Dual basis explained

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} 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 ⋅} v _{j} = \delta ^{i} _{j} = \begin{cases} 1 & if i = j\\ 0 & if i \ne j , \end{cases} where \delta ^{i} _{j} 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,

x = x ^{1} i _{1} + x ^{2} i _{2} + x ^{3} i _{3}

where

\{ i _{1,} i _{2,} i _{3\}} is the basis in a Cartesian frame. The components of x can be found byx ^{k} = x ⋅ i _{k.}

However, in a non-Cartesian frame, we do not necessarily have

e _{i ⋅ e} _{j=0} for all i ≠ j . However, it is always possible to find vectors e ^{i} in the dual space such thatx ^{i} = e ^{i(x)} (i = 1 , 2 , 3 ).

The equality holds when the

e ^{i} s are the dual basis of 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 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 $w=\backslash sum\_\backslash alpha\_iv^i$ for a finite subset K of I, then for any j not in K, $w(v\_j)=\backslash left(\backslash sum\_\backslash alpha\_iv^i\backslash right)\backslash left(v\_j\backslash right)=0$, 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,...,e} _{n\}} and B ^{*=\{e} ^{1,...,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 ^{i} _{j.}

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 - Mod ). Then we define the dual space of A , denoted A ^{\ast} , to be 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 Hom ^{\ast} _{R(A} ,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\ne 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 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

\R ^{2} (the Cartesian plane) are\left\{ e _{1,} 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

(\R ^{2)} ^{*} are\left\{ e ^{1,} e ^{2\right} \} = \left\{ \begin{pmatrix} 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \end{pmatrix} \right\} .

In 3-dimensional Euclidean space, for a given basis

\{ e _{1,} e _{2,} e _{3\}} , the biorthogonal (dual) basis \{ e ^{1,} e ^{2,} e ^{3\}} can be found by formulas below:e ^{1} = \left( (e _{2} x e _{3})/(V) \right) ^{T,} e ^{2} = \left( (e _{3} x e _{1})/(V) \right) ^{T,} e ^{3} = \left( (e _{1} x e _{2})/(V) \right) ^{T.}

where denotes the transpose and

V = \left( e _{1;e} _{2;e} _{3\right)} = e _{1 ⋅ (e} _{2 x e} _{3)} = e _{2 ⋅ (e} _{3 x e} _{1)} = e _{3 ⋅ (e} _{1 x e} _{2)}

is the volume of the parallelepiped formed by the basis vectors

e _{1,e} _{2} and 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{align} F & = \begin{bmatrix}f _{1} & … & f _{n} \end{bmatrix} \\ G & = \begin{bmatrix}f ^{1} & … & f ^{n} \end{bmatrix} \end{align}

Then the defining property of the dual basis states that

G ^{TF} = I

Hence the matrix for the dual basis

G can be computed asG = \left(F ^{-1} \right) ^{T}

See also

• Reciprocal lattice
• Miller index
• Zone axis

References

• Book: Tensor Analysis With Applications to Mechanics . Lebedev . Leonid P. . Cloud . Michael J. . Eremeyev . Victor A. . 2010 . World Scientific . 978-981431312-4 .
• Web site: Finding the Dual Basis . . May 27, 2012 .