# I Rank of the Jacobian matrix

1. Oct 17, 2016

### spaghetti3451

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. Oct 17, 2016

### andrewkirk

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.

3. Oct 17, 2016

### spaghetti3451

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.