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I am trying to figure out why a result invoked in a proof might be proved.

Basically, the following equation holds \begin{equation*} \boldsymbol{c}\left(\boldsymbol{\theta}\right)'\mathbf{M}\left(\boldsymbol{\theta}\right)\boldsymbol{c}\left(\boldsymbol{\theta}\right)=0,\; \boldsymbol{\theta}\in \mathcal{O}_{\boldsymbol{\theta}_0} \end{equation*} where $\mathbf{M}:\mathbb{R}^p\longmapsto \mathbb{R}^{p \times p}$ is a matrix-valued function, $\boldsymbol{c}:\mathbb{R}^p \longmapsto \mathbb{R}^p$ is a vector-valued function, and the equation holds in an open neighbourhood of $\boldsymbol{\theta}_0$, where, also, importantly, the rank of the matrix map is constant in the open neighbourhood of $\boldsymbol{\theta}_0$ and the matrix is singular at $\boldsymbol{\theta}_0$.

The claim made is that if the matrix-map is continuous in the open neighbourhood, then the vector-valued map, $\boldsymbol{c}$, can be chosen to be continuous in the same open neighbourhood. Which then leads to an inverse function with the usual properties - $d\boldsymbol{\theta}(t)/dt=\boldsymbol{c}\left(\boldsymbol{\theta}\right)$ and $\boldsymbol{\theta}(0)=\boldsymbol{\theta}_0$.

This seems intuitive but I can't prove it. Any hints as to a proof or a textbook treatment will be appreciated. Thanks.

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It seems like something is a little off with your question. Do you possibly mean $M:\mathbb{R} \rightarrow \mathbb{R}^{p \times p}$ (i.e. $\theta \in \mathbb{R}$ -- not a vector)? Also, what conditions exactly are you looking to place on the map ${\bf c}$? If you're not placing any conditions on ${\bf c}$ other than it satisfies the matrix equation, then ${\bf c}(\theta)=0$ works. Finally, ${\bf c}(\theta)'$ means ${\bf c}(\theta)$ transpose, correct? – Bill Cook Oct 2 '11 at 18:49
I do mean matrix valued functions of vector parameters. My particular application is statistical theory, so the information matrix of a finite dimensional family is a specific example. The example that you give ($\boldsymbol{c}=\mathbf{0}$) is does not admit an inverse function, so is not valid. Basically, $\mathbf{c}$ needs to be continuous (a fortiori invertible). – Praetoria Oct 2 '11 at 19:48
Also, yes, prime ($'$) denotes transpose. – Praetoria Oct 2 '11 at 19:51

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