In linear algebra, given a vector space ${\displaystyle V}$ with a basis ${\displaystyle B}$ of vectors indexed by an index set ${\displaystyle I}$ (the cardinality of ${\displaystyle I}$ is the dimension of ${\displaystyle V}$), the dual set of ${\displaystyle B}$ is a set ${\displaystyle B^{*))$ of vectors in the dual space ${\displaystyle V^{*))$ with the same index set ${\displaystyle I}$ such that ${\displaystyle B}$ and ${\displaystyle B^{*))$ form a biorthogonal system. The dual set is always linearly independent but does not necessarily span ${\displaystyle V^{*))$. If it does span ${\displaystyle V^{*))$, then ${\displaystyle B^{*))$ is called the dual basis or reciprocal basis for the basis ${\displaystyle B}$.

Denoting the indexed vector sets as ${\displaystyle B=\{v_{i}\}_{i\in I))$ and ${\displaystyle B^{*}=\{v^{i}\}_{i\in I))$, 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 ${\displaystyle V^{*))$ on a vector in the original space ${\displaystyle V}$:

${\displaystyle v^{i}\cdot v_{j}=\delta _{j}^{i}={\begin{cases}1&{\text{if ))i=j\\0&{\text{if ))i\neq j{\text{,))\end{cases))}$

where ${\displaystyle \delta _{j}^{i))$ 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.^[1] For example,

${\displaystyle \mathbf {x} =x^{1}\mathbf {i} _{1}+x^{2}\mathbf {i} _{2}+x^{3}\mathbf {i} _{3))$

where ${\displaystyle \{\mathbf {i} _{1},\mathbf {i} _{2},\mathbf {i} _{3}\))$ is the basis in a Cartesian frame. The components of ${\displaystyle \mathbf {x} }$ can be found by

${\displaystyle x^{k}=\mathbf {x} \cdot \mathbf {i} _{k}.}$

However, in a non-Cartesian frame, we do not necessarily have ${\displaystyle \mathbf {e} _{i}\cdot \mathbf {e} _{j}=0}$ for all ${\displaystyle i\neq j}$. However, it is always possible to find vectors ${\displaystyle \mathbf {e} ^{i))$ in the dual space such that

${\displaystyle x^{i}=\mathbf {e} ^{i}(\mathbf {x} )\qquad (i=1,2,3).}$

The equality holds when the ${\displaystyle \mathbf {e} ^{i))$s are the dual basis of ${\displaystyle \mathbf {e} _{i))$s. Notice the difference in position of the index ${\displaystyle i}$.

Existence and uniqueness

The dual set always exists and gives an injection from V into V^∗, namely the mapping that sends v_i to vⁱ. 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 vⁱ, say ${\textstyle w=\sum _{i\in K}\alpha _{i}v^{i))$ for a finite subset K of I, then for any j not in K, ${\textstyle w(v_{j})=\left(\sum _{i\in K}\alpha _{i}v^{i}\right)\left(v_{j}\right)=0}$, 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 ${\displaystyle B=\{e_{1},\dots ,e_{n}\))$ and ${\displaystyle B^{*}=\{e^{1},\dots ,e^{n}\))$. If one denotes the evaluation of a covector on a vector as a pairing, the biorthogonality condition becomes:

${\displaystyle \left\langle e^{i},e_{j}\right\rangle =\delta _{j}^{i}.}$

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 ${\displaystyle A}$ be a module defined over the ring ${\displaystyle R}$ (that is, ${\displaystyle A}$ is an object in the category ${\displaystyle R{\text{-))\mathbf {Mod} }$). Then we define the dual space of ${\displaystyle A}$, denoted ${\displaystyle A^{\ast ))$, to be ${\displaystyle {\text{Hom))_{R}(A,R)}$, the module formed of all ${\displaystyle R}$-linear module homomorphisms from ${\displaystyle A}$ into ${\displaystyle R}$. Note then that we may define a dual to the dual, referred to as the double dual of ${\displaystyle A}$, written as ${\displaystyle A^{\ast \ast ))$, and defined as ${\displaystyle {\text{Hom))_{R}(A^{\ast },R)}$.

