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Assume $$g: R^n \times R^m \rightarrow R^n$$ $$h: R^n \times R^m \rightarrow R$$ $$(x,y) \in R^n \times R^m$$

I would like to show that the following vectors are linearly independent: \begin{equation} \left[ \begin{array}{c} \frac{\partial g_1}{\partial x_1}\\ \vdots\\ \frac{\partial g_1}{\partial x_n}\\ \frac{\partial g_1}{\partial y_1}\\ \vdots\\ \frac{\partial g_1}{\partial y_m}\\ \end{array} \right] \dots \left[ \begin{array}{c} \frac{\partial g_n}{\partial x_1}\\ \vdots\\ \frac{\partial g_n}{\partial x_n}\\ \frac{\partial g_n}{\partial y_1}\\ \vdots\\ \frac{\partial g_n}{\partial y_m}\\ \end{array} \right] \left[ \begin{array}{c} \frac{\partial h}{\partial x_1}\\ \vdots\\ \frac{\partial h}{\partial x_n}\\ \frac{\partial h}{\partial y_1}\\ \vdots\\ \frac{\partial h}{\partial y_m}\\ \end{array} \right] \end{equation}

I know, that the matrix with the columns \begin{equation} \left[ \begin{array}{c} \frac{\partial g_1}{\partial x_1}\\ \vdots\\ \frac{\partial g_1}{\partial x_n} \end{array} \right] \dots \left[ \begin{array}{c} \frac{\partial g_n}{\partial x_1}\\ \vdots\\ \frac{\partial g_n}{\partial x_n} \end{array} \right] \end{equation} already is invertible. Because of that, I know that the first $n$ vectors are linearly independent.

To show that all $n+1$ vectors are linearly independent, I have to show that \begin{equation} \left[ \begin{array}{c} \frac{\partial g_1}{\partial y_1}\\ \vdots\\ \frac{\partial g_1}{\partial y_m}\\ \end{array} \right] \dots \left[ \begin{array}{c} \frac{\partial g_n}{\partial y_1}\\ \vdots\\ \frac{\partial g_n}{\partial y_m}\\ \end{array} \right] \left[ \begin{array}{c} \frac{\partial h}{\partial y_1}\\ \vdots\\ \frac{\partial h}{\partial y_m}\\ \end{array} \right] \end{equation}

also is linearly independent, right?

This is equivalent to $\nabla_y h= g_y^\top a$ with $a \in R^n$ not having any solutions. How can I find conditions for which this is true?

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  • $\begingroup$ "To show that all $n+1$ vectors are linearly independent, I have to show that ... right?" No. The statement you made is not necessary. It is sufficient. But if $m \leq n$ the condition you wrote is impossible, but it is possible that the overall linear independent is true. $\endgroup$ – Willie Wong Dec 2 '13 at 16:11
  • $\begingroup$ To give a simple example, let $n = m = 1$ and let $g(x,y) = x + y$ and $h(x,y) = y$. $\endgroup$ – Willie Wong Dec 2 '13 at 16:12
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Given that the vectors $\frac{\partial g_k}{\partial x_i}$ for fixed $k$ are linearly independent, there exists a unique set of coefficients $a_1, \ldots a_n$ such that

$$ \begin{bmatrix} \partial_{x_1} h \\ \vdots \\ \partial_{x_n} h\end{bmatrix} = a_1 \begin{bmatrix} \partial_{x_1} g_1 \\ \vdots \\ \partial_{x_n} g_1\end{bmatrix} + \cdots + a_n \begin{bmatrix} \partial_{x_1} g_n \\ \vdots \\ \partial_{x_n} g_n \end{bmatrix} $$

The necessary and sufficient condition for your set of $n+1$ vectors to be linearly independent is then

$$ \begin{bmatrix} \partial_{y_1} h \\ \vdots \\ \partial_{y_m} h\end{bmatrix} \color{red}{\neq} a_1 \begin{bmatrix} \partial_{y_1} g_1 \\ \vdots \\ \partial_{y_m} g_1\end{bmatrix} + \cdots + a_n \begin{bmatrix} \partial_{y_1} g_n \\ \vdots \\ \partial_{y_m} g_n \end{bmatrix} $$

This is much less restrictive then what you have written.

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