Affine equivalent for [[Fixed Points]] in [[Linear Transformation|Linear Transformations]]. $\huge \augmented{c|c}{A-I & -\vec b} $ *Same as the formula for fixed points in Linear Transformations because linear transformations are a subset of Affine transformations, where $\vec b = \vec 0$* $\huge\begin{align*} \let T &: \Rn3\to\Rn3 \\ T: \mat{x\\y\\z} &\mapsto\mat{2x+z+5\\-3x-y-3\\2x+2y-2} \\ T\pa{\vec x} &= \mat{ 2 & 0 & 1\\ -3 & -1 & 0 \\ 2 & 2 & 0 } \mat{x\\y\\z} + \mat{5\\-3\\-2} \\ \\ \let \vec x_{0} &\text{ be a fixed point of } T: \\ \mat{x_{0}\\ y_{0} \\ z_{0}} &= \mat{2x_0+z_0+5\\-3x_0-y_0-3\\2x_0+2y_0-2 }\\ \vec x_{0} &= \mat{ 2 & 0 & 1\\ -3 & -1 & 0 \\ 2 & 2 & 0 } \mat{x_0\\y_0\\z_0} + \mat{5\\-3\\-2} \\ \rowechelon{ 1 & 0 & 1 & 5\\ -3 & -2 & 0 & -3 \\ 1 & 1 & -1 & -2 \\ } &\sim \cdots \rowechelon{ 1 & 0 & 1 & -5 \\ 0 & 1 & -\frac{3}{2} & 6 \\ 0 & 0 & 0 & 0 } \\ \sim \vec x_0 \in \cases{ x+z &= -5 \\ y - \frac{3}{2}z &= 6 \\ 0 &= 0\\ } \end{align*}$ #### Using fixed points to find translation If we have our matrix $A$ and fixed point $\vec x_{0}$, $\huge \vec b = -\pa{A - I}\vec x_{0} \\ $ For any standard affine transformation, $\huge \vec b = \frac{c}{\norms{\vec n}^{2}}\vec n$