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If $f\in C^1(\mathbb{R},M_n(\mathbb{R}))$ such that $f(0)=0$ and $f'(0)=I$, show that the image of f contains a regular matrix.

While trying to prove something (elementary) from representation theory, I came to a stop. This fact would complete the proof. Can anyone prove it or find a counterexample? Thanks!

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    @NateEldredge You supposed correctly, under "regular" I meant "invertible".2012-09-21

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Here I assume that "regular" means "invertible".

The fact that $f(0)=0$ and $f'(0)=I$ means that as $t \to 0$, $\frac{1}{t} f(t) \to I$. The set of invertible matrices is open and of course contains $I$. Therefore, for all sufficiently small nonzero $t$, $\frac{1}{t} f(t)$ is invertible, and thus so is $f(t)$.

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    Thank you very much! So simple, yet (to me) so elusive...2012-09-21