I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
The following proposition is in this book:
3.59 Dimension shows whether vector spaces are isomorphic
Two finite-dimensional vector spaces over $\mathbb{F}$ are isomorphic if and only if they have the same dimension.
The author proved if two finite-dimensional vector spaces over $\mathbb{F}$ are isomorphic, then they have the same dimension.
Two isomorphic vector spaces have the same algebraic structure.
Do we really need to prove if two finite-dimensional vector spaces over $\mathbb{F}$ are isomorphic, then they have the same dimension?
If we consider two vector spaces $U$ and $V$ are the same vector space (we consider two canonically isomorphic vector spaces are the same vector space), then any property that holds for $U$ also must hold for $V$ because $U=V$.
If we consider two vector spaces $U$ and $V$ are not the same vector space, but they are isomorphic, then I think any algebraic property that holds for $U$ also holds for $V$ because $U$ and $V$ have the same algebraic structure.
When we can declare that we consider $U$ and $V$ are the same vector space, then we don't need to prove that any property that holds for $U$ also holds for $V$.
But when we don't consider $U$ and $V$ are the same vector space, but they are isomorphic, then do we need to prove that any algebraic property that holds for $U$ also holds for $V$?
Best Answer
Like you my instinct was that this was very obvious and that a proof was scarcely necessary.
However, on reflection, perhaps something does need to be said. I would interpret "have the same algebraic structure" to mean that there is a bijection between the elements of the underlying sets which preserves the algebraic operations, in this case the $0$ element, the addition $+$, the negation$-$, and the scalar multiplications $\lambda\cdot$. There's no mention of "basis" here, I am not sure that it is part of the "algebraic structure" strictly understood. So I'd maybe feel obliged to prove that the bijection carries bases to bases - but it's very obvious.