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Let $\Omega$ be an open set in $\mathbb{R}^n,$ and $f:\Omega\rightarrow \mathbb{R}^n $ be in $C^{1}(\Omega)$.

Why:

If $\forall x\in\Omega$, we have det $D_{f}(x)\ne0$, then $[D_{f}(x)]^{-1}$ is a continuous function of $x$ on $\Omega$.

The linear mapping $D_{f}:\Omega \rightarrow Hom(\mathbb{R^{n}},\mathbb{R^{n}})$ is a continuous on $\Omega$. What does mean to say that $D_{f}:\Omega \rightarrow Hom(\mathbb{R^{n}},\mathbb{R^{n}}) $ is continuous?

Elliot
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1 Answers1

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  1. The maps $D_f$ and $A\rightarrow A^{-1}$ (taking inverses) are both continuous (see this M.SE question for a proof of the latter fact). So $[D_f(x)]^{-1}$ is continuous because it is a composition of continuous maps.

  2. The set $\operatorname{Hom}(\mathbb R^n, \mathbb R^n)$ is just the set of $n\times n$ matrices, which can be viewed in a metric space. Think of it as being equivalent to $\mathbb R^{n\times n}$, for example. Then we say $D_f$ is continuous if it satisfies the usual definition for continuous maps between metric spaces.

Community
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Potato
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  • I am puzzled that why $A\rightarrow A^{-1}$ is continuous ? I need some clarifying explaination about this .Thanks! – Elliot Apr 30 '15 at 06:02
  • @Teddy See this question: http://math.stackexchange.com/questions/810670/continuity-of-the-inverse-matrix-function – Potato Apr 30 '15 at 06:03