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$F(x,y)=(x^2+2y^2,2x^2+y^2)$, 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

Bob
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  • Can you find a pair $(x,y)$ such that $F(x,y) = (4,2)$? If you cannot, then you can't choose $u=4, v=2$. – Jonny Apr 03 '15 at 01:17
  • @Jonny You make an excellent point. So to show this for the general case do I set u to be $x^2+2y^2$, and v to be $2x^2+y^2$? – Bob Apr 03 '15 at 01:27
  • Yes. Then you must demonstrate that a pair is in $F(A)$ if and only if it is in ${(u,v): 0<u/2:v:2u}$ – Jonny Apr 03 '15 at 01:47
  • @Jonny Can I trouble you with the second question? I am guessing I need to find an inverse function, but haven't ever attempted this in $\mathbb R^2$. Any hints you can give me? – Bob Apr 03 '15 at 02:00
  • I think a lot of the work in solving question 1 will help with question 2. – Jonny Apr 03 '15 at 02:29

1 Answers1

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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 $G$

Frieder
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