$F(x,y)=(x^2+2y^2,2x^2+y^2)$, and $A=\{(x,y):x>0,y>0\}$
I need to show $F(A)=\{(u,v):0<u/2<v<2u\}$
I also need to find what is $G(=F^{-1}):B\rightarrow A$
For the first question I misunderstand what it is asking me.. I think if I choose u = 4, v=2 the inequality does not hold. What is the meaning of this?
For the second question, do I need to find the inverse functions, where B is the new domain of the inverse function?
Many thanks
Given $$\begin{gathered} F:{\mathbb{R}^2} \to {\mathbb{R}^2} \hfill \\ F(x,y) = \left( {\begin{array}{*{20}{c}} 1&2 \\ 2&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{x^2}} \\ {{y^2}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{x^2} + 2{y^2}} \\ {2{x^2} + {y^2}} \end{array}} \right) \hfill \\ \end{gathered}$$
Then: $$\left( {\begin{array}{*{20}{c}} {{x^2}} \\ {{y^2}} \end{array}} \right) = \frac{1}{3}\left( {\begin{array}{*{20}{c}} { - 1}&2 \\ 2&{ - 1} \end{array}} \right)\left( {\begin{array}{*{20}{c}} u \\ v \end{array}} \right) = \frac{1}{3}\left( {\begin{array}{*{20}{c}} { - u + 2v} \\ {2u - v} \end{array}} \right)$$ That is: $$\begin{gathered} x = \frac{1}{{\sqrt 3 }}\sqrt { - u + 2v} \hfill \\ y = \frac{1}{{\sqrt 3 }}\sqrt {2u - v} \hfill \\ \end{gathered}$$
Now we have a map: $$\begin{gathered} G:{\mathbb{R}^2} \to {\mathbb{R}^2} \hfill \\ G(u,v) = \frac{1}{{\sqrt 3 }}\left( {\begin{array}{*{20}{c}} {\sqrt { - u + 2v} } \\ {\sqrt {2u - v} } \end{array}} \right) \hfill \\ \end{gathered} $$ with property: $$\begin{gathered} G(F(x,y)) = \frac{1}{{\sqrt 3 }}\left( {\begin{array}{*{20}{c}} {\sqrt { - ({x^2} + 2{y^2}) + 2(2{x^2} + {y^2})} } \\ {\sqrt {2({x^2} + 2{y^2}) - (2{x^2} + {y^2})} } \end{array}} \right) \hfill \\ = \frac{1}{{\sqrt 3 }}\left( {\begin{array}{*{20}{c}} {\sqrt { - {x^2} - 2{y^2} + 4{x^2} + 2{y^2}} } \\ {\sqrt {2{x^2} + 4{y^2} - 2{x^2} - {y^2})} } \end{array}} \right) \hfill \\ = \frac{1}{{\sqrt 3 }}\left( {\begin{array}{*{20}{c}} {\sqrt {3{x^2}} } \\ {\sqrt {3{y^2}} } \end{array}} \right) = \frac{1}{{\sqrt 3 }}\left( {\begin{array}{*{20}{c}} {\sqrt 3 x} \\ {\sqrt 3 y} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) \hfill \\ \end{gathered} $$
But that means:$$(G \circ F)(x,y) = (x,y)$$
$G$ is called the inverse map for $F$.
And the other truth: $$(F \circ G)(u,v) = (u,v)$$ works the same way.$F$ is called the inverse map for $G$