The Unapologetic Mathematician

Mathematics for the interested outsider

The Implicit Function Theorem I

Let’s consider the function F(x,y)=x^2+y^2-1. The collection of points (x,y) so that F(x,y)=0 defines a curve in the plane: the unit circle. Unfortunately, this relation is not a function. Neither is y defined as a function of x, nor is x defined as a function of y by this curve. However, if we consider a point (a,b) on the curve (that is, with F(a,b)=0), then near this point we usually do have a graph of x as a function of y (except for a few isolated points). That is, as we move y near the value b then we have to adjust x to maintain the relation F(x,y)=0. There is some function f(y) defined “implicitly” in a neighborhood of b satisfying the relation F(f(y),y)=0.

We want to generalize this situation. Given a system of n functions of n+m variables

\displaystyle f^i(x;t)=f^i(x^1,\dots,x^n;t^1,\dots,t^m)

we consider the collection of points (x;t) in n+m-dimensional space satisfying f(x;t)=0.

If this were a linear system, the rank-nullity theorem would tell us that our solution space is (generically) m dimensional. Indeed, we could use Gauss-Jordan elimination to put the system into reduced row echelon form, and (usually) find the resulting matrix starting with an n\times n identity matrix, like


This makes finding solutions to the system easy. We put our n+m variables into a column vector and write


and from this we find


Thus we can use the m variables t^j as parameters on the space of solutions, and define each of the x^i as a function of the t^j.

But in general we don’t have a linear system. Still, we want to know some circumstances under which we can do something similar and write each of the x^i as a function of the other variables t^j, at least near some known point (a;b).

The key observation is that we can perform the Gauss-Jordan elimination above and get a matrix with rank n if and only if the leading n\times n matrix is invertible. And this is generalized to asking that some Jacobian determinant of our system of functions is nonzero.

Specifically, let’s assume that all of the f^i are continuously differentiable on some region S in n+m-dimensional space, and that (a;b) is some point in S where f(a;b)=0, and at which the determinant

\displaystyle\det\left(\frac{\partial f^i}{\partial x^j}\bigg\vert_{(a;t)}\right)\neq0

where both indices i and j run from 1 to n to make a square matrix. Then I assert that there is some k-dimensional neighborhood T of b and a uniquely defined, continuously differentiable, vector-valued function g:T\rightarrow\mathbb{R}^n so that g(b)=a and f(g(t);t)=0.

That is, near (a;b) we can use the variables t^j as parameters on the space of solutions to our system of equations. Near this point, the solution set looks like the graph of the function x=g(t), which is implicitly defined by the need to stay on the solution set as we vary t. This is the implicit function theorem, and we will prove it next time.

November 19, 2009 - Posted by | Analysis, Calculus


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