# The role of the Jacobian in the Implicit Function Theorem

1. Aug 23, 2012

### Fractal20

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?).

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?

Hopefully this wasn't too rambling and unspecific. Any insight would be appreciated! Thanks!

2. Aug 23, 2012

### chiro

Hey Fractal20.

The Jacobian is basically a way of quantifying the volumetric change at a particular point for a n-dimensional mapping to an n-dimensional mapping.

If the Jacobian is zero, it means that there is no change whatsoever, and this means you get an overall change of zero at that point (with respect to the rate of change with respect to the expansion and contraction with respect to the entire volume).

Since you have no-expansion or contraction at that point it means that you won't get an inverse because an inverse function requires that you are able to use the rate of change of the function to get the rate of change of the inverse.

As a really brief example, consider y = 2 for all x on the real line. dy/dx = 0. Now you can't get an inverse in this instance because there is no dependency between y and x and for an inverse you need a dependency at the minimum before you can even think about getting an inverse.

The other to think about the inverse function is to think about branch cuts when you get turning points. Take for example f(x) = sin(x). You get a max at pi/2 and a minimum at -pi/2 (and all additions of n*pi). So in this case you get a principal branch of -pi/2 to pi/2. But for an inverse function to exist it must be a 1-1 mapping which means you have to look at the parts in-between the points when the derivative is 0.

This isn't a formal argument but it should give some intuition behind the justification of the inverse function theorem.

As for the Jacobian being square, you have to be aware that a transformation must go from one dimension to the same dimension and this ends up in a square matrix.

If you went from a higher dimension to a lower one you would be losing information and it would be not a substitution but a projection and a substitution that preserves the equivalence of the representation preserves the dimension and in terms of a transformation (at least a linear one) this means the matrix is a square one.

3. Aug 23, 2012

### Fractal20

Great, that clears up things about the squareness. About that bit towards the end of my original post. Are there 1x1 Jacobians, or does that not really apply? Maybe it is translating area to area or just length to length. And lastly, again in the case of a situation like g(x,y) = 0. Do both gx and gy need to not be equal to 0 in order to invoke the implicit function theorem, or is it just that in order to write a variable (let's say u) as a function of another (v) then the derivative g with respect to u must not be 0?

4. Aug 23, 2012

### micromass

Staff Emeritus
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.

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 $\frac{\partial f}{\partial x}\neq 0$.

5. Aug 23, 2012

### chiro

A 1x1 Jacobian is basically dy/dx or whatever your dependent/independent variables are.

The other thing that will help you is to think of what the Jacobian matrix (not just the determinant) actually is with respect to the derivatives.

Once you have done that, think about what the inverse matrix (i.e. the matrix above but inverted) represents.

This generalization is made very clear in the theory of tensors where you deal with arbitrary co-ordinate systems and you want to go from one system to another and back again. If you can only go one way then it's really pointless when you need to go back.

Then take this all together and consider the transformations as 'co-ordinate systems' in the tensor way and it should start to make a lot more sense.