Editing Dual space (section)

=== Finite-dimensional case ===
{{See also|Dual basis}}
If <math>V</math> is finite-dimensional, then <math>V^*</math> has the same dimension as <math>V</math>. Given a [[basis of a vector space|basis]] <math>\{\mathbf{e}_1,\dots,\mathbf{e}_n\}</math> in <math>V</math>, it is possible to construct a specific basis in <math>V^*</math>, called the [[dual basis]]. This dual basis is a set <math>\{\mathbf{e}^1,\dots,\mathbf{e}^n\}</math> of linear functionals on <math>V</math>, defined by the relation
: <math> \mathbf{e}^i(c^1 \mathbf{e}_1+\cdots+c^n\mathbf{e}_n) = c^i, \quad i=1,\ldots,n </math>
for any choice of coefficients <math>c^i\in F</math>. In particular, letting in turn each one of those coefficients be equal to one and the other coefficients zero, gives the system of equations
: <math> \mathbf{e}^i(\mathbf{e}_j) = \delta^{i}_{j} </math>
where <math>\delta^{i}_{j}</math> is the [[Kronecker delta]] symbol. This property is referred to as the ''bi-orthogonality property''.

{{Collapse top | Proof}}
Consider <math>\{\mathbf{e}_1,\dots,\mathbf{e}_n\}</math> the basis of V. Let <math>\{\mathbf{e}^1,\dots,\mathbf{e}^n\}</math> be defined as the following:

<math> \mathbf{e}^i(c^1 \mathbf{e}_1+\cdots+c^n\mathbf{e}_n) = c^i, \quad i=1,\ldots,n </math>.

These are a basis of <math>V^*</math> because:

# The <math>\mathbf{e}^i , i=1, 2, \dots, n, </math> are linear functionals, which map <math> x,y \in V </math> such as <math> x=  \alpha_1\mathbf{e}_1 + \dots + \alpha_n\mathbf{e}_n </math> and <math> y = \beta_1\mathbf{e}_1 + \dots + \beta_n \mathbf{e}_n </math> to scalars <math> \mathbf{e}^i(x)=\alpha_i </math> and <math> \mathbf{e}^i(y)=\beta_i</math>. Then also, <math> x+\lambda y=(\alpha_1+\lambda \beta_1)\mathbf{e}_1 + \dots + (\alpha_n+\lambda\beta_n)\mathbf{e}_n </math> and <math> \mathbf{e}^i(x+\lambda y)=\alpha_i+\lambda\beta_i=\mathbf{e}^i(x)+\lambda \mathbf{e}^i(y) </math>. Therefore, <math> \mathbf{e}^i  \in V^* </math> for <math> i= 1, 2, \dots, n </math>.
# Suppose <math> \lambda_1 \mathbf{e}^1 + \cdots + \lambda_n \mathbf{e}^n =0  \in V^*</math>. Applying this functional on the basis vectors of <math> V </math> successively, lead us to <math> \lambda_1=\lambda_2= \dots=\lambda_n=0 </math> (The functional applied in <math> \mathbf{e}_i </math> results in <math> \lambda_i </math>). Therefore, <math>\{\mathbf{e}^1,\dots,\mathbf{e}^n\}</math> is linearly independent on <math>V^*</math>.
#Lastly, consider <math> g \in V^* </math>. Then
:<math>
g(x)=g(\alpha_1\mathbf{e}_1 + \dots + \alpha_n\mathbf{e}_n)=\alpha_1g(\mathbf{e}_1) + \dots + \alpha_ng(\mathbf{e}_n)=\mathbf{e}^1(x)g(\mathbf{e}_1) + \dots + \mathbf{e}^n(x)g(\mathbf{e}_n)
</math>
and <math>\{\mathbf{e}^1,\dots,\mathbf{e}^n\}</math> generates <math>V^*</math>. Hence, it is a basis of <math> V^*</math>. 
{{Collapse bottom}}

