A finite-dimensional vector space is useful because its elements can be named by scalars. The difficulty is that the naming system is not built into the space. In $\mathbb{R}^2$, the vector usually written as $(3,2)$ already looks like two numbers, but those numbers depend on having silently chosen the horizontal and vertical directions. Change the directions, and the same vector receives a different coordinate record.

The central tension is that two opposite mistakes are easy to make. If we choose too few vectors, some elements cannot be reached. If we choose too many, the same element has several descriptions. A basis is exactly the point where spanning and uniqueness meet.

[example: A Spanning Set with Ambiguous Coordinates] Let $V=\mathbb{R}^2$ over $\mathbb{R}$, and set $v_1=(1,0)$, $v_2=(0,1)$, and $v_3=(1,1)$. The set $\{v_1,v_2,v_3\}$ spans $V$ because every vector $(a,b)\in \mathbb{R}^2$ can be written using $v_1$ and $v_2$ alone: \begin{align*} av_1+bv_2+0v_3=a(1,0)+b(0,1)+0(1,1)=(a,0)+(0,b)+(0,0)=(a,b). \end{align*} Now consider the particular vector $(1,1)$. Using $v_1$ and $v_2$ gives \begin{align*} 1v_1+1v_2+0v_3=1(1,0)+1(0,1)+0(1,1)=(1,0)+(0,1)+(0,0)=(1,1). \end{align*} Using $v_3$ instead gives \begin{align*} 0v_1+0v_2+1v_3=0(1,0)+0(0,1)+1(1,1)=(0,0)+(0,0)+(1,1)=(1,1). \end{align*} The two coefficient triples are different, since $(1,1,0)\ne (0,0,1)$, but they produce the same vector. Thus this spanning set gives existence of coordinates without uniqueness; the extra vector $v_3$ is redundant because $v_3=v_1+v_2$. [/example]

This example shows why a basis is not just a convenient list of generators. It is a coordinate system with no redundancy. The whole theory explains how such lists exist, how large they are, how they change, and what breaks when we leave vector spaces for general modules.

## Definition

A basis is the exact compromise promised by the opening example: enough vectors to reach every element, but not so many that coordinates become ambiguous. We state the main definition first, then unpack the two ingredients that make it work.

[definition: Basis] Let $V$ be a vector space over a field $k$. A basis of $V$ is a subset $B \subset V$ such that $B$ is linearly independent and $B$ is a spanning set for $V$. [/definition]

The definition hides two separate demands. To understand why both are necessary, we first isolate reachability, then uniqueness.

### Spanning

A set of vectors is useful only if the vectors we care about can be assembled from it using the allowed scalar operations. This is the algebraic version of having enough directions.

[definition: Linear Combination] Let $V$ be a vector space over a field $k$, and let $S \subset V$. A linear combination of elements of $S$ is a vector of the form \begin{align*} a_1v_1 + \cdots + a_nv_n, \end{align*} where $n \in \mathbb{N}$, $v_1, \ldots, v_n \in S$, and $a_1, \ldots, a_n \in k$. [/definition]

Linear combinations are finite by definition. This matters because vector spaces do not carry a default notion of infinite summation. Infinite expansions require additional topology, as in Hilbert space or Banach space theory, and are not part of the algebraic notion of basis. We also allow the empty linear combination, whose value is $0$; with this convention, the span of the empty set is $\{0\}$.

Once linear combinations are available, the natural next question is what part of the ambient space they reach. We need a name for the set generated by a chosen family before we can ask whether the family reaches everything.

[definition: Span] Let $V$ be a vector space over a field $k$, and let $S \subset V$. The span of $S$, denoted $\operatorname{span}_k(S)$, is the set of all linear combinations of elements of $S$. [/definition]

Knowing the span of a set still leaves an existence problem: some vectors of the ambient space may lie outside all finite combinations of the chosen vectors. The useful case is when this obstruction disappears, so every vector in the space can be reached from the chosen set.

[definition: Spanning Set] Let $V$ be a vector space over a field $k$. A subset $S \subset V$ is a spanning set for $V$ if \begin{align*} \operatorname{span}_k(S) = V. \end{align*} [/definition]

A spanning set answers the existence question: can every vector be described using the chosen vectors? It does not answer the uniqueness question. The ambiguity in the opening example came from a nonzero relation among the chosen vectors.

### Independence

The second half of the problem is uniqueness. If a combination of chosen vectors can vanish without all its coefficients vanishing, then the same vector can be written in more than one way. We first name the algebraic witness to that ambiguity.

[definition: Linear Relation] Let $V$ be a vector space over a field $k$, and let $v_1, \ldots, v_n \in V$. A linear relation among $v_1, \ldots, v_n$ is an equality \begin{align*} a_1v_1 + \cdots + a_nv_n = 0, \end{align*} where $a_1, \ldots, a_n \in k$. [/definition]