The notion of dual space

Linear maps: Dual vector spaces

The notion of dual space

According to Linearity of sum and scalar multiples of linear maps we can add linear maps from one vector space to another and multiply them by scalars. These operations comply with the rules for a vector space:

Let $V$ and $W$ be vector spaces. The set $L(V,W)$ of linear images from $V$ to $W$ is a vector space with the known addition and scalar multiplication.

If $V$ has finite dimension $n$ and $W$ has finite dimension $m$ , then the vector space $L(V,W)$ is isomorphic to the vector space $M_{m\times n}$ of all $(m\times n)$ -matrices. More precisely, if we choose a basis $\alpha$ of $V$ and a basis $\beta$ of $W$ , then the map $\varphi : L(V,W)\to M_{m\times n}$ determined by $\varphi(L) =\left(\beta L\alpha^{-1}\right)_\varepsilon$ is an isomorphism.

The fact that $L(V,W)$ is a vector space is easy to verify, with the obvious zero vector and opposite.

Let $L$ , $M$ , $N$ lie in $L(V,W)$ and let $\lambda$ , $\mu$ be numbers. We run by the eight rules of a vector space:

Commutativity: $L+M =M+L$ . This follows from the commutativity of addition in $W$ because, for $\vec{v}$ in $V$ , we have $\begin{array}{rcl}(L+M)(\vec{v})& =& L(\vec{v})+M(\vec{v}) \\&&\phantom{xx}\color{blue}{\text{definition of addition in }L(V,W)}\\ &=& M(\vec{v})+L(\vec{v})\\&&\phantom{xx}\color{blue}{\text{commutativity of addition in }W}\\ &=& (M+L)(\vec{v})\\&&\phantom{xx}\color{blue}{\text{definition of addition in }L(V,W)} \end{array}$
Associativity of addition: $(L+M)+N =L+(M+N)$ . This follows from the associativity of the addition in $W$ : $\begin{array}{rcl}((L+M)+N)(\vec{v})& =& (L+M)(\vec{v})+N(\vec{v}) \\&&\phantom{xx}\color{blue}{\text{definition of addition in }L(V,W)}\\& =& \left(L(\vec{v})+M(\vec{v})\right)+N(\vec{v}) \\&&\phantom{xx}\color{blue}{\text{definition of addition in }L(V,W)}\\ &=& L(\vec{v})+\left(M(\vec{v})+N(\vec{v}))\right)\\&&\phantom{xx}\color{blue}{\text{associativity of addition in }W}\\ &=& L(\vec{v})+(M+N)(\vec{v})\\&&\phantom{xx}\color{blue}{\text{definition of addition in }L(V,W)} \\&=& (L+(M+N))(\vec{v})\\&&\phantom{xx}\color{blue}{\text{definition of addition in }L(V,W)} \end{array}$
Zero vector: The function $0$ which assigns to each vector in $V$ the vector $\vec{0}$ of $W$ satisfies the property $L+0=L$ .
Opposite: the opposite $-L$ of $L$ is the function that assigns to $\vec{v}$ in $V$ the vector $-L(\vec{v})$ in $W$ .
Scalar one: the number $1$ satisfies $1\cdot L=L$ because if $\vec{v}$ in $V$ , then $(1\cdot L)(\vec{v}) =1\cdot L(\vec{v}) = L(\vec{v})$
Associativity of the scalar multiplication: $(\lambda \cdot\mu )\cdot L=\lambda \cdot(\mu\cdot L)$ . This follows from the associativity of the scalar multiplication in $W$ .
Distributivity of scalar multiplication over scalar addition: $(\lambda +\mu )\cdot L=\lambda\cdot L +\mu \cdot L$ . This follows from the distributivity of scalar multiplication over scalar addition in $W$ .
Distributivity of scalar multiplication over vector addition: $\lambda\cdot (L+M)=\lambda\cdot L +\lambda\cdot M$ . This follows from the distributivity of scalar multiplication over vector addition in $W$ .

For proof of the second statement, we choose a basis $\alpha$ of $V$ and a basis $\beta$ of $W$ . Let $L$ be an element of $L(V,W)$ . Then $\beta L\alpha^{-1}$ is the linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$ determined by the matrix $\left(\beta L\alpha^{-1}\right)_\varepsilon$ in $M_{m\times n}$ .

We prove that the map $\varphi : L(V,W)\to M_{m\times n}$ given by $\varphi(L) =\left(\beta L\alpha^{-1}\right)_\varepsilon$ is an isomorphism.

