A (Hamel) basis of a vector space is a linearly independent spanning set.

It can be proved that any two bases of the same vector space must have the same cardinality. This introduces the notion of dimension of a vector space, which is precisely the cardinality of the basis, and is denoted by $\operatorname{dim}(V)$ , where $V$ is the vector space.

The fact that every vector space has a Hamel basis is an important consequence of the axiom of choice (in fact, that proposition is equivalent to the axiom of choice.)

Examples.

• $\beta = \{e_i\}$ , $1\le i \le n$ , is a basis for $\mathbb{R}^n$ (the $n$ -dimensional vector space over the reals). For $n=4$ ,

  $$ \beta = \left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} , \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} , \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\} $$

• $ \beta = \{ 1, x , x^2 \} $ is a basis for the vector space of polynomials with degree at most 2, over a division ring.
• The set

  $$ \beta = \left\{ \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix} , \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix} , \begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix} , \begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix} \right\} $$

  is a basis for the vector space of $2 \times 2$ matrices over a division ring, and assuming that the characteristic of the ring is not 2, then so is

  $$ \beta' = \left\{ \begin{bmatrix}2 & 0 \\ 0 & 0 \end{bmatrix} , \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix} , \begin{bmatrix}0 & 0 \\ 0 & 4 \end{bmatrix} , \begin{bmatrix}0 & 0 \\ \frac{1}{2} & 0 \end{bmatrix} \right\}. $$

• The empty set is a basis for the trivial vector space which consists of the unique element $0$ .

Remark. More generally, for any (left) right module $M$ over a ring $R$ , one may define a (left) right basis for $M$ as a subset $B$ of $M$ such that $B$ spans $M$ and is linearly independent. However, unlike bases for a vector space, bases for a module may not have the same cardinality.