How to prove the implicit function theorem fails Define $$F(x,y,u,v)= 3x^2-y^2+u^2+4uv+v^2$$ $$G(x,y,u,v)=x^2-y^2+2uv$$
Show that there is no open set in the $(u,v)$ plane such that $(F,G)=(0,0)$ defines $x$ and $y$ in terms of $u$ and $v$.
If (F,G) is equal to say (9,-3) you can just apply the Implicit function theorem and show that in a neighborhood of (1,1) $x$ and $y$ are defined in terms of $u$ and $v$. But this question seems to imply that some part of the assumptions must be necessary for such functions to exist?
I believe that since the partials exist and are continuous the determinant of $$\pmatrix{
\frac{\partial F}{\partial x}&\frac{\partial F}{\partial y}\cr
\frac{\partial G}{\partial x}&\frac{\partial G}{\partial y} }$$ must be non-zero in order for x and y to be implicitly defined on an open set near any point (u,v) but since the above conditions require x=y=0 the determinant of the above matrix is =0.
I have not found this in an analysis text but this paper http://www.u.arizona.edu/~nlazzati/Courses/Math519/Notes/Note%203.pdf claims it is necessary.
 A: Hint number 2: Forget about the implicit function theorem and do it the pedestrian way. The system $F=0$, $G=0$ is linear in $x^2$ and $y^2$, so go and solve it for these auxiliary variables. 
A: To say $(F,G) = (0,0)$ is to say that $y^2 - 3x^2 = u^2 + 4uv + v^2$ and $y^2 - x^2 = 2uv$. By some algebra, this is equivalent to $x^2 = -{1 \over 2}(u + v)^2$ and $y^2 = -{1 \over 2}(u - v)^2$. So you are requiring the nonnegative quantities on the left to be equal to the nonpositive quantities on the right. Hence the solution set is only $(x,y,u,v) = (0,0,0,0)$, where everything is zero.
Suppose on the other hand you had equations $x^2 = {1 \over 2}(u + v)^2$ and $y^2 = {1 \over 2}(u - v)^2$. Then you could solve them, but there is no uniqueness now; you could take $(x,y) = (\pm {1 \over \sqrt{2}}(u + v),\pm {1 \over \sqrt{2}}(u - v))$ obtaining four distinct smooth solutions that come together at $(0,0,0,0)$.
So these are good examples showing that if the determinant is zero at $(0,0)$ you don't have to have existence of solutions, nor uniqueness when you do have existence.
