# Question about basis and finite dimensional vector space

I have seen the statement "Every finite dimensional vector space has a basis." (Here on page 5)

I'm confused about what this tells me. It seems to tell me nothing: by definition, the dimension of a vector space is the number of elements in a basis of it. Then saying a vector space is finite dimensional is the same as saying that it has a basis.

Or are there any other definitions of dimension than the number of basis elements?

• Nice question. I think the statment "Every finite dimensional vector space has a basis." is used to distinguish the infinite dimensional case, since "Not Every infinite dimensional vector space has a basis." May 3, 2012 at 12:22
• You could define a vector space to be "finite dimensional" without reference to a specific dimension or to a basis - a finite dimensional vector space is one which has a finite spanning set. This is stated on page 1 of the link. May 3, 2012 at 12:25
• @Sunni, every vector space has a basis, even infinite dimensional ones. This is equivalent to the Axiom of Choice.
– lhf
May 3, 2012 at 12:39
• @Sunni,ihf,Mark Every vector space has a basis whether it's finite or infinite dimensional,but that doesn't really explain to the OP what's different in going from the finite to the infinite dimensional case. The main difference-from which most of the other properties that distinguish the 2 cases are derived-is that since infinite sets can be put in one to one correspondence without having the same number of elements (for example, the natural numbers and the integers), infinite dimensional vector spaces don't have a unique dimension! Feb 21, 2013 at 9:00
• @Mathemagician1234 : The statement "without having the same number of elements" is either meaningless or false; indeed, under any reasonable interpretation of "the same number of elements", there are as many natural numbers as integers. Feb 22, 2013 at 18:29

A more general definition of dimension is the maximal number of nested subspaces: a vector space $V$ has dimension $d$ if and only if $d$ is the largest number such that there exist subspaces.

$$\{0\}\subsetneq V_1\subsetneq V_2\subsetneq\cdots\subsetneq V_d=V$$

This is useful in defining dimension for more general modules.

You can in fact drop the "finite-dimensional" to get the statement that every vector space has a basis (this requires the Axiom of Choice), and then there's no problem. Or indeed, as Mark points out in the comments, the definition of "finite-dimensional" being used is that there is a finite spanning set, so the statement is telling you that the existence of a finite spanning set implies the existence of a finite basis.

Perhaps a broader context will be valuable. Let $R$ be a commutative ring (noncommutativity isn't relevant to the point I'm trying to make) and $M$ a module over it. We say that $M$ is finitely generated if there exist $m_1, ... m_n \in M$ such that every element of $M$ can be written in the form $$m = r_1 m_1 + ... + r_n m_n.$$

When $R$ is a field $k$, an $R$-module is precisely a $k$-vector space, and a finitely generated $R$-module is precisely a finite-dimensional $k$-vector space. (Note that I can define "finite-dimensional" without defining "dimension.")

We say that $m_1, ... m_n$ is a basis of $M$ if it is a generating set which is linearly independent in the sense that if $$0 = r_1 m_1 + ... + r_n m_n$$

then $r_1 = ... = r_n = 0$. This is equivalent to saying that each element of $M$ is uniquely expressible as a sum of the $m_i$. Abstractly, it says that the natural map $$R^n \to M$$

given by sending $(r_1, ... r_n)$ to $r_1 m_1 + ... + r_n m_n$ is an isomorphism of $R$-modules. When $M$ has this property in module theory, we say that $M$ is a free module.

To say that every finite-dimensional vector space has a basis is to say that every finitely generated module over a field is free. It may be useful to see why this is false for more general rings: for example, a $\mathbb{Z}$-module is just an abelian group, and a $\mathbb{Z}$-module such as $\mathbb{Z}/p\mathbb{Z}$ ($p$ a prime) cannot be free. Indeed, no generating set can be linearly independent since $pm = 0$ for all $m$.

More generally, if $R$ is a commutative ring and $I$ a nontrivial ideal of it, then $R/I$ is never free. Thus the only commutative rings $R$ for which every finitely generated module is free are those with no nontrivial ideals, and this is one of several equivalent ways to define a field.

When $R$ is noncommutative, a more astonishing thing that can go wrong is that free modules can fail to have invariant basis number.

• An the key to the invariant basis number property for commutative rings is the existence of maximal ideals (also equivalent to the axiom of choice) and the invariant basis number property for fields.
– lhf
May 3, 2012 at 14:49
• Some very interesting module theory,Quaochu-but I'm not sure what it has to do with the OP's question. I think generalizing the base ring of the module beyond fields kind of loses the point of the question in this case. Feb 21, 2013 at 18:47
• The point is to demonstrate that the statement has content by showing why a more general statement is false. Feb 21, 2013 at 19:53

In the document you're linking to, the following is used as the definition of finitely dimensional space.

If a vector space $V$ is spanned by a finite number of vectors, we say that it is finite dimensional.

So we do not know a priori that every finite-dimensional space (w.r.t. this definition) has a basis, we need to show this.

Only after you show that every such space has a finite basis and that any two bases have the same number elements, you can use $V$ has a basis consisting of finitely many elements as an equivalent definition of finite-dimensional space.

I think that the definition of finite-dimensional space as the space having a finite spanning set is used quite often in introductory texts on linear algebra. (Among other things, you can speak about finite-dimensional spaces before defining the notion of basis.)

IIRC I have even seen some texts where basis is assumed to have at least one element, so if you use that definition, then $\{0\}$ is an example of finite dimensional space which does not have a basis. (It is finite dimensional because it is spanned by the finite set $\{0\}$.)

Of course, if you've already mastered linear algebra, you will use any of equivalent definitions of finite-dimensional space almost automatically, details like this are important only in the introductory course, when the students encounter these notions for the first time.

EDIT: Now I noticed that this was explained already in Mark Bennet's comment above.