When do two vector spaces with same dimension have the same basis?

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.

• How do you claim that $a_i \in F$? Mar 19 '19 at 20:04
• Better argue by dimension. Mar 19 '19 at 20:05
• Also, a Basis of $F^n$ is a Basis of $S^n$. But the converse is trivially wrong. Mar 19 '19 at 20:06
• 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 $\mathbb R^2$.
– amd
Mar 19 '19 at 20:06
• @user251257 Yes, but I am not saying that it works both way Mar 19 '19 at 20:12

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.

• 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 $\mathbb{R}$ Mar 19 '19 at 22:07
• These are sometimes called "coordinate spaces". Mar 19 '19 at 22:15

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.

• 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? Mar 19 '19 at 21:12
• 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($F^n$). Mar 19 '19 at 21:41