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I am struggling through Rudin's proof of the rank theorem (9.32) in the baby Rudin book. There is a part in the proof where he claims that for a finite-dimensional linear operator A, if the set V is open, then A(V) is an open subset of the range of A. I have seem things about the open mapping theorem involving Banach spaces, but I am not on that level yet and I don't see why the justification of this statement could possibly involve Banach spaces, considering this book does not talk about those. How does Rudin justify this statement, at the level of this book?

Thanks!

nickodel
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2 Answers2

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Pick any $x_0 \in V$. We will show that $Ax_0 $ is an interior point of $A(V)$.

By translating (i.e. consider $V - x_0$ instead of $V$), we can assume $x_0 = 0$.

Let $y_1, \dots, y_n$ be a basis of $\rm{Range}(A)$ and choose $x_1, \dots, x_n$ with $y_i = Ax_i$ for each $i$.

As $V$ is open with $0 \in V$, there is some $\varepsilon > 0$ such that $\sum_i \alpha_i x_i \in V$ holds for all $\alpha_1, \dots, \alpha_n$ with $|\alpha_i| < \varepsilon$ for all $i$.

This implies that $A(V)$ contains the set

$$ \bigg\{ \sum_i \alpha_i y_i \mid |\alpha_1|, \dots, |\alpha_n| < \varepsilon\bigg\}. $$

Why does that imply your claim?

PhoemueX
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  • Thanks for the help! But I am still confused... I think I see what you are trying to explain to me, that A(V) contains the neighborhood at the point 0 with the radius epsilon. But are the x_i's a basis of V? And why is it that if you show A0 is an interior point of A(V), every other point Ax_0 is? – nickodel Aug 01 '14 at 14:14
  • The following shows why it is enough to show that $0 = A0$ is an interior point of $A(V)$. If you know that, you can just apply this to the set $W := V - x_0 = { v - x_0 \mid v \in V }$ instead of $V$, to see that $0$ is an interior point of $A(W) = A(V) - A x_0$. This shows that $A x_0$ is an interior point of $A(V)$ (why?). For your other question, the $x_i$ are not necessarily a basis of $V$. What is relevant is that the $y_i$ are a basis of $\rm{Range}(A)$. This implies that the "large" set at the end of my answer contains an open ball around $0$ (with respect to $\rm{Range}(A)$) (why?). – PhoemueX Aug 01 '14 at 14:41
  • ok! I think I'm with you now. Essentially what you have proven is that is a neighborhood of A0 in A(V), every point of which is a linear combination of vectors that are elements of A(V), and therefore also elements of A(V). Therefore, this neighborhood is contained in A(V). Yes? The fact that the y_i's are a basis implies the set you described in the last equation generates a neighborhood? Thanks again for the help, you saved me many hours of lost sleep. :) – nickodel Aug 01 '14 at 19:42
  • Yes, that is correct. Essentially, the claim that the "large" set is a unit neighbourhood in $A(V)$, because the $(y_i)_i$ form a basis of $A(V)$, boils down to equivalence of norms on finite dimensional spaces. One possible norm ist $\sum \alpha_i y_i \mapsto \max{|\alpha_i| \mid i =1, \dots, n}$ and with this norm, the "large" set is just the $\varepsilon$ ball around the origin. – PhoemueX Aug 01 '14 at 19:47
  • Thanks a lot. I was kind of messed up with the same problem and you saved me. – Henry Choi Apr 15 '20 at 14:32
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This appears to be true only when range of A have dimension larger than 0. Otherwise, $A(V)=0$, which is not open.

Daniel Li
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