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Statement

Let be $\Bbb{M}_{m\times n}(\Bbb{R})$ the set of the matrices of order $m\times n$ with coefficients in $\Bbb{R}$. Let be $\text{Hom}(\Bbb{R}^n,\Bbb{R}^m)$ the vector space of the linear transformation between $\Bbb{R}^n$ and $\Bbb{R}^m$. So this two space are homeomorphic to $\Bbb{R}^{m\cdot n}$.

First of all we structure the set $\Bbb{M}_{m\times n}(\Bbb{R})$ to metric space and so in particolar we define a distance $|\cdot|:\Bbb{M}_{m\times n}(\Bbb{R})\rightarrow\Bbb{R}$ through the position $$ |A|=\text{max}\{a_{i,j}\in A:i=1,...,m\wedge j=1,...,n\} $$ for any $A\in\Bbb{M}_{m\times n}(\Bbb{R})$. So now we prove that $\Bbb{M}_{m\times n}(\Bbb{R})$ is homeomorphic to $\Bbb{R}^{m\cdot n}$. For starters we remember that it is possible to define in $\Bbb{R}^k$ an analogous distance to that we define above in matrix space and this distance is equivalent to euclidean distace and so the corresponding topologies are equivalent too: here a better explanation. Now we consider the canonical isomorphism $\phi$ between $\Bbb{M}_{m\times n}(\Bbb{R})$ and $\Bbb{R}^{m\cdot n}$, that is $$ \phi(A)=(a_{_{1,1}},...,a_{_{m,1}},...,a_{_{1,n}},...,a_{_{m,n}}) $$ for any $A\in\Bbb{M}_{m\times n}(\Bbb{R})$ and we prove that it is an homeomorphism too. Now we observe that obviously $$ |A|=|\phi(A)| $$ for any $A\in\Bbb{M}_{m\times n}(\Bbb{R})$ and so if $B\in\Bbb{M}_{m\times n}(\Bbb{R})$ is such that $|B-A|<\epsilon$ for some $\epsilon>0$ then by linearity of $\phi$ it follows that $$ |\phi(B)-\phi(A)|=|\phi(B-A)|=|B-A|<\epsilon $$ and so we conclude that $\phi$ is continuous and open too: indeed for any cube $\mathcal{C}(\phi(A),\epsilon)$ we can consider the cube $\mathcal{C'}(A,\epsilon)$ such that $\phi[\mathcal{C'(A,\epsilon)}]\subseteq\mathcal{C}(\phi(A),\epsilon)$ and so $\phi$ is continuous; then for any $x\in\mathcal{C}(\phi(A),\epsilon)$ by linearity of $\phi^{-1}$ we observe that $$ |\phi^{-1}(x)-A|=|\phi^{-1}(x)-\phi^{-1}(\phi(A))|=|\phi^{-1}\big(x-\phi(A)\big)|=\Big|\phi\Big(\phi^{-1}\big(x-\phi(A)\big)\Big)\Big|=|x-\phi(A)|<\epsilon$$ and so $\phi^{-1}(x)\in\mathcal{C'}(A,\epsilon)$ and so $\phi[\mathcal{C'(A,\epsilon)}]=\mathcal{C}(\phi(A),\epsilon)$. So we conclude that $\phi$ is an homeomorphism. Now we prove that $\Bbb{M}_{m\times n}(\Bbb{R})$ and $\text{Hom}(\Bbb{R}^n,\Bbb{R}^m)$ are homeomorphic but to do this we have to define a topology in $\text{Hom}(\Bbb{R}^n,\Bbb{R}^m)$ and so we remember that these two spaces are isomorphic through the canonical isomorphism $\psi$ that sed any linear transformation $f$ in the corresponding matrix $M_f$, provided that we have chosen a basis in $\Bbb{R}^n$ and in $\Bbb{R}^m$: so if we consider the initial topology corresponding to $\psi$ we have proved the statemet, since by defition of initial topology $\psi$ is continuous and then open too, since it is a bjiection.

So I ask if what I have wrote is correct: in particular I ask if $\phi$ and $\psi$ are actually the canonical isomorphism and if the proof is correct in any its step. So could someone help me, please?

Antonio Maria Di Mauro
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  • Your proof idea is correct, but it can definitely be simplified if you appeal to the theorem about equivalence of norms on finite-dimensional vector spaces. Anyway, if you don't want to appeal to stronger theorems, note that essentially your argument boils down to the following: $\phi$ and $\phi^{-1}$ are norm-preserving linear maps, hence they are Lipschitz, and therefore continuous. – peek-a-boo May 12 '20 at 13:16
  • @peek-a-boo Could you enunciate the theorem about the equivalence of norms on finite-dimensional vector spaces? Unfortunately I don't know it. – Antonio Maria Di Mauro May 12 '20 at 13:23
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    Every finite-dimensional vector space can be given a norm (this is easy to prove using a basis). Every norm on a vector space gives rise to a topology (also simple to prove). The theorem is that every norm on a finite-dimensional vector space gives rise to the same topology (this is a classic result, and there are several proofs available online, and several questions about this on the site). Also, it is easy to deduce from this that every linear transformation between finite-dimensional normed vector spaces is continuous (it is also Lipschitz-continuous, or more commonly called "bounded"). – peek-a-boo May 12 '20 at 13:26
  • @peek-a-boo Okay, could you indicate to me a book where I can find the proof? – Antonio Maria Di Mauro May 12 '20 at 13:29
  • @ Take a look at Henri Cartan's Differential Calculus, Section 1.6. This text might be slightly difficult to get hold of, but this is a very standard result so it shouldn't be too hard to find a proof of it online. Another reference is Loomis and Sternberg's Advanced Calculus book (available freely online): Take a look at Theorem 4.6 (page 208) and Theorem 4.2 (page 133) – peek-a-boo May 12 '20 at 13:37

1 Answers1

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This can be proven much easier:

Use the following two theorems:

(i) If $\dim V = \dim W < \infty$, then $V \cong W$ as vector spaces.

(ii) A linear map between finite dimensional normed vector spaces is automatically continuous.

Corollary: Two finite dimensional normed spaces of the same dimension are automatically homeomorphic.

J. De Ro
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  • Okay, however I don't understand the step ii. Could you explain it better, please? – Antonio Maria Di Mauro May 12 '20 at 15:44
  • Then I don't know if $\text{Hom}(V,W)$ is a normed space when $V$ and $W$ are normed space. – Antonio Maria Di Mauro May 12 '20 at 15:45
  • How do you consider $Hom(V,W)$ as a topological space/metric space? – J. De Ro May 12 '20 at 16:11
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    For your other question, see https://math.stackexchange.com/questions/112985/every-linear-mapping-on-a-finite-dimensional-space-is-continuous – J. De Ro May 12 '20 at 16:12
  • As you can see in the question I don't know if $\text{Hom}(V,W)$ is a topological space or metric space but I discover that if $\Bbb{M}{m\times n}(\Bbb{R})$ is a metric space then through initial topology of the canonical isomorphism between $\text{Hom}(V,W)$ and $\Bbb{M}{m\times n}(\Bbb{R})$ we can structure $\text{Hom}(V,W)$ as a topological space. – Antonio Maria Di Mauro May 12 '20 at 16:16
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    One possible way to make $Hom(V,W)$ into a normed space is via $\Vert T \Vert := \sup {\Vert Tv \Vert : v \in V, \Vert v \Vert \leq 1}$, there are many other ways though. As you said, you can just fix bases for $V,W$ and use the canonical isomorphism to lift a norm of the matrices to a norm on $Hom(V,W)$. – J. De Ro May 12 '20 at 16:22