What are the implications of the fact that vector spaces are isomorphic?

Question

Two theorems in my linear algebra textbook state that if we have a basis $B=\{ b_1,...,b_n\} $ of vector space $V$, then there exists a unique one-to-one linear coordinate mapping $x \implies[x]_B$ from $V$ onto $\Bbb R^n$.

It further states that such a coordinate mapping is an isomorphism, which implies that "vector space calculations in $V$ are accurately reproduced in $\Bbb R^n$". Now, I know the term isomorphism from mathematical logic, and I've learned about the so called "Isomorphism Lemma", which states:

Isomorphism Lemma. If mathematical structures $\mathfrak A$ and $\mathfrak B$ are isomorphic, then all first-order $S$-sentences are true in $\mathfrak A$ iff they are true in $\mathfrak B$.

Where a map $\pi:A\implies B$ is called an isomorphism from $S$-structures $\mathfrak A$ to $\mathfrak B$, if all first-order functions and relations in the language $S$ are preserved. For functions: for any $n$-parameter function $f$ in $S$, $\pi(f^{\mathfrak A}(a_1,...,a_n)=f^\mathfrak > B(\pi(a_1),...,\pi(a_n)).$

My question is: What can we actually conclude about vector spaces from this Lemma, given that there is an isomorphism for all vector spaces of dimension $n$?

Sure, all theorems that are derivable from vector space axioms hold for all vector spaces, but we didn't need an isomorphism to conclude that. What does the fact that there is an isomorphism from an n-dimensional vector space onto $\Bbb R ^n$ actually imply in for us? Surely it doesn't mean "everything" that is true in $\Bbb R^n$ is also true in $n$-dimensional vector space $V$?

I understand the Isomorphism Lemma in the abstract, but I don't yet concretely see what this implies for a practical example like vector spaces. Perhaps some examples would help.

Answer 1

One example on vector spaces. Let $U$ and $W$ two vector spaces with the same field $K$. Then a function

$$T:U\to W$$

is $K$-linear if

$$T(\lambda u+\mu v)=\lambda T(u)+\mu T(v)$$

for all $u,v\in U$ and $\lambda,\mu\in K$. This mean that $T$ is an homomorphism between the vector spaces. If $T$ is bijective then it is an isomorphism.

This mean that $T$ respect the operations of each vector space, the addition of vectors and the scalar multiplication.

In general an isomorphism is a bijective function between spaces that respect the common algebraic structure of both, for the case of vector spaces this structure is composed by the addition of vectors and the scalar multiplication.

The consequence of this is that the vectors spaces $U$ and $V$, if exists some bijective $T$ as described above, are equivalent. And for practical purposes are the same space respect to it use as vector space.

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I will add a concrete example that exemplify what mean "for practical purposes are the same".

We know that all $K$-vector spaces of dimension $n$ are isomorphic to $K^n$. This mean that exists some bijective map such that

$$f:V\to K^n,\quad v\mapsto \langle k_0,k_1,...,k_n\rangle$$

and

$$f(\lambda v+\mu u)=\lambda f(v)+\mu f(u),\quad \forall\lambda,\mu\in K,\forall u,v\in V$$

where $V$ is a $K$-vector space of dimension $n$ (then $f$ is called an isomorphism between $V$ and $K^n$). Then for any computation or problem where $V$ is involved as a vector space we can change $V$ by $K^n$, because they are basically the same thing at the level of vector spaces (they are "the same" vector space).

It is possible that $V$ is not only a vector space, it is possible that it had other algebraic structures, but if the problem where it is involved is using only it vector space operations then we can change it by $K^n$.

After solved the problem we can change again the solution based in $K^n$ to it image in $V$ because as $f$ is bijective then the inverse function $f^{-1}$ exists.