Implicit function theorem for several complex variables

  • Thread starter Kalidor
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  • #1
Kalidor
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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 [tex] \det \{\frac{\partial f_j}{\partial w_k}\}^m_{j,k=1} \neq 0
[/tex]
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.
 
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Answers and Replies

  • #2
Kalidor
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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
xaos
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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.
 

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