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

\displaystyle\begin{pmatrix}1&0&0&2&1\\{0}&1&0&3&0\\{0}&0&1&-1&1\end{pmatrix}

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

\displaystyle\begin{pmatrix}1&0&0&2&1\\{0}&1&0&3&0\\{0}&0&1&-1&1\end{pmatrix}\begin{pmatrix}x^1\\x^2\\x^3\\t^1\\t^2\end{pmatrix}=\begin{pmatrix}x^1+2t^1+t^2\\x^2+3t^1\\x^3-t^1+t^2\end{pmatrix}=\begin{pmatrix}0\\{0}\\{0}\end{pmatrix}

and from this we find

\displaystyle\begin{aligned}x^1&=-2t^1-t^2\\x^2&=-3t^1\\x^3&=t^1-t^2\end{aligned}

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.

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November 19, 2009 - Posted by | Analysis, Calculus

4 Comments »

  1. [...] Implicit Function Theorem II Okay, today we’re going to prove the implicit function theorem. We’re going to think of our function as taking an -dimensional vector and a -dimensional [...]

    Pingback by The Implicit Function Theorem II « The Unapologetic Mathematician | November 20, 2009 | Reply

  2. [...] extremely difficult. At least we do know that such a parameterization will often exist. Indeed, the implicit function theorem tells us that if we have continuously differentiable constraint functions whose zeroes describe a [...]

    Pingback by Extrema with Constraints I « The Unapologetic Mathematician | November 25, 2009 | Reply

  3. [...] this, we’ll write , so we can write the point as and particularly . Now we can invoke the implicit function theorem! We find an -dimensional neighborhood of and a unique continuously differentiable function so [...]

    Pingback by Extrema with Constraints II « The Unapologetic Mathematician | November 27, 2009 | Reply

  4. [...] can also recall the implicit function theorem. This is less directly generalizable to manifolds, since talking about a function is effectively [...]

    Pingback by The Implicit Function Theorem « The Unapologetic Mathematician | April 15, 2011 | Reply


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