I was thinking of this case: two vector spaces – $\mathbb{R}^2$ over $\mathbb{R}$ and $\mathbb{C}^2$ over $\mathbb{C}$. Every basis in the first vector space is also a basis for the second vector space.
I am only starting linear algebra (also with proofs), so not sure if it makes any sense, but I will just give it a shot.
Suppose we have two vector spaces $F^n$ over $F$ and $S^n$ over $S$,
then the vector space $S^n$ has the same basis as $F^n$ iff
$\dim{S^n}$ = $\dim{F^n}$ and $F\subseteq S$.
This is my idea of the proof:
$V= { \{\vec{v_1}, \vec{v_2}, ..,\vec{v_n}\} }$ is some basis of $F^n$ and by the definiton $\vec{u} = (c_1\vec{v_1} + c_2\vec{v_2} + …+c_n\vec{v_3})$, for every $\vec{u}\in F^n$ and $c_i \in F$. Because $F \subseteq S$, $\vec{u'} = (a_1\vec{v_1} + a_2\vec{v_2} + …+a_n\vec{v_3})$ for every $\vec{u'}\in S^n$ and $a_i\in F \in S $.
I apologize if I have made any logical mistakes.
Best Answer
Specifically for the case that the field $F$ is a subfield of the field $S$, and we're comparing $F^n$ to $S^n$, any basis of $F^n$ is also a basis of $S^n$.
To show this, it's enough to show that any basis of $F^n$ spans $S^n$. (Then it's a basis by a dimension argument.) We can do so by working through the standard basis $\{\vec e^1, \vec e^2, \dots, \vec e^n\}$: these vectors are contained in both $F^n$ and $S^n$ since both $F$ and $S$ contain $0$ and $1$.
But your argument is not valid: in particular, you're claiming that every element of $S^n$ is an $F$-linear combination of vectors in the basis of $F^n$, which is false. (Any element of $S^n$ that is not in $F^n$ is a counterexample.)
Also, you shouldn't say "$S^n$ has the same basis as $F^n$" because both $F^n$ and $S^n$ have many different bases. You're making it sound like there is only one basis, and it is the same for both.