# Dual basis

In linear algebra, given a vector space $V$ with a basis $B$ of vectors indexed by an index set $I$ (the cardinality of $I$ is the dimension of $V$), the dual set of $B$ is a set $B^{*}$ of vectors in the dual space $V^{*}$ with the same index set $I$ such that $B$ and $B^{*}$ form a biorthogonal system. The dual set is always linearly independent but does not necessarily span $V^{*}$. If it does span $V^{*}$, then $B^{*}$ is called the dual basis or reciprocal basis for the basis $B$.
Denoting the indexed vector sets as $B=\{v_{i}\}_{i\in I}$ and $B^{*}=\{v^{i}\}_{i\in I}$, being biorthogonal means that the elements pair to have an inner product equal to 1 if the indexes are equal, and equal to 0 otherwise. Symbolically, evaluating a dual vector in $V^{*}$ on a vector in the original space $V$:
    $v^{i}\cdot v_{j}=\delta _{j}^{i}={\begin{cases}1&{\text{if }}i=j\\0&{\text{if }}i\neq j{\text{,}}\end{cases}}$
where $\delta _{j}^{i}$ is the Kronecker delta symbol.

## Introduction

To perform operations with a vector, we must have a straightforward method of calculating its components. In a Cartesian frame the necessary operation is the dot product of the vector and the base vector. For example,
    $\mathbf {x} =x^{1}\mathbf {i} _{1}+x^{2}\mathbf {i} _{2}+x^{3}\mathbf {i} _{3}$
where $\{\mathbf {i} _{1},\mathbf {i} _{2},\mathbf {i} _{3}\}$ is the basis in a Cartesian frame. The components of $\mathbf {x}$ can be found by
    $x^{k}=\mathbf {x} \cdot \mathbf {i} _{k}.$
However, in a non-Cartesian frame, we do not necessarily have $\mathbf {e} _{i}\cdot \mathbf {e} _{j}=0$ for all $i\neq j$. However, it is always possible to find vectors $\mathbf {e} ^{i}$ in the dual space such that
    $x^{i}=\mathbf {e} ^{i}(\mathbf {x} )\qquad (i=1,2,3).$
The equality holds when the $\mathbf {e} ^{i}$s are the dual basis of $\mathbf {e} _{i}$s. Notice the difference in position of the index $i$.

## Existence and uniqueness

The dual set always exists and gives an injection from V into V$^∗$, namely the mapping that sends v$_i$ to v$^i$. This says, in particular, that the dual space has dimension greater or equal to that of V.
However, the dual set of an infinite-dimensional V does not span its dual space V$^∗$. For example, consider the map w in V$^∗$ from V into the underlying scalars F given by w(v$_i$) = 1 for all i. This map is clearly nonzero on all v$_i$. If w were a finite linear combination of the dual basis vectors v$^i$, say w=∑$_{i∈K}$α$_i$v$^i${\textstyle w=\sum _{i\in K}\alpha _{i}v^{i}} for a finite subset K of I, then for any j not in K, w(v$_j$)=(∑$_{i∈K}$α$_i$v$^i$)(v$_j$)=0{\textstyle w(v_{j})=\left(\sum _{i\in K}\alpha _{i}v^{i}\right)\left(v_{j}\right)=0}, contradicting the definition of w. So, this w does not lie in the span of the dual set.
The dual of an infinite-dimensional space has greater dimension (this being a greater infinite cardinality) than the original space has, and thus these cannot have a basis with the same indexing set. However, a dual set of vectors exists, which defines a subspace of the dual isomorphic to the original space. Further, for topological vector spaces, a continuous dual space can be defined, in which case a dual basis may exist.

### Finite-dimensional vector spaces

In the case of finite-dimensional vector spaces, the dual set is always a dual basis and it is unique. These bases are denoted by $B=\{e_{1},\dots ,e_{n}\}$ and $B^{*}=\{e^{1},\dots ,e^{n}\}$. If one denotes the evaluation of a covector on a vector as a pairing, the biorthogonality condition becomes:
    $\left\langle e^{i},e_{j}\right\rangle =\delta _{j}^{i}.$
The association of a dual basis with a basis gives a map from the space of bases of V to the space of bases of V$^∗$, and this is also an isomorphism. For topological fields such as the real numbers, the space of duals is a topological space, and this gives a homeomorphism between the Stiefel manifolds of bases of these spaces.

