Why must $A(V)$ be an open subset of $A$'s range?

Asked Aug 13 '15 at 12:41

Active Aug 13 '15 at 13:03

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Suppose $A$ is a linear transformation from $\Bbb R^n$ to $\Bbb R^m$ with $\text{rank}A=r>0$, $V$ is an open subset of $\Bbb R^n$, then is $A(V)$ an open (relative to $\text{Im}(A)$) subset of $\text{Im} (A)$?

I don't know how to prove this. I cannot construct a diffeomorphism from $V$ to $A(V)$ because there is a dimension shrinkage.

Can you help me with this? Thanks a lot!

edited Aug 13 '15 at 13:03

asked Aug 13 '15 at 12:41

Vim

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This is only going to be true if rank$(A) = m$. – Ben Grossmann Aug 13 '15 at 12:55
1

@Omnomnomnom That's it! Thank you very much. – Vim Aug 13 '15 at 13:13

Why must $A(V)$ be an open subset of $A$'s range?

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