Given that $T(x,y)= (y-2,x+3)$ is a rigid transformation how can i express $T = S \circ U$ where S is a translation and $U$ is an orthogonal transformation.
It may be a little simple, but it´s been a long time since i dive into geometry and i´m trying to be on the track again. Thanks :)
It's indeed simple:
Let $U$ be the reflection through the diagonal line $x=y$, that is, $$U(x,y)=(y,x)$$ and let $S$ be the translation by vector $(-2,3)$, i.e. $$S(a,b)= (a-2,\,b+3)\,.$$ Then $$S\circ U(x,y)=S(y,x)=(y-2,\,x+3)=T(x,y)$$ for any vector $(x,y)\in\Bbb R^2$.
Bonus: can you find the other translation $S'$ that makes $T=U\circ S'$?