## A categorical and algebraic construction of the dual space

Another way to introduce the dual space of a vector space (module) is by introducing it in a categorical sense. To do this, let $A$ be a module defined over the ring $R$ (that is, $A$ is an object in the category $R{\text{-}}\mathbf {Mod}$). Then we define the dual space of $A$, denoted $A^{\ast }$, to be ${\text{Hom}}_{R}(A,R)$, the module formed of all $R$-linear module homomorphisms from $A$ into $R$. Note then that we may define a dual to the dual, referred to as the double dual of $A$, written as $A^{\ast \ast }$, and defined as ${\text{Hom}}_{R}(A^{\ast },R)$.
To formally construct a basis for the dual space, we shall now restrict our view to the case where $F$ is a finite-dimensional free (left) $R$-module, where $R$ is a ring with unity. Then, we assume that the set $X$ is a basis for $F$. From here, we define the Kronecker Delta function $\delta _{xy}$ over the basis $X$ by $\delta _{xy}=1$ if $x=y$ and $\delta _{xy}=0$ if $x\neq y$. Then the set $S=\lbrace f_{x}:F\to R\;|\;f_{x}(y)=\delta _{xy}\rbrace$ describes a linearly independent set with each $f_{x}\in {\text{Hom}}_{R}(F,R)$. Since $F$ is finite-dimensional, the basis $X$ is of finite cardinality. Then, the set $S$ is a basis to $F^{\ast }$ and $F^{\ast }$ is a free (right) $R$-module.

## Examples

For example, the standard basis vectors of $\mathbb {R} ^{2}$ (the Cartesian plane) are
    $\left\{\mathbf {e} _{1},\mathbf {e} _{2}\right\}=\left\{{\begin{pmatrix}1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\end{pmatrix}}\right\}$
and the standard basis vectors of its dual space $(\mathbb {R} ^{2})^{*}$ are
    $\left\{\mathbf {e} ^{1},\mathbf {e} ^{2}\right\}=\left\{{\begin{pmatrix}1&0\end{pmatrix}},{\begin{pmatrix}0&1\end{pmatrix}}\right\}{\text{.}}$
In 3-dimensional Euclidean space, for a given basis $\{\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}\}$, the biorthogonal (dual) basis $\{\mathbf {e} ^{1},\mathbf {e} ^{2},\mathbf {e} ^{3}\}$ can be found by formulas below:
    $\mathbf {e} ^{1}=\left({\frac {\mathbf {e} _{2}\times \mathbf {e} _{3}}{V}}\right)^{\mathsf {T}},\ \mathbf {e} ^{2}=\left({\frac {\mathbf {e} _{3}\times \mathbf {e} _{1}}{V}}\right)^{\mathsf {T}},\ \mathbf {e} ^{3}=\left({\frac {\mathbf {e} _{1}\times \mathbf {e} _{2}}{V}}\right)^{\mathsf {T}}.$
where $^T$ denotes the transpose and
    $V\,=\,\left(\mathbf {e} _{1};\mathbf {e} _{2};\mathbf {e} _{3}\right)\,=\,\mathbf {e} _{1}\cdot (\mathbf {e} _{2}\times \mathbf {e} _{3})\,=\,\mathbf {e} _{2}\cdot (\mathbf {e} _{3}\times \mathbf {e} _{1})\,=\,\mathbf {e} _{3}\cdot (\mathbf {e} _{1}\times \mathbf {e} _{2})$
is the volume of the parallelepiped formed by the basis vectors $\mathbf {e} _{1},\,\mathbf {e} _{2}$ and $\mathbf {e} _{3}.$
In general the dual basis of a basis in a finite-dimensional vector space can be readily computed as follows: given the basis $f_{1},\ldots ,f_{n}$ and corresponding dual basis $f^{1},\ldots ,f^{n}$ we can build matrices
    ${\begin{aligned}F&={\begin{bmatrix}f_{1}&\cdots &f_{n}\end{bmatrix}}\\G&={\begin{bmatrix}f^{1}&\cdots &f^{n}\end{bmatrix}}\end{aligned}}$
Then the defining property of the dual basis states that
    $G^{\mathsf {T}}F=I$
Hence the matrix for the dual basis $G$ can be computed as
    $G=\left(F^{-1}\right)^{\mathsf {T}}$