# The Unapologetic Mathematician

## 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.

November 19, 2009 Posted by | Analysis, Calculus | 4 Comments