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Suppose a vector space $V$ has infinite dimension, $\dim V = \infty$. Is the cardinality of the basis of $V$ neccesarily countable infinite?

If a vector space has infinite dimension does $V$ then have a basis ? If one can specify a set of infinitely many linearly independent vectors is this set then regarded as a basis for $V$ ?

Are there results that can be proven for vector spaces with finite dimension, but not for vector spaces with infinite dimension ?

Why does most books on linear algebra only mention finite dimension vector spaces ?

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    $\begingroup$ The vector space of real functions on $\mathbb{R}$ has an uncountable basis. In finite dimensional spaces you can prove, for instance, that $U+U^{\perp}=V$, where $U$ is any subspace of the euclidean vector space $V$. This is not true in infinite dimensional euclidean space. $\endgroup$
    – egreg
    May 9, 2014 at 13:19
  • $\begingroup$ Given any set one can define a vector space with this set as it's basis. Moreover given an infinite dimensional vector space, the dual space always has a basis of strictly larger cardinality. $\endgroup$
    – Seth
    May 9, 2014 at 13:22
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    $\begingroup$ No. In infinite dimensional Banach spaces, no Hamel basis can be countable. See this. $\endgroup$
    – David Mitra
    May 9, 2014 at 13:32
  • $\begingroup$ A lot of basic linear algebra carries over quite well to the infinite-dimensional case, assuming you accept the axiom of choice. You get bases, (infinite-dimensional) matrices, scalar products, ... One thing that doesn't really carry over I think is spectral theory, i.e. eigenvalues. You get something similar in (some) topological spaces, but not with a strictly algebraic approach. $\endgroup$
    – fgp
    May 9, 2014 at 13:51

3 Answers 3

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  1. Not all infinite dimensional vector spaces have countable basis. In fact, most spaces you come across in "real life" don't. The simplest example would be a sequence space like $$ \ell^2 = \{(x_n) \subset \mathbb{C} : \sum_{n=1}^{\infty} |x_n|^2 < \infty\} $$ This is an example of a Banach space - a vector space that comes with a nice notion of distance (a norm) and one that is complete with respect to that norm.

    It is a general fact that an infinite dimensional Banach space cannot have a countably infinite basis (due to the Baire Category theorem).

  2. However, the space $$ c_{00} := \{(x_n) \in \mathbb{C} : x_n \neq 0 \text{ for only finitely many } n\} $$ is an example of an infinite dimensional vector space whose basis is countable (the "standard" basis).

  3. Most books on Linear Algebra mention only finite dimensional vector spaces because they are easy to visualize (just extend your notion of a vector in $\mathbb{R}^2$), but they are also deep enough to prove some rather interesting results (for instance the spectral theorem). Furthermore, infinite dimensional vector spaces are best analysed with some topology in mind - this is what functional analysis studies.

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  • $\begingroup$ A somewhat simpler example for (2) is the vector space of polynomials over $\mathbb{R}$ or $\mathbb{C}$. $\endgroup$
    – fgp
    May 9, 2014 at 13:53
  • $\begingroup$ The usual basis of $\ell^2$ is not a countable set? $\endgroup$
    – Vincenzo Tibullo
    May 9, 2014 at 15:42
  • $\begingroup$ If by "usual" basis, you mean the standard basis, then it is not a basis. The sequence $(1/n) \in \ell^2$ cannot be written as a finite linear combination of these basis elements. $\endgroup$
    – Prahlad Vaidyanathan
    May 9, 2014 at 15:44
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Let $X$ be any set with cardinality $\alpha$, then consider the set of formal linear combinations $$\{a_1x_1+\cdots+a_nx_n: a_i\in K, x_i \in X, n\in\mathbb{N}\}$$ When equipped with usual addition and multiplication, this set forms a vector space with basis $X$ and dimension $\alpha$. So we have vector spaces of arbitrary dimensions.

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If you accept the axiom of choice, then every vector space has a basis. This result can be further improved to $R$-modules over division rings I guess.

Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. What's the cardinality of a basis for $\mathbb{R}$ over $\mathbb{Q}$? Think about it.

Think of a basis as a maximal linearly independent set. Can you somehow use Zorn's lemma to prove that every vector space has a basis now?

Why does most books on linear algebra talk about only the finite dimensional case? Because the infinite dimensional case requires some knowledge of Analaysis I guess. And most tools that we employ in linear algebra work because we have matrices. We don't have such tools in the infinite-dimensional case, so we need to extend our tools and theory.

Addendum: Read this page. It should answer a lot of your questions.

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