# Implicit function theorem for several complex variables

1. Sep 8, 2013

### Kalidor

This is the statement, in case you're not familiar with it.
Let $f_j(w,x), \; j=1, \ldots, m$ be analytic functions of $(w,z) = (w_1, \ldots, w_m,z_1,\ldots,z_n)$ in a neighborhood of $w^0,z^0$ in $\mathbb{C}^m \times \mathbb{C}^n$ and assume that $f_j(w^0,z^0)=0, \, j=1,\ldots,m$ and that $$\det \{\frac{\partial f_j}{\partial w_k}\}^m_{j,k=1} \neq 0$$
at $(w^0,z^0)$.
Then the equations $f_j(w,z)=0 \; j=1,\ldots,m$, have a uniquely determined analytic solution $w(z)$ in a neighborhood of $z_0$, such that $w(z_0) = w_0$.
In the proof of this statement I find in Hormander's book he claims that in order to apply the usual implicit function theorem one must first prove that the equations $df_j = 0$ and $dz_k=0$ for $j =1, \ldots, m$ and $k = 1, \ldots, n$ imply $dw_j = 0$ for $j = 1, \ldots, m$. I don't understand what this condition means and why it is needed.

Last edited: Sep 8, 2013
2. Sep 10, 2013

### Kalidor

I can't edit anymore, but of course the x in in $f_j(w,x)$ is a typo. It should read $f_j(w,z).$

3. Oct 22, 2013

### xaos

you have an Mx(M+N) system. if the last N columns are all zero, then the first M columns are linearly independent if the first M rows are. you need an invertible MxM submatrix to solve for an M-vector of coefficients from this system. this an intermediate step and not necessary if one already knows that a nonzero jacobian determinant implies invertibility. the jacobian submatrix is not invertible if its columns are not linearly independent. hopefully this isn't too abstract.