Fractal20 said:
I am trying to understand the role of the Jacobian in the Implicit Function Theorem. However, I have had a hard time finding any discussions that use the Jacobian and are accessible for my level.
http://mathworld.wolfram.com/ImplicitFunctionTheorem.html has been the best thing I have found.
More specifically I am unsure about maintaining that the Jacobian be square. So for the theorem to apply book is it true that given m equations with n variables (m < n) the number of dependent variables
needs to be equal to the number of equations? If so, then I think I see why the Jacobian will be square (maybe it shows my ignorance to even suggest that something not square could be Jacobian?).
If you got too few equations, then you might not have a unique solution.
The goal of the implicit function theorem is to come up with a function g:\mathbb{R}^n\rightarrow \mathbb{R}^k. For example, let's take n=1 and k=2.
Now, the implicit function theorem will come up with a function g:\mathbb{R}\rightarrow \mathbb{R}^2 that satisfies some equation f(x,g(x))=0. Now, let's say that f:\mathbb{R}\times \mathbb{R}^2\rightarrow \mathbb{R}. Then the level set f(x,y,z)=0 will have "dimension 2". For example, given f(x,y,z)=x^2+y^2+z^2-1, then we have the level set x^2+y^2+z^2=1.
Given an element (a,b,c) on the level set, we wish to find g:U\rightarrow V such that g(a)=(b,c) and such that f(x,g(x))=0 for all x. But we see that x^2+y^2+z^2=1 is the entire sphere. If we want a function such that (for example) g(0)=(0,1), then there are multiple ways to do so. For example, we can pick g(x)=(0,\sqrt{1-x^2}), or we can pick g(x)=(x,\sqrt{1-2x^2}). There are an infinite number of g that we can pick under our conditions.
Likewise, if we take the equation f(x,y,z)=y, then we will have the level set y=0. If we want g(0)=(0,1), then there are again multiple g that satisfy our equations. For example, everything of the form g(x)=(0,h(x)) for h:\mathbb{R}\rightarrow \mathbb{R}.
In our first case, we wanted a function from \mathbb{R} to the sphere. This was impossible to do uniquely because the sphere has dimension 2, while the line has dimension 1. In the second case, we want a function from \mathbb{R} to a plane. This is impossible to do uniquely because the plane again has dimension 2, while the line has dimension 1.
Now, what if we intersect our plane and our sphere?? This wil give something of dimension 1. So can we demand a function from \mathbb{R} to the intersection of our plane and our sphere? This comes down to take a function f:\mathbb{R}\times\mathbb{R}^2\rightarrow \mathbb{R}^2. Indeed, we want the function f(x,y,z)=(x^2+y^2+z^2-1,y). If we want a function now that satisfies g(0)=(0,1), then this is possible to do in a unique way: g(x)=(0,\sqrt{1-x^2}).
Now, what if we wanted to be malicious and intersect our sphere with the same sphere. Then we end up with something of dimension 2, which is too large. So the implicit function theorem should break down. So, take f(x,y,z)=(x^2+y^2+z^2-1,x^2+y^2+z^2-1). It can now be checked that the determinant of our Jacobian is zero, so indeed, the implicit function theorem is not applicable. So, the determinant condition is partly there to ensure us that our dimension is sufficiently reduced.
Also, in the case of something like f(x,y) = 0 and we are interested in finding y as a function of x, does the Jacobian still play a role? I want to say the Jacobian is a 1x1: \frac{\deltad}{\deltay}. Is it okay to have 1x1 Jacobians and is the determinant of a 1x1 matrix simply the single element? Furthermore, if in such a case the partial deriv with respect to x is 0, does the theorem still apply? Or does it just mean dy/dx = 0?
Such a thing can again be solved by the implicit function theorem. For example, given f:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}:(x,y)\rightarrow x^2+y^2-1. Let's say we want a function g(y)=x. The implicit function theorem does not apply directly, but we can make it apply by introducing a function f:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}:(x,y)\rightarrow f(y,x). The implicit function theorem appied to this, will yield a function such that f^\prime(x,g(x))=0. This is of course the same as f(g(x),x). So we have found a function going from the second variable to the first.
Of course, the condition for such a function to exist, is now of course that \frac{\partial f^\prime}{\partial y}\neq 0. This is the same as asking that \frac{\partial f}{\partial x}\neq 0.