First, $\varphi$ is linear because, if $L$ and $M$ belong to $L(V,W)$ and $\lambda$ , $\mu$ are scalars , then

$\begin{array}{rcl}\varphi(\lambda\,L+\mu\, M) &=&\left(\beta (\lambda\,L+\mu\, M)\alpha^{-1}\right)_\varepsilon \\ &=& \left(\left(\lambda\,\beta (L)+\mu\, \beta(M)\right)\alpha^{-1}\right)_\varepsilon \\ &=& \left(\lambda\,\beta (L)\alpha^{-1}+\mu\, \beta(M)\alpha^{-1}\right)_\varepsilon \\ &=&\lambda\, \left(\beta (L)\alpha^{-1}\right)_\varepsilon+\mu\, \left(\beta(M)\alpha^{-1}\right)_\varepsilon \\ &=&\lambda\, \varphi(L)+\mu\, \varphi( M)\end{array}$

Second, $\varphi$ is invertible because the map $\psi: M_{m\times n}\to L(V,W)$ given by $\psi(A)=\beta^{-1} L_A \alpha$ is the inverse of $\varphi$ : $\begin{array}{rcl}\varphi(\psi(A)) &=& \varphi(\beta^{-1} L_A \alpha) \\ &=& \left(\beta(\beta ^{-1}L_A \alpha) \alpha^{-1}\right)_\varepsilon \\ &=& \left(L_A \right)_\varepsilon \\ &=& A\end{array}$

The map $\varphi$ assigns to the linear mapping $L:V\to W$ the matrix of the linear map $L$ with respect to the basis $\alpha$ of $V$ and the basis $\beta$ of $W$ .

Since isomorphic vector spaces have the same dimension and $\dim{M_{m\times n}}=m\cdot n$ , we have

$\dim{L(V,W)} = m\cdot n$ if $\dim{V} = n$ and $\dim{W} = m$ .

A special case occurs if $W=\mathbb{R}$ . We then speak of linear functions or linear functionals on $V$ . We now focus on the space of linear functions.

Let $V$ be a real vector space. The vector space $L(V,\mathbb{R})$ of linear maps $V\rightarrow \mathbb{R}$ is called the dual space of $V$ . This vector space is denoted as $V^\star$ .

The "sum of the first and second coordinate'' is a linear function $L:\mathbb{R}^3\to\mathbb{R}$ , that is to say, an element of $(\mathbb{R}^3)^*$ . In a formula, $L(\rv{x_1, x_2, x_3})= x_1 + x_2$ . The function $M:\mathbb{R}^3 \rightarrow \mathbb{R}$ given by $M(\rv{x_1, x_2, x_3}) = x_1^2$ is not linear, and so does not belong to $(\mathbb{R}^3)^*$ .

Let $F$ be the vector space of all functions from $\mathbb{R}$ to itself. The image $A : F \rightarrow \mathbb{R}, \quad A(f)= f(0)$ is linear because $A (\lambda \,f+\mu\, g)=(\lambda \,f+\mu\, g)(0) = \lambda\, f(0) + \mu\, g(0) = \lambda\, A( f )+\mu\, A (g)$ . So $A$ belongs to $F^\star$ .
Let $W$ be the space of infinitely often differentiable functions $\mathbb{R}$ . The image $D_0 : W \rightarrow \mathbb{R}, \quad D_0(f) = f'(0)$ is linear as well. As a consequence, $D_0$ belongs to $W^\star$ .

By the above theorem The linear space of linear maps, the dimension of the dual space $V^\star$ of $V$ is equal to that of $V$ as the dimension of $V$ is finite. In that case, $V^\star$ is isomorphic to $V$ . This is not the case if $V$ is infinite-dimensional. To illustrate this, consider the infinite-dimensional vector space $P$ of all polynomials in $x$ . Let $\delta_i$ be the linear function $P$ that assigns the coefficient of $x^i$ to each polynomial, and consider the set $Q$ of all infinite sums $\sum_{i=0}^\infty d_i\delta_i \quad \text{ for }d_i\in \mathbb{R}$ Here we allow $d_i\ne0$ for an infinite number of $i$ . Each element $q = \sum_{i=0}^\infty d_i\delta_i$ of $Q$ can be seen as a function on $P$ because, for every polynomial $p(x) = \sum_{j= 0}^m a_ix^i$ the right hand side of the expression

$q\left(p(x)\right) = \sum_{j= 0}^m d_i\cdot a_i$

is the sum of a finite number of real numbers. Moreover, $Q$ is a vector space with the usual addition and scalar multiplication. We are not concerned about convergence; we just deal with abstract symbols. The newly defined function $q(p(x))$ is a linear function on $P$ and thus an element of $P^\star$ . Since the map that assigns to $q$ the function $q(p(x))$ is injective, we can view $Q$ as a linear subspace of $P^\star$ . But the space $Q$ is much larger than the space $P$ because it has the cardinality of all infinite sequences of real numbers $(d_i)$ , unlike $P$ , which has the cardinality of all finite sequences of real numbers.

In the case of a vector space $V$ of finite dimension, we can use an inner product to construct an isomorphism between the dual space $V^\star$ and $V$ .

Inner product with fixed vector is linear mapping

If $V$ is a real inner product space and $\vec{a} \in V$ , then the map $L_{\vec{a}} : V \rightarrow \mathbb{R}$ , defined by $L_{\vec{a}}( \vec{