A basis promises that every vector can be written uniquely as a linear combination of chosen vectors, but uniqueness is not the same as readability. In the standard basis of $k^n$, the coordinates of $(a_1, \ldots, a_n)$ stare back at us. In a skew basis, or in a shifted polynomial basis such as $(1, X-1, (X-1)^2)$, the coefficients are no longer visible from the notation alone. The dual basis is the device that makes coordinates into functions.

The guiding question is this: once a basis has been chosen, can we build linear functionals that extract the individual coordinates? If $v = a_1 e_1 + \cdots + a_n e_n$, we want a functional $e_i^*$ satisfying $e_i^*(v)=a_i$. The content of the theory is that these coordinate-reading maps exist, are unique, form a basis of the [dual space](/page/Dual%20Space), and transform contravariantly when the original basis changes.

The first surprise is that the coordinate maps are not an extra Euclidean structure. They do not require an [inner product](/page/Inner%20Product), a length, or an angle. They come only from the algebra of a basis. This is why dual bases appear in linear algebra, tensor calculus, differential geometry, representation theory, and commutative algebra: whenever vectors are described by coordinates, there are dual coordinate functionals silently doing the reading.

Before naming the construction, it helps to see the failure it repairs. In a nonstandard basis, the coefficient of a vector cannot be recovered by looking at a component of the ambient coordinate tuple.

[example: Hidden Coordinates in a Skew Basis] Let $V=\mathbb R^2$ and take the ordered basis $\mathcal B=(e_1,e_2)$ with $e_1=(1,1)$ and $e_2=(1,-1)$. To find the $\mathcal B$-coordinates of $v=(5,1)$, write \begin{align*} (5,1)=a_1(1,1)+a_2(1,-1). \end{align*} Expanding the right-hand side gives \begin{align*} a_1(1,1)+a_2(1,-1)=(a_1+a_2,a_1-a_2). \end{align*} Thus the coefficients must satisfy \begin{align*} a_1+a_2=5 \quad \text{and} \quad a_1-a_2=1. \end{align*} Adding the two equations gives $2a_1=6$, so $a_1=3$. Substituting into $a_1+a_2=5$ gives $3+a_2=5$, hence $a_2=2$. Therefore \begin{align*} (5,1)=3(1,1)+2(1,-1), \end{align*} so the standard coordinates $(5,1)$ and the $\mathcal B$-coordinates $(3,2)$ are different. The same calculation for a general vector $(x,y)$ gives \begin{align*} (x,y)=a_1(1,1)+a_2(1,-1)=(a_1+a_2,a_1-a_2). \end{align*} Hence \begin{align*} a_1+a_2=x \quad \text{and} \quad a_1-a_2=y. \end{align*} Adding these equations gives $2a_1=x+y$, so $a_1=(x+y)/2$. Subtracting the second equation from the first gives $2a_2=x-y$, so $a_2=(x-y)/2$. The coordinate-reading functionals for the basis $\mathcal B$ are therefore \begin{align*} e_1^*(x,y)=\frac{x+y}{2} \end{align*} and \begin{align*} e_2^*(x,y)=\frac{x-y}{2}. \end{align*} Applying them to $v=(5,1)$ gives $e_1^*(5,1)=3$ and $e_2^*(5,1)=2$, so these two linear functionals recover the hidden coordinates. [/example]

The example shows the whole idea in miniature. A dual basis is not obtained by reusing the original basis vectors; it is built from the coordinate extraction problem attached to that basis.

## Definition

### Coordinate Tests

The construction we want is a whole family of scalar-valued tests, one test for each basis vector. Each test should return $1$ on its own basis vector and $0$ on the others, so that applying the tests to a vector reveals exactly the coefficients in its basis expansion.

[definition: Dual Basis] Let $V$ be a finite-dimensional [vector space](/page/Vector%20Space) over a field $k$, and let $\mathcal B=(e_1,\ldots,e_n)$ be an ordered basis of $V$. The dual basis of $\mathcal B$ is the ordered tuple $\mathcal B^*=(e_1^*,\ldots,e_n^*)$ in $V^*$ such that \begin{align*} e_i^*(e_j)=\delta_{ij} \end{align*} for all $i,j\in\{1,\ldots,n\}$. [/definition]

The defining equations say that each $e_i^*$ recognises one basis vector and annihilates the others. The notation $e_i^*$ depends on the whole basis $\mathcal B$, not only on the single vector $e_i$. If $e_i$ is kept fixed but the other basis vectors change, the functional labelled $e_i^*$ may change.

### The Ambient Dual Space

To make the definition fully precise, we unpack the two pieces of language it uses. First, the dual basis lives in the space of all linear scalar measurements on $V$.

[definition: Dual Space] Let $V$ be a vector space over a field $k$. The dual space of $V$ is \begin{align*} V^* := \operatorname{Hom}_k(V,k), \end{align*} the vector space of all $k$-linear maps $f:V\to k$. [/definition]

### Ordering and Coordinate Notation

Second, the phrase "the $i$-th coordinate" requires an ordering. A basis as a set spans the space, but a coordinate list has positions: a first vector, a second vector, and so on. Without this extra ordering data, the symbol $e_i^*$ has no well-defined index to read, so we distinguish an ordered basis from an unordered spanning list.

[definition: Ordered Basis] Let $V$ be a finite-dimensional vector space over a field $k$. An ordered basis of $V$ is a tuple $\mathcal B=(e_1,\ldots,e_n)$ such that $\{e_1,\ldots,e_n\}$ is a basis of $V$. [/definition]

The ordering lets us write every vector $v\in V$ in the form

\begin{align*} v=\sum_{i=1}^n a_i e_i, \end{align*}

where the scalar $a_i\in k$ is the $i$-th coordinate of $v$ with respect to $\mathcal B$. The next task is to turn the rule $v\mapsto a_i$ into an actual member of the dual space.