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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.

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  • $\begingroup$ How do you claim that $a_i \in F$? $\endgroup$
    – little o
    Mar 19, 2019 at 20:04
  • $\begingroup$ Better argue by dimension. $\endgroup$
    – user251257
    Mar 19, 2019 at 20:05
  • $\begingroup$ Also, a Basis of $F^n$ is a Basis of $S^n$. But the converse is trivially wrong. $\endgroup$
    – user251257
    Mar 19, 2019 at 20:06
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    $\begingroup$ Vectors can be objects other than tuples of scalars. For instance, the set of polynomials in $x$ of degree $\le 2$ with real coefficients forms a vector space of dimension $3$, but it makes no sense for any basis of this space to be a basis of $\mathbb R^3$ since the elements of the first space are polynomials, not triples of real numbers. Less abstrusely, the $x$-$y$ plane in $\mathbb R^3$ is a two-dimensional vector space (that’s a subspace of $\mathbb R^3$), but its elements are not elements of $\mathbb R^2$. $\endgroup$
    – amd
    Mar 19, 2019 at 20:06
  • $\begingroup$ @user251257 Yes, but I am not saying that it works both way $\endgroup$
    – Bek Masharipov
    Mar 19, 2019 at 20:12

2 Answers 2

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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 $1$.

  • If we're given a basis of $F^n$, then we can express each $\vec e^i$ as an $F$-linear combination of vectors of that basis.
  • We can express each element of $S^n$ as an $S$-linear combination of vectors in the standard basis: $(s_1, \dots, s_n) = s_1 \vec e^1 + \dots + s_n \vec e^n$.
  • Putting these together, we can express each element of $S^n$ as an $S$-linear combination of vectors in the basis of $F^n$ we started with.

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.

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  • $\begingroup$ Thank you! If I understand this correctly, it means that we take the basis from $F^n$ and use scalars from $S$. Also, do we have any name or description for vector spaces defined like $F^n$ over $F$, where $F$ is the same set? Like $\mathbb{R}^n$ over $\mathbb{R}$ $\endgroup$
    – Bek Masharipov
    Mar 19, 2019 at 22:07
  • $\begingroup$ These are sometimes called "coordinate spaces". $\endgroup$
    – Misha Lavrov
    Mar 19, 2019 at 22:15
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Assume your hypothesis was true; then it would lead to $F=S$. Also, note that $a_i$ is not necessarily in $S^n$.

The problem is, that you want to think of vector spaces as some $F^n$. Let me illustrate this problem by referring on the said space of polynomials of the form $a_0+a_1x+a_2x^2$, where $a_0,a_1,a_2\in F$ for some field $F$. So what is the difference between this space and $F^3$? The answer is in my opinion: The way you write your vectors down. In the case of polynomials, you write them seperated by symbols as $x$ or $x^2$ and a weird $+$-symbol that basically does nothing and is just a seperation symbol. In the case of $F^3$, the vector $(a_0,a_1,a_2)$ has his coefficients seperated by commas and sometimes people like to write them in a column instead of a row. It is up to you, wether you want to interpret your vectors $(a_0,a_1,a_2)$ as a point in space or as polynomial. But I could easily think of the vector (0,3,2) as a two dimensional line that has slope 3 and intersects the $y$-axis at 2. That is basically what people consider, if they say two vector spaces are isomorphic: They are es equal as two vector spaces can get.

So in a sense, you are right, two vector spaces with same dimension have the same basis. The basis vectors are just written differently.

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  • $\begingroup$ I think you are right. I will look at isomorphic vector spaces . As @amd mentioned it doesn't work for the set of polynomials. Anyway is there any definition for vector spaces that are defined like F^N over F, where F is the same set? $\endgroup$
    – Bek Masharipov
    Mar 19, 2019 at 21:12
  • $\begingroup$ I edited my comment. Still, I want to add: Even if you only work with $F^n$ and $S^n$, once your fields are different (like $\mathbb{Q},\mathbb{R},\mathbb{C}$, or many other fields, like $\mathbb{Z}$/$p\mathbb{Z}$ for some prime $p$), they will never have the same basis, even though dim($S^n$)=dim($F^n$). $\endgroup$
    – Mahdimatika
    Mar 19, 2019 at 21:41

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