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I'm reading through Rudin's Functional Analysis text, where, as far as I can tell, the fact that "every linear map between finite-dimensional spaces is continuous" is not explicitly stated anywhere.

I was wondering if it could be proved using the following facts, which are proved in the text:

Lemma: If $X$ is a complex topological vector space and $f: \mathbf{C}^n \rightarrow X$ is linear then $f$ continuous.

Theorem: If $n$ is positive and $Y$ is an $n$-dimensional subspace of a complex topological vector space $X$, then

a) every isomorphism of $\mathbf{C}^n$ into $Y$ is a homeomorphism

b) $Y$ is closed

Can someone shed some light on this? The fact I'm asking about seems pretty fundamental, so I'm surprised that I haven't seen it stated in the text. Am I missing something?

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    $\begingroup$ By part a) of the theorem, every [real or complex] (Hausdorff) topological vector space of finite dimension is isomorphic to $\mathbb{C}^n$ (or $\mathbb{R}^n$) as a topological vector space. Thus it's just the observation that every linear map $\mathbb{C}^n \to \mathbb{C}^m$ is continuous. Which is easy to see from the matrix representation. $\endgroup$ – Daniel Fischer Jan 11 '18 at 19:27
  • $\begingroup$ "The fact I'm asking about seems pretty fundamental, so I'm surprised that I haven't seen it stated in the text." Rudin is probably assuming you already know how to prove linear maps are continuous using undergraduate analysis arguments. $\endgroup$ – fourierwho Jan 11 '18 at 20:42
  • $\begingroup$ I gave an answer using the lemma and the theorem. To be honest, though, I can't figure out what you mean by a "complete topological vector space." Is this the exact wording in Rudin? Do you mean "complex topological vector space?" $\endgroup$ – fourierwho Jan 11 '18 at 20:59
  • $\begingroup$ Yes, that should be complex, not complete. fixed now. $\endgroup$ – theQman Jan 12 '18 at 0:54
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Suppose $T : \mathbb{C}^{n} \to \mathbb{C}^{k}$ is linear. Recall that $\mathbb{C}^{k}$ is a Banach space: thus, it's a complex topological vector space that is a complete metric space. Therefore, by the lemma, $T$ is continuous. Note that this is also clear by the triangle inequality since $$|T(x)| \leq |x_{1}| |T(e_{1})| + \dots + |x_{n}| |T(e_{n})| \leq \max\{|T(e_{1})|,\dots,|T(e_{n})|\} \sqrt{n} \|x\|$$ if $\|x\| = \left(|x_{1}|^{2} + \dots + |x_{n}|^{2}\right)^{\frac{1}{2}}$.

In the general case, if $E$ and $F$ are finite dimensional complex TVS and $T : E \to F$ is linear, let $\Phi : \mathbb{C}^{n} \to E$ and $\Psi : \mathbb{C}^{k} \to F$ be any linear isomorphisms. (These exist by the assumption about finite dimension.) Now $\Psi^{-1} \circ T \circ \Phi : \mathbb{C}^{n} \to \mathbb{C}^{k}$ is linear. By the previous paragraph, it is continuous. By the theorem, $\Phi$ and $\Psi$ are homeomorphisms. Therefore, $\Phi^{-1}$ and $\Psi$ are both continuous. By composition, $T = \Psi \circ (\Psi^{-1} \circ T \circ \Phi) \circ \Phi^{-1}$ is continuous.

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