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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.^[1] 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 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 $B=\{e_{1},\dots ,e_{n}\}$ 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}}

Notes

^ Lebedev, Cloud & Eremeyev 2010, p. 12.

References

Lebedev, Leonid P.; Cloud, Michael J.; Eremeyev, Victor A. (2010). Tensor Analysis With Applications to Mechanics. World Scientific. ISBN 978-981431312-4.
"Finding the Dual Basis". Stack Exchange. May 27, 2012.

[FOOTNOTELebedevCloudEremeyev201012-1] Lebedev, Cloud & Eremeyev 2010, p. 12.

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