ATRACTOR

Sejam \(A=\begin{pmatrix}-1 & 2\\ 1 & -1 \end{pmatrix}\) e \(Q=\left\{ (x,y)\in\mathbb{R}^{2}:x,y>0\right\}\).

Diagonalizando a matriz \(A\), vem: \[\begin{array}{cll} A & = & \begin{pmatrix}-1 & 2\\ 1 & -1 \end{pmatrix}\\ & = & \begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix}-1-\sqrt{2} & 0\\ 0 & -1+\sqrt{2} \end{pmatrix}\begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}^{-1} \end{array}\]

Então, as sucessivas potências de \(A\) são dadas por: \[\begin{array}{cll} A^{n} & = & \begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix}-1-\sqrt{2} & 0\\ 0 & -1+\sqrt{2} \end{pmatrix}^{n}\begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}^{-1}\\ & = & \begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix}\left(-1-\sqrt{2}\right)^{n} & 0\\ 0 & \left(-1+\sqrt{2}\right)^{n} \end{pmatrix}\begin{pmatrix}-\frac{1}{2\sqrt{2}} & \frac{1}{2}\\ \frac{1}{2\sqrt{2}} & \frac{1}{2} \end{pmatrix}\\ & = & \begin{pmatrix}\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2} & \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}\\ \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}} & \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2} \end{pmatrix} \end{array}\]

Assim, vem: \[\begin{array}{cll} P_{n} & = & A^{n}P_{0}\\ & = & \begin{pmatrix}\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2} & \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}\\ \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}} & \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2} \end{pmatrix}\begin{pmatrix}x_{0}\\ y_{0} \end{pmatrix}\\ & = & \begin{pmatrix}\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}x_{0}+\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}y_{0}\\ \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}x_{0}+\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}y_{0} \end{pmatrix} \end{array}\]

Se \(P_{n}\in\mathbb{Q}\), temos: \[\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}x_{0}+\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}y_{0}>0\] \[\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}x_{0}+\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}y_{0}>0\]

Se \(n\) é par, então \[\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}=\frac{\left(1+\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}>0\] e \[\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}=\frac{-\left(1+\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}<0,\] dado que \(1+\sqrt{2}>-1+\sqrt{2}>0\Rightarrow\left(1+\sqrt{2}\right)^{n}>\left(-1+\sqrt{2}\right)^{n}>0\). Logo, vem:
\[\begin{array}{cl} & \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}x_{0}+\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}y_{0}>0\\ \Leftrightarrow & x_{0}>\frac{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}\sqrt{2}y_{0} \end{array}\]
\[\begin{array}{cl} & \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}x_{0}+\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}y_{0}>0\\ \Leftrightarrow & x_{0}<\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}\sqrt{2}y_{0} \end{array}\]
\[\begin{array}{ccl} \frac{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}} & = & \frac{1-\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}{1+\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}\\ & = & \frac{1-\left(2\sqrt{2}-3\right)^{n}}{1+\left(2\sqrt{2}-3\right)^{n}}\\ & = & \frac{1-\left(3-2\sqrt{2}\right)^{n}}{1+\left(3-2\sqrt{2}\right)^{n}}\\ & = & \frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1 \end{array}\]
\[\begin{array}{ccl} \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}} & = & \frac{1+\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}{1-\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}\\ & = & \frac{1+\left(2\sqrt{2}-3\right)^{n}}{1-\left(2\sqrt{2}-3\right)^{n}}\\ & = & \frac{1+\left(3-2\sqrt{2}\right)^{n}}{1-\left(3-2\sqrt{2}\right)^{n}}\\ & = & \frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1 \end{array}\]

Logo, temos: \[P_{n}\in\mathbb{Q}\Leftrightarrow\left(\frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}<x_{0}<\left(\frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}\]

Se \(n\) é ímpar, então \[\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}=\frac{-\left(1+\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}<0\] e \[\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}>0,\] dado que \(1+\sqrt{2}>-1+\sqrt{2}>0\Rightarrow\left(1+\sqrt{2}\right)^{n}> \left(-1+\sqrt{2}\right)^{n}>0\).

Logo, vem:
\[\begin{array}{cl} & \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}x_{0}+\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}y_{0}> 0\\ \Leftrightarrow & x_{0}<\frac{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}\sqrt{2}y_{0} \end{array}\]
\[\begin{array}{cl} & \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}x_{0}+\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}y_{0}>0\\ \Leftrightarrow & x_{0}>\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}\sqrt{2}y_{0} \end{array}\]
\[\begin{array}{ccl} \frac{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}} & = & \frac{1-\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}{1+\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}\\ & = & \frac{1-\left(2\sqrt{2}-3\right)^{n}}{1+\left(2\sqrt{2}-3\right)^{n}}\\ & = & \frac{1+\left(3-2\sqrt{2}\right)^{n}}{1-\left(3-2\sqrt{2}\right)^{n}}\\ & = & \frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1 \end{array}\]
\[\begin{array}{ccl} \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}} & = & \frac{1+\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}{1-\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}\\ & = & \frac{1+\left(2\sqrt{2}-3\right)^{n}}{1-\left(2\sqrt{2}-3\right)^{n}}\\ & = & \frac{1-\left(3-2\sqrt{2}\right)^{n}}{1+\left(3-2\sqrt{2}\right)^{n}}\\ & = & \frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1 \end{array}\]

Logo, temos: \[P_{n}\in\mathbb{Q}\Leftrightarrow\left(\frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}<x_{0}<\left(\frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}\]

Em ambos os casos, temos: \(P_{n}\in\mathbb{Q}\) se e só se \(x_{0}\) está entre \(\left(\frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}\) e \(\left(\frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}\), com \(y_{0}>0\)

Notemos agora que, como \(0<3-2\sqrt{2}<1\), \(\left(\frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1\right)_{n\in\mathbb{N}}\) é uma sucessão estritamente crescente cujo limite é \[\lim\left(\frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1\right)=\frac{2}{1+\lim\left(3-2\sqrt{2}\right)^{n}}-1=\frac{2}{1+0}-1=1\]

Analogamente, \(\left(\frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1\right)_{n\in\mathbb{N}}\) é uma sucessão estritamente decrescente que converge para \(1\) (de facto, trata-se da sucessão inversa da anterior).

Portanto, \(P_{n}\in\mathbb{Q}\) para todo o \(n\in\mathbb{N}\) se e só se \(x_{0}=\sqrt{2}y_{0}\) e \(y_{0}>0\). Podemos então concluir que os sucessivos iterados do ponto inicial pertencem ao primeiro quadrante apenas quando o ponto inicial é um ponto da recta \(x=\sqrt{2}y\) situado no primeiro quadrante, ou seja, as suas coordenadas correspondem ao valor da diagonal e do lado de um quadrado.

Suponhamos agora que, em vez da matriz \(A\), considero a sua inversa. O que acontecerá com as sucessivas imagens de um ponto do plano pela aplicação dessa matriz?

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