\ matrix}$ and $\ds{\sigma_{x}}$ is a Pauli Matrix.

\begin{equation} \bbx{\ds{\mbox{Note that}\quad \sigma_{x}^{2} = \sigma_{0}}} \label{3}\tag{3} \end{equation}

\begin{align} \pars{\begin{array}{cc} \ds{t} & \ds{1 - t} \\ \ds{1 - t} & \ds{t} \end{array}}^{n} & = t^{n}\,n!\bracks{z^{n}}\exp\pars{\bracks{\sigma_{0} + \Lambda\sigma_{x}}z} = t^{n}\,n!\bracks{z^{n}}\bracks{\exp\pars{z\sigma_{0}}\exp\pars{\Lambda z\sigma_{x}}} \\[5mm] & = t^{n}\,n!\bracks{z^{n}}\braces{\exp\pars{z\sigma_{0}} \bracks{\cosh\pars{\Lambda z}\sigma_{0} + \sinh\pars{\Lambda z}\sigma_{x}}} \qquad\pars{~\mbox{see}\ \eqref{3}~} \\[5mm] & = t^{n}\,n!\bracks{z^{n}}\bracks{% {\exp\pars{\bracks{1 + \Lambda}z} \over 2}\,\pars{\sigma_{0} + \sigma_{x}} + {\exp\pars{\bracks{1 - \Lambda}z} \over 2}\,\pars{\sigma_{0} - \sigma_{x}}} \\[5mm] & = {1 \over 2}\,t^{n}\bracks{% \pars{1 + \Lambda}^{n} \pars{\begin{array}{cc}\ds{1} & \ds{1} \\ \ds{1} & \ds{1}\end{array}} + \pars{1 - \Lambda}^{n}\pars{\begin{array}{rr}\ds{1} & \ds{-1} \\ \ds{-1} & \ds{1}\end{array}}} \\[5mm] & = {1 \over 2}\,\bracks{% \pars{\begin{array}{cc}\ds{1} & \ds{1} \\ \ds{1} & \ds{1}\end{array}} + \pars{2t - 1}^{n}\pars{\begin{array}{rr}\ds{1} & \ds{-1} \\ \ds{-1} & \ds{1}\end{array}}}\qquad\pars{~\mbox{see}\ \eqref{2}~} \\[5mm] & = \bbx{\ds{{1 \over 2} \pars{\begin{array}{cc} \ds{1 + \bracks{2t - 1}^{n}} & \ds{1 - \bracks{2t - 1}^{n}} \\[2mm] \ds{1 - \bracks{2t - 1}^{n}} & \ds{1 + \bracks{2t - 1}^{n}} \end{array}}}} \end{align}