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Prove or Disprove:

Vector spaces $\mathbb{R^{N}}$ and $\mathbb{R^{N \times N}}$ are isomorphic.

It is generally known that two real vector spaces are isomorphic to one another if and only if they have the same dimensions (see also this question). This if and only if relationship holds even when their dimensions are infinite.

Vector spaces $\mathbb{R^{N}}$ and $\mathbb{R^{N \times N}}$ both appear to have dim $\mathbb{R^{N}}$ = $\infty$, and dim $\mathbb{R^{N \times N}}$ = $\infty$, which appears to suggest that both vector spaces are isomorphic to one another - that is, there exists an invertible, linear map between $\mathbb{R^{N}}$ and $\mathbb{R^{N \times N}}$.

But specific elements from both vector spaces, such as $\mathbb{R^{2}}$ and $\mathbb{R^{2 \times 2}}$, have different dimensions - dim $\mathbb{R^{2}}$ = 2 and dim $\mathbb{R^{2 \times 2}}$ = 4. This means $\nexists \;\; T : \mathbb{R^{2}} \longrightarrow \mathbb{R^{2 \times 2}}$ such that T is both linear and invertible.

Is the above line of reasoning correct?

Greg Martin
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user9487
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    The title and the body doesn't match. And it is not clear to me why $\Bbb R^\Bbb N$ and $\Bbb R^{\Bbb N \times \Bbb N}$ have different dimensions. Can you elaborate? – azif00 Apr 01 '24 at 02:33
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    Why do you think they have different dimensions? – jjagmath Apr 01 '24 at 02:40
  • @koala13, $\mathbb N \times \mathbb N$ is the Cartesian product, it's elements are pairs $(n,n')$. I think you are confusing this one with the multiplication $nn'$. – MAS Apr 01 '24 at 03:01
  • @azif00 Sorry for the careless error - I revised the question and added more elaboration. – user9487 Apr 01 '24 at 03:02
  • Two vector spaces over a field $k$ are isomorphic iff they have the same dimension; that is, iff they have bases with the same cardinality. Just being infinite dimensional is not sufficient; the real vectors with bases $\mathbb{Z}$ and $\mathbb{R}$ are not isomorphic, for example. – anomaly Apr 01 '24 at 03:26
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    OTOH, $\mathbb N$ and $\mathbb N \times \mathbb N$ are both countable, so there’s a bijection there which can be used to construct an isomorphism. – Eric Apr 01 '24 at 04:04
  • @Eric How would one do that exactly? For example, considering the example bijection provided here: https://math.stackexchange.com/questions/444447/bijection-between-mathbbn-and-mathbbn-times-mathbbn – user9487 Apr 01 '24 at 07:07
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    @koala13 Choose a basis of each space. Use the bijection you quoted to get a bijection of the bases. This extends to an isomorphism. – user Apr 01 '24 at 08:25

1 Answers1

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Your reasoning is almost correct. Indeed, vector spaces are isomorphic when they are characterized by the same dimension. In other words, their respective bases possess the same cardinality, which itself translates as the existence of bijection between those bases. The spaces $\Bbb{R}^\Bbb{N}$ and $\Bbb{R}^{\Bbb{N}\times\Bbb{N}}$ are isomorphic not because their dimensions are both infinite, but because the sets $\Bbb{N}$ and $\Bbb{N}\times\Bbb{N}$ are related by a bijection.

Concretely, let's call $\phi : \Bbb{N} \to \Bbb{N}\times\Bbb{N}$ the said bijection, whatever its form. Let's also name the basis of $\Bbb{R}^\Bbb{N}$ as $\{e_i\}_{i\in\Bbb{N}}$ and the one of $\Bbb{R}^{\Bbb{N}\times\Bbb{N}}$ as $\{\varepsilon_{(n,m)}\}_{(n,m)\in\Bbb{N}\times\Bbb{N}}$. Then, the linear map $\psi : \Bbb{R}^\Bbb{N} \to \Bbb{R}^{\Bbb{N}\times\Bbb{N}}$ such that $\psi(e_i) = \varepsilon_{\phi(i)}$ is an isomorphism.

Abezhiko
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