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

Question

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!

@NateEldredge You supposed correctly, under "regular" I meant "invertible". — sonjcy, Sep 21 '12 at 20:50

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

4

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 $f(t)$.

edited Apr 13 '17 at 12:20

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answered Sep 21 '12 at 16:59

Nate Eldredge

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Thank you very much! So simple, yet (to me) so elusive... – sonjcy Sep 21 '12 at 20:53

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

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