Find a formula for $\phi^{-1}(u,v)$ and show that $\phi\circ\phi^{-1}(u,v) = (u,v)$ and $\phi^{-1}\circ\phi(x,y)=(x,y)$

Question

Consider $\phi:\mathbb{R}^2\rightarrow\mathbb{R}^2$ given by $\phi(x,y) = (x+y,2x-y)$, which is an isomorphism.

Find a formula for $\phi^{-1}(u,v)$ and show that $\phi\circ\phi^{-1}(u,v) = (u,v)$ and $\phi^{-1}\circ\phi(x,y)=(x,y)$

---

How does one find those with the given values and where is $(u,v)$ coming from?

Answer 1

I like to use matrices. Note that $\phi$ is of the form $\phi(\vec x)=A\vec x$ where $$ A= \left[\begin{array}{rr} 1 & 1 \\ 2 & -1 \end{array}\right] $$ Since $$ A^{-1}= \left[\begin{array}{rr} \frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{array}\right] $$ we have $$ \phi^{-1}(\vec x)=A^{-1}\vec x=\frac{1}{3}(x+y,2\,x-y) $$

Answer 2

"(u,v)" is simply the name of the point that $\phi$ maps (x, y) into. You are told that $\phi$ maps (x, y) into (x+ y, 2x- y)= (u, v). The inverse function $\phi^{-1}$ maps (u, v) into (x, y) so you want to solve x+ y= u, 2x- y= v for x and y as functions of u and v.