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Finite-dimensional vector spaces, Paul R. Halmos, reprint of 2nd edition, paragraph 9, "Isomorphism":

Definition. Two vector spaces $\cal{U}$ and $\cal{V}$ (over the same field) are isomorphic if there is a one-to-one correspondence between the vectors $x$ of $\cal{U}$ and the vectors $y$ of $\cal{V}$, say $y = T(x)$, such that

$$ T(\alpha_1 x_1 + \alpha_2 x_2) = \alpha_1 T(x_1) + \alpha_2 T(x_2). $$

It is easy to see that isomorphic finite-dimensional vector spaces have the same dimension; to each basis in one space there corresponds a basis in the other space.

But what if I define two vector spaces $\cal{U} = \{(1, 0), \; (0, 1)\}$, and $\cal{V} = \{(1, 0, 0), \; (2, 0, 0)\}$ and a bijection $((1, 0) \mapsto (1, 0, 0), (0, 1) \mapsto (2, 0, 0))$. Then $\cal{U}$ has two elements in its basis, while $\cal{V}$ has one.

Sorry for my misconception of vector spaces. I leave it as a warning for others. In case you are interested in a very well formulated proof, read this post of bfff.

Max Herrmann
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    First, your $U$ and $V$ are not vector spaces, but potential bases (assuming based on how you wrote them). Secondly, the vectors in $V$ are not independent. Thirdly, your map is not an isomorphism. – Randall Sep 05 '21 at 13:18
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    This is usually proved like this: suppose $\varphi:V\to W$ is an isomorphism of finite dimensional vector spaces. Let $B$ be a basis of $V$. Then show $\varphi(B)$ is a basis for $W$. Since $\varphi$ is a bijection, the size of the basis is preserved, so their dimensions are the same. – morrowmh Sep 05 '21 at 13:20
  • @Randall: I understand that $\cal{U}$ and $\cal{V}$ are no vector spaces, since $x+y \notin \cal{U}, \cal{V}$. And I understand that the map is not an isomorphism, since it is no bijection. But what's the problem with the elements in $V$ being linearly dependent? – Max Herrmann Sep 06 '21 at 08:39
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    @MaxHerrmann Because I was inferring that you were treating $\mathcal{U}$ and $\mathcal{V}$ as bases, but $\mathcal{V}$ cannot be since it has linearly dependent vectors. The bigger mistake was treating them as vector spaces. – Randall Sep 07 '21 at 13:43

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The answer to the question in the title is somewhat philosophical.

The dimension of a vector space is defined as the the number of elements in a basis. That makes sense as a definition only after you have proved directly from the vector space axioms that any two bases have the same number of elements.

When two vector spaces are isomorphic - that is, when there is a linear bijection between them - then that bijection translates any theorem proved in one of the spaces directly into the same theorem in the other. Since the dimension of a vector space is determined by reasoning from the vector space axioms it must be the same for isomorphic spaces. (See What is an Homomorphism/Isomorphism "Saying"?)

The particular problem with you "example" is that it does not define a bijection between the spans of $\{(1,0),(0,1)\}$ and $\{(1,0,0),(2,0,0)\}$. The vectors $2(1,0)$ and $(0,1)$ both map to the same thing.

Ethan Bolker
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If what you mean by $\mathcal{U}$ and $\mathcal{V}$ is the vector space over $\mathbb{R}$ created by spanning off of those sets, then $\mathcal{U} = \mathbb{R}^2$ while $\mathcal{V}$ is the subspace of all elements $(x,0,0)$ in $\mathbb{R}^3$. These two vector spaces are not isomorphic: the first has dimenson $2$, the second dimension $1$.

The map $f$ you gave is therefore not capable of being an isomorphism. Suppose that $f$ was an isomorphism, so that it is linear. For one, $f(0, \frac{1}{2}) = f(\frac{1}{2}(0,1))$ is the same as $\frac{1}{2}f(0,1)=(1,0,0)=f(1,0)$ and so $f$ isn't one-to-one. So, your counterexample is not a counterexample.

Randall
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