For example, if <math>V</math> is <math>\R^2</math>, let its basis be chosen as <math>\{\mathbf{e}_1=(1/2,1/2),\mathbf{e}_2=(0,1)\}</math>. The basis vectors are not orthogonal to each other. Then, <math>\mathbf{e}^1</math> and <math>\mathbf{e}^2</math> are [[one-form]]s (functions that map a vector to a scalar) such that <math>\mathbf{e}^1(\mathbf{e}_1)=1</math>, <math>\mathbf{e}^1(\mathbf{e}_2)=0</math>, <math>\mathbf{e}^2(\mathbf{e}_1)=0</math>, and <math>\mathbf{e}^2(\mathbf{e}_2)=1</math>. (Note: The superscript here is the index, not an exponent.) This system of equations can be expressed using matrix notation as
:<math>
\begin{bmatrix}
e^{11} & e^{12} \\
e^{21} & e^{22}
\end{bmatrix}
\begin{bmatrix}
e_{11} & e_{21} \\
e_{12} & e_{22}
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}.
</math>
Solving for the unknown values in the first matrix shows the dual basis to be <math>\{\mathbf{e}^1=(2,0),\mathbf{e}^2=(-1,1)\}</math>. Because <math>\mathbf{e}^1</math> and <math>\mathbf{e}^2</math> are functionals, they can be rewritten as <math>\mathbf{e}^1(x,y)=2x</math> and <math>\mathbf{e}^2(x,y)=-x+y</math>.

In general, when <math>V</math> is <math>\R^n</math>, if <math>E=[\mathbf{e}_1|\cdots|\mathbf{e}_n]</math> is a matrix whose columns are the basis vectors and <math>\hat{E}=[\mathbf{e}^1|\cdots|\mathbf{e}^n]</math> is a matrix whose columns are the dual basis vectors, then
:<math>\hat{E}^\textrm{T}\cdot E = I_n,</math>
where <math>I_n</math> is the [[identity matrix]] of order <math>n</math>. The biorthogonality property of these two basis sets allows any point <math>\mathbf{x}\in V</math> to be represented as
:<math>\mathbf{x} = \sum_i \langle\mathbf{x},\mathbf{e}^i \rangle \mathbf{e}_i = \sum_i \langle \mathbf{x}, \mathbf{e}_i \rangle \mathbf{e}^i,</math>
even when the basis vectors are not orthogonal to each other. Strictly speaking, the above statement only makes sense once the inner product <math>\langle \cdot, \cdot \rangle</math> and the corresponding duality pairing are introduced, as described below in ''{{section link||Bilinear_products_and_dual_spaces}}''.

In particular, <math>\R^n</math> can be interpreted as the space of columns of <math>n</math> [[real number]]s, its dual space is typically written as the space of ''rows'' of <math>n</math> real numbers. Such a row acts on <math>\R^n</math> as a linear functional by ordinary [[matrix multiplication]]. This is because a functional maps every <math>n</math>-vector <math>x</math> into a real number <math>y</math>. Then, seeing this functional as a matrix <math>M</math>, and <math>x</math> as an <math>n\times 1</math> matrix, and <math>y</math> a <math>1\times 1</math> matrix (trivially, a real number) respectively, if <math>Mx=y</math> then, by dimension reasons, <math>M</math> must be a <math>1\times n</math> matrix; that is, <math>M</math> must be a row vector.

If <math>V</math> consists of the space of geometrical [[Vector (geometric)|vector]]s in the plane, then the [[level set|level curves]] of an element of <math>V^*</math> form a family of parallel lines in <math>V</math>, because the range is 1-dimensional, so that every point in the range is a multiple of any one nonzero element.
So an element of <math>V^*</math> can be intuitively thought of as a particular family of parallel lines covering the plane. To compute the value of a functional on a given vector, it suffices to determine which of the lines the vector lies on. Informally, this "counts" how many lines the vector crosses.
More generally, if <math>V</math>  is a vector space of any dimension, then the [[level sets]] of a linear functional in <math>V^*</math>  are parallel hyperplanes in <math>V</math>, and the action of a linear functional on a vector can be visualized in terms of these hyperplanes.<ref>{{harvnb|Misner|Thorne|Wheeler|1973|loc=§2.5}}</ref>