Dual basis

 ( Linear Algebra ) 

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 Linear 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 $B\; =\; \backslash \{v\_i\backslash \}\_\{i\backslash in\; I\}$ and $B^\{*\}\; =\; \backslash \{v^i\backslash \}\_\{i\; \backslash in\; I\}$, 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^i_j = \begin{cases}

 1 & \text{if } i = j\\
 0 & \text{if } i \ne j\text{,}
     

\end{cases} where $\backslash delta^i\_j$ 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,

$\backslash mathbf\{x\}\; =\; x^1\; \backslash mathbf\{i\}\_1\; +\; x^2\; \backslash mathbf\{i\}\_2\; +\; x^3\; \backslash mathbf\{i\}\_3$

where $\backslash \{\backslash mathbf\{i\}\_1,\; \backslash mathbf\{i\}\_2,\; \backslash mathbf\{i\}\_3\backslash \}$ is the basis in a Cartesian frame. The components of $\backslash mathbf\{x\}$ can be found by

$x^k\; =\; \backslash mathbf\{x\}\; \backslash cdot\; \backslash mathbf\{i\}\_k.$

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

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

The equality holds when the $\backslash mathbf\{e\}^i$s are the dual basis of $\backslash mathbf\{e\}\_i$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 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 $w=\backslash sum\_\{i\backslash in\; K\}\backslash alpha\_iv^i$ for a finite subset K of I, then for any j not in K, $w(v\_j)=\backslash left(\backslash sum\_\{i\backslash in\; K\}\backslash alpha\_iv^i\backslash right)\backslash left(v\_j\backslash right)=0$, 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 $B=\backslash \{e\_1,\backslash dots,e\_n\backslash \}$ and $B^*=\backslash \{e^1,\backslash dots,e^n\backslash \}$. If one denotes the evaluation of a covector on a vector as a pairing, the biorthogonality condition becomes:

$\backslash left\backslash langle\; e^i,\; e\_j\; \backslash right\backslash rangle\; =\; \backslash delta^i\_j.$

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 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 $R\backslash text\{-\}\backslash mathbf\{Mod\}$). Then we define the dual space of $A$, denoted $A^\{\backslash ast\}$, to be $\backslash text\{Hom\}\_R(A,R)$, 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 $A^\{\backslash ast\backslash ast\}$, and defined as $\backslash text\{Hom\}\_R(A^\{\backslash ast\},R)$.

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 $\backslash delta\_\{xy\}$ over the basis $X$ by $\backslash delta\_\{xy\}=1$ if $x=y$ and $\backslash delta\_\{xy\}=0$ if $x\backslash ne\; y$. Then the set $S\; =\; \backslash lbrace\; f\_x:F\; \backslash to\; R\; \backslash ;\; |\; \backslash ;\; f\_x(y)=\backslash delta\_\{xy\}\; \backslash rbrace$ describes a linearly independent set with each $f\_x\; \backslash in\; \backslash text\{Hom\}\_R(F,R)$. Since $F$ is finite-dimensional, the basis $X$ is of finite cardinality. Then, the set $S$ is a basis to $F^\backslash ast$ and $F^\backslash ast$ is a free (right) $R$-module.

---

Examples For example, the standard basis vectors of $\backslash R^2$ (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 $(\backslash R^2)^*$ 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 $\backslash \{\backslash mathbf\{e\}\_1,\; \backslash mathbf\{e\}\_2,\; \backslash mathbf\{e\}\_3\backslash \}$, the biorthogonal (dual) basis $\backslash \{\backslash mathbf\{e\}^1,\; \backslash mathbf\{e\}^2,\; \backslash mathbf\{e\}^3\backslash \}$ 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 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 $\backslash mathbf\{e\}\_1,\backslash ,\backslash mathbf\{e\}\_2$ and $\backslash mathbf\{e\}\_3.$

In general the dual basis of a basis in a finite-dimensional vector space can be readily computed as follows: given the basis $f\_1,\backslash ldots,f\_n$ and corresponding dual basis $f^1,\backslash ldots,f^n$ we can build matrices

$$

\begin{align} F &= \begin{bmatrix}f_1 & \cdots & f_n \end{bmatrix} \\ G &= \begin{bmatrix}f^1 & \cdots & f^n \end{bmatrix} \end{align}

Then the defining property of the dual basis states that

$G^\backslash mathsf\{T\}F\; =\; I$

Hence the matrix for the dual basis $G$ can be computed as

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

---

See also

• Reciprocal lattice
• Miller index
• Zone axis

---

Notes

• (2025). 9789814313124, World Scientific. ISBN 9789814313124

Post Comment