# Math Help - Implicit function theorem: what if for a given x_i=x^0_i, there are more y_j=y^k_j?

1. ## Implicit function theorem: what if for a given x_i=x^0_i, there are more y_j=y^k_j?

In the case of m+n variables we have n equations $F_i(x_1,...,x_m;y_1,...y_n)=0$ which must be satisfied for $x_i=x^0_i, y_j=y^0_j$ for the theorem to be applicable, althought (I understand that) there might be more than one solutions. But if for a given $x_i=x^0_i$ there are more than one $y_j=y^k_j$ satisfying the n equations (and I do not find any statement in the theorem preventing this) how can the theorem ensure that there is one and only one set of solutions $y_j=f_j(x_1,...,x_m)$ which are continuous, satisfy $F_j=0$ and for which $y^0_j=f_j(x^0_1,...,x^0_m)$ ? I mean how can f be a function?...Of course I must be missing something...

thanks a lot

2. ## Re: Implicit function theorem: what if for a given x_i=x^0_i, there are more y_j=y^k_

let's look at a very simple example: when m = n = 1. define:

F(x,y) = x2+y2 - 1.

consider now (a,b) such that F(a,b) = 0. suppose that a = √2/2, for example. well, we have two choices for b, both of which work: b = √2/2, and b = -√2/2.

if we use (a,b) = (√2/2,√2/2) we get the function (defined only in a neighborhood of √2/2 in R) f(x) = √(1 - x2).

if we use (a,b) = (√2/2,-√2/2), we get the function f(x) = -√(1 - x2).

if we try to define f in a neighborhood of a = 1, we have a problem: neither half of the circle works. why is this so? let's look at the Jacobian matrix DF:

$DF(a,b) = \begin{bmatrix} \frac{\partial F}{\partial x}(a,b)&\frac{\partial F}{\partial y}(a,b) \end{bmatrix} = [2a\ 2b]$

at the point (1,0), the square (1x1) matrix 2b is not invertible, we have b = 0, and the inverse of the 1x1 matrix [2b] does not exist, since det([2b]) = 2b = 2(0) = 0.

this is the important part you're missing about the implicit function theorem, there is a condition on how F is changing with respect to the y-variables, the nxn matrix (∂Fm+i/∂yj) MUST be invertible.

the second key point is "which solution" of F(x1,....,xm,y1,....,yn) = 0 we choose (and there could be several), depends on which point:

(a1,...,am,b1,....,bn) we're at, the implicitly defined function f is only defined LOCALLY (it may be impossible to find a function f that works everywhere).

here is a more "grown-up example":

let F(x,y,z,w) = (x2+y2-z2-w2,x2+y2+z+w) this is a function from R4 to R2.

clearly F(1,1,-1,-1) = (0,0). let's check the following matrix determinant:

$\begin{vmatrix} \dfrac{\partial F_1}{\partial z}(1,1,-1,1)& \dfrac{\partial F_1}{\partial w}(1,1,-1,-1)\\ \ &\ \\ \dfrac{\partial F_2}{\partial z}(1,1,-1,-1)& \dfrac{\partial F_2}{\partial w}(1,1,-1,-1) \end{vmatrix}$

$= \begin{vmatrix} -2z&-2w\\1&1 \end{vmatrix}_{(1,1,-1,-1)} = \begin{vmatrix}2&2\\1&1 \end{vmatrix} = 0$

this tells us that (1,1,-1,-1) is a "bad point" to try to implicitly define a function f with F(x,y,f(x,y)) = (0,0).

on the other hand, we also have F(√2/2,√2/2,-1,0) = (0,0) and for this point the matrix above is:

$\begin{bmatrix}2&0\\1&1 \end{bmatrix}$, which has non-zero determinant, so this is a "good point".