Given a square image, each pixel can be treated as a position vector, and the modulo is taken to be the length of any side of the image. This is equivalent to considering the image as a square of area \(1\) whose pixels are located at positions \((0,0) \leq (x,y) \leq (1,1)\). Arnold's cat map will take each pixel and move it to a different location, or different position vector, under the transformation:

One might expect that the larger the image size, the longer the iterative period. However, although the period increased from 60 for the \(100\space x\space 100\) image to 216 for the \(162\space x\space 162\) image, the period length dropped dramatically down to 24 for the \(220\space x\space 220\) image. In fact, although the topic has been explore, there is no known formula to determine the Arnold's cat map period of an image based upon its size or number of pixels.

$$
\begin{bmatrix}
1 & 1 \\
1 & 2 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1-\lambda & 1 \\
1 & 2-\lambda \\
\end{bmatrix}
$$

$$ (1 - \lambda )(2 - \lambda ) - 1 = 0 $$

$$ \lambda^{2} - 3\lambda + 1 = 0 $$

$$ \lambda = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{5}}{2} $$

$$ (1 - \lambda )(2 - \lambda ) - 1 = 0 $$

$$ \lambda^{2} - 3\lambda + 1 = 0 $$

$$ \lambda = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{5}}{2} $$

_{cat}:

$$ \lambda_1 = \frac{3 + \sqrt{5}}{2} \space \space \space \space \space \space \space \lambda_2 = \frac{3 - \sqrt{5}}{2} $$

$$
(A_{cat}-\lambda_{1}I)\vec{x_{1}} = \vec{0}
$$

$$ \begin{bmatrix} 1-\lambda_1 & 1 \\ 1 & 2-\lambda_1 \\ \end{bmatrix} \begin{bmatrix} x_{1} \\ y_{1} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

$$ \begin{bmatrix} 1-\frac{3 + \sqrt{5}}{2} & 1 \\ 1 & 2-\frac{3 + \sqrt{5}}{2} \\ \end{bmatrix} \begin{bmatrix} x_{1} \\ y_{1} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

$$ (1-\frac{3 + \sqrt{5}}{2})x_{1} + (1)y_{1} = 0 \rightarrow (\frac{-1 - \sqrt{5}}{2})x_{1} + y_{1} = 0 \rightarrow y_{1} = (\frac{1 + \sqrt{5}}{2})x_{1} $$

So, for \(\lambda_{1}\), the associated eigenvector is:

\begin{bmatrix} 1 \\ \frac{1 + \sqrt{5}}{2} \\ \end{bmatrix}

$$ \begin{bmatrix} 1-\lambda_1 & 1 \\ 1 & 2-\lambda_1 \\ \end{bmatrix} \begin{bmatrix} x_{1} \\ y_{1} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

$$ \begin{bmatrix} 1-\frac{3 + \sqrt{5}}{2} & 1 \\ 1 & 2-\frac{3 + \sqrt{5}}{2} \\ \end{bmatrix} \begin{bmatrix} x_{1} \\ y_{1} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

$$ (1-\frac{3 + \sqrt{5}}{2})x_{1} + (1)y_{1} = 0 \rightarrow (\frac{-1 - \sqrt{5}}{2})x_{1} + y_{1} = 0 \rightarrow y_{1} = (\frac{1 + \sqrt{5}}{2})x_{1} $$

\begin{bmatrix} 1 \\ \frac{1 + \sqrt{5}}{2} \\ \end{bmatrix}

$$
(A_{cat}-\lambda_{2}I)\vec{x_{2}} = \vec{0}
$$

$$ \begin{bmatrix} 1-\lambda_2 & 1 \\ 1 & 2-\lambda_2 \\ \end{bmatrix} \begin{bmatrix} x_{2} \\ y_{2} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

$$ \begin{bmatrix} 1-\frac{3 - \sqrt{5}}{2} & 1 \\ 1 & 2-\frac{3 - \sqrt{5}}{2} \\ \end{bmatrix} \begin{bmatrix} x_{2} \\ y_{2} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

$$ (1-\frac{3 - \sqrt{5}}{2})x_{2} + (1)y_{2} = 0 \rightarrow (\frac{-1 + \sqrt{5}}{2})x_{2} + y_{2} = 0 \rightarrow y_{2} = -(\frac{1 + \sqrt{5}}{2})^{-1}y_{2} $$

So, for \(\lambda_{2}\), the associated eigenvector is:

\begin{bmatrix} 1 \\ -(\frac{1 + \sqrt{5}}{2})^{-1} \\ \end{bmatrix}

$$ \begin{bmatrix} 1-\lambda_2 & 1 \\ 1 & 2-\lambda_2 \\ \end{bmatrix} \begin{bmatrix} x_{2} \\ y_{2} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

$$ \begin{bmatrix} 1-\frac{3 - \sqrt{5}}{2} & 1 \\ 1 & 2-\frac{3 - \sqrt{5}}{2} \\ \end{bmatrix} \begin{bmatrix} x_{2} \\ y_{2} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$

$$ (1-\frac{3 - \sqrt{5}}{2})x_{2} + (1)y_{2} = 0 \rightarrow (\frac{-1 + \sqrt{5}}{2})x_{2} + y_{2} = 0 \rightarrow y_{2} = -(\frac{1 + \sqrt{5}}{2})^{-1}y_{2} $$

\begin{bmatrix} 1 \\ -(\frac{1 + \sqrt{5}}{2})^{-1} \\ \end{bmatrix}

And so the two eigenvalues are \(1 + \varphi\) and \(2 - \varphi\), with eigenvetors \(\begin{bmatrix} 1 \\ \varphi \\ \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ -\varphi^{-1} \\ \end{bmatrix}\) respetively.