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Theorem: Two finite-dimensional vector spaces over $F$ are isomorphic if and only if they have the same dimension.

$F$ denotes either $\mathbb{R}$ or $\mathbb{C}$

Below is proof of the theorem. Where in the proof do we assume that the vector spaces are over $F$? It seems to me like that assumption is not used. Can the theorem be restated as: "Two finite-dimensional vector spaces over the same field are isomorphic if and only if they have the same dimension"?

Proof of theorem:

jenny9
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    Exactly, this works over any field – Riemann'sPointyNose Sep 05 '22 at 21:14
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    Some authors unnecessarily limit the field to be either the reals or the complex numbers. – John Douma Sep 05 '22 at 21:17
  • Does it not really make sense to talk about two vector spaces being isomorphic when each vector space is over a different field? Like if vector space one is over field A and vector space two if over field B, this theorem and related theorems don't really apply? – jenny9 Sep 05 '22 at 21:25
  • Yes Jenny, it does not make sense for vector spaces over two different fields. Btw, I find it ugly to use the rank–nullity theorem in the first part of the proof. Better say that an isomorphism $V\to W$ sends any basis of $V$ to a basis of $W$. – Anne Bauval Sep 05 '22 at 21:41
  • @jenny9 see https://math.stackexchange.com/questions/1192347/mapping-vector-spaces-over-two-different-fields – utobi Sep 05 '22 at 21:44

2 Answers2

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On page $10$ of the book the author states that you can think of $F$ denoting arbitrary fields throughout chapter $1,2$ and $3$.

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Seeker
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As suggested in the comments, the assumption of the two vector spaces $V,W$ being defined over the same field $F$, whatever is $F$, is actually relevant. In fact, when we say

there is a linear mapping $T:V\to W$ then bla, bla, bla...

this is meaningful only if $V$ and $W$ are built over the same field of scalars.

utobi
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