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I Rank of the Jacobian matrix

  1. Oct 17, 2016 #1
    Let the matrix of partial derivatives ##\displaystyle{\frac{\partial y^{j}}{\partial y^{i}}}## be a ##p \times p## matrix, but let the rank of this matrix be less than ##p##.

    Does this mean that some given element of this matrix, say ##\displaystyle{\frac{\partial y^{1}}{\partial u^{2}}}##, can be written as

    ##\displaystyle{\frac{\partial y^{1}}{\partial u^{2}}=A_{1k_{1}}M_{k_{1}2}}##,

    where ##A## is a ##p\times p## matrix of rank less than ##p## and ##M## is an arbitrary matrix?
     
  2. jcsd
  3. Oct 17, 2016 #2

    andrewkirk

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    The answer to that is trivially yes. Given the element ##\frac{\partial y^1}{\partial u^2}## we define the matrix ##A## to have all zero entries except for
    ##A_{1k_1}=\frac{\partial y^1}{\partial u^2}## and the matrix ##M## is all zeros except that ##M_{k_12}=1##.

    It sounds like you were trying to ask something different and less trivial, but it's not clear what that is.
     
  4. Oct 17, 2016 #3
    Indeed, I am trying to ask something different and less trivial.

    Consider the following expression:

    ##\displaystyle{\alpha_{j_{1}\dots j_{p}}\frac{\partial y^{j_{1}}}{\partial u^{i_{1}}} \dots \frac{\partial y^{j_{p}}}{\partial u^{i_{p}}}du^{i_{1}}\wedge \dots \wedge du^{i_{p}}}##.

    My goal is to show that this is zero, if the matrix of partial derivatives is of rank less than the dimension ##p## of the matrix. My approach is to try and rewrite every factor of partial derivatives in the form

    ##\displaystyle{\frac{\partial y^{j_{1}}}{\partial u^{i_{2}}}=A_{j_{1}k_{1}}M_{k_{1}i_{2}}}##

    and play around with indices.
     
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