Could anyone help me in this :
put into normal form the isometry of $u$ in $\mathbb{R}^2$ that maps $(0,0),(1,0),(0,1)\to (2,-1),(1,-1),(2,0)$ respectively, and the hint is given that $u=r^{\epsilon}st$ is a normal form. thanks for helping.
put into normal form the isometry of $u$ in $\mathbb{R}^2$
0
$\begingroup$
linear-algebra
isometry
-
0I assume you mean an **affine** isometry as opposed to a linear isometry. – 2017-02-16
-
0So, what are $r$, $s$ and $t$? – 2017-02-16