To formally construct a basis for the dual space, we shall now restrict our view to the case where ${\displaystyle F}$ is a finite-dimensional free (left) ${\displaystyle R}$-module, where ${\displaystyle R}$ is a ring with unity. Then, we assume that the set ${\displaystyle X}$ is a basis for ${\displaystyle F}$. From here, we define the Kronecker Delta function ${\displaystyle \delta _{xy))$ over the basis ${\displaystyle X}$ by ${\displaystyle \delta _{xy}=1}$ if ${\displaystyle x=y}$ and ${\displaystyle \delta _{xy}=0}$ if ${\displaystyle x\neq y}$. Then the set ${\displaystyle S=\lbrace f_{x}:F\to R\;|\;f_{x}(y)=\delta _{xy}\rbrace }$ describes a linearly independent set with each ${\displaystyle f_{x}\in {\text{Hom))_{R}(F,R)}$. Since ${\displaystyle F}$ is finite-dimensional, the basis ${\displaystyle X}$ is of finite cardinality. Then, the set ${\displaystyle S}$ is a basis to ${\displaystyle F^{\ast ))$ and ${\displaystyle F^{\ast ))$ is a free (right) ${\displaystyle R}$-module.

Examples

For example, the standard basis vectors of ${\displaystyle \mathbb {R} ^{2))$ (the Cartesian plane) are

${\displaystyle \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 ${\displaystyle (\mathbb {R} ^{2})^{*))$ are

${\displaystyle \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 ${\displaystyle \{\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}\))$, the biorthogonal (dual) basis ${\displaystyle \{\mathbf {e} ^{1},\mathbf {e} ^{2},\mathbf {e} ^{3}\))$ can be found by formulas below:

${\displaystyle \mathbf {e} ^{1}=\left({\frac {\mathbf {e} _{2}\times \mathbf {e} _{3)){V))\right)^{\mathsf {T)),\ \mathbf {e} ^{2}=\left({\frac {\mathbf {e} _{3}\times \mathbf {e} _{1)){V))\right)^{\mathsf {T)),\ \mathbf {e} ^{3}=\left({\frac {\mathbf {e} _{1}\times \mathbf {e} _{2)){V))\right)^{\mathsf {T)).}$

where ^T denotes the transpose and

${\displaystyle V\,=\,\left(\mathbf {e} _{1};\mathbf {e} _{2};\mathbf {e} _{3}\right)\,=\,\mathbf {e} _{1}\cdot (\mathbf {e} _{2}\times \mathbf {e} _{3})\,=\,\mathbf {e} _{2}\cdot (\mathbf {e} _{3}\times \mathbf {e} _{1})\,=\,\mathbf {e} _{3}\cdot (\mathbf {e} _{1}\times \mathbf {e} _{2})}$

is the volume of the parallelepiped formed by the basis vectors ${\displaystyle \mathbf {e} _{1},\,\mathbf {e} _{2))$ and ${\displaystyle \mathbf {e} _{3}.}$

In general the dual basis of a basis in a finite-dimensional vector space can be readily computed as follows: given the basis ${\displaystyle f_{1},\ldots ,f_{n))$ and corresponding dual basis ${\displaystyle f^{1},\ldots ,f^{n))$ we can build matrices

${\displaystyle {\begin{aligned}F&={\begin{bmatrix}f_{1}&\cdots &f_{n}\end{bmatrix))\\G&={\begin{bmatrix}f^{1}&\cdots &f^{n}\end{bmatrix))\end{aligned))}$

Then the defining property of the dual basis states that

${\displaystyle G^{\mathsf {T))F=I}$

Hence the matrix for the dual basis ${\displaystyle G}$ can be computed as

${\displaystyle G=\left(F^{-1}\right)^{\mathsf {T))}$