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Let $X$ be a vector space over $\mathbb{R}$ with dimension $n.$ Then I could prove that there is a bijective linear transformation $\phi$ from $X$ onto $\mathbb{R}^n$ by choosing an ordered basis $\mathcal{B}=\{\alpha_1,...,\alpha_n\}$ and using the fact that the coordinate matrix with respect to $\mathcal{B}$ of any given vector $\alpha\in X$ is unique.

I do not understand the significance of this result. Does this mean that instead of studying a vector space of dimension $n$ we can study $\mathbb{R}^n$ given the corresponding fields considered are the same? I mean does a property of $\mathbb{R}^n$ hold in $X$ as well?

I know that in group theory isomorphic groups bear same properties. Can I see a bijective linear map between two vector spaces as an isomorphism between two algebraic structures? Maybe the word linear transformation is the cause of my confusion. If I can see a linear transformation as a homomorphism, I think my doubts are cleared. The thing is that I read a little bit about modules and they mentioned something called $R-$module homomorphisms and I am inclined to think that I can consider the linear transformation $\phi$ as an $\mathbb{R}-$module isomorphism and be free from my doubts. But can I really?

Thanks.

Cameron Williams
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Janitha357
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  • There's no difference between a linear map and a morphism in the category of vector spaces over a given field. It's also a module homomorphism in the case the base ring is a field. – Bernard Jul 30 '17 at 23:26

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In general, if $R$ is a ring, a (left) module over $R$ is an abelian group $M$ and an operation $\cdot : R \times M \to R$ that satisfies certain properties listed here. There is the definition of an $R$-module homomorphism over $R$-modules.

A vector space is just a module over a field (a commutative ring where every element has a multiplicative inverse), in this case over $\mathbb{R}.$ So a linear transformation over $\mathbb{R}$-vector spaces is just, as you said, an $\mathbb{R}$-module homomorphism, and if it is bijective then it is an $\mathbb{R}$-module isomorphism.

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positron0802
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  • So in analysis, can we restrict the study of $\mathbb{R}$-vector spaces of dimension $n$ to the study of $\mathbb{R}^n$? I'm asking because in my real analysis notes the aforementioned result is given as an exercise. – Janitha357 Jul 30 '17 at 23:40
  • Isomorphic vector spaces share algebraic properties, but in real analysis we study other concepts such are metric and distance, which a vector space may not have. – positron0802 Jul 30 '17 at 23:46
  • Conversely, many metric spaces do not have a vector space structure and we cannot even add or subtract points. – gary Jul 30 '17 at 23:58
  • Vector spaces with a norm i.e. normed vector spaces are studied in analysis right? True that a vector space alone may not have a topological structure but with a norm they do. – Janitha357 Jul 31 '17 at 01:23
  • You're right, a finite-dimensional vector space $V$ share most properties of $\mathbb{R}^n.$ For example, a subspace of $V$ is compact iff it is closed and bounded, all norms on $V$ are equivalent, all linear functions defined on $V$ are continuous... But you should also look at this question: https://math.stackexchange.com/questions/2072977/why-study-finite-dimensional-vector-spaces-in-the-abstract-if-they-are-all-isomo – positron0802 Aug 01 '17 at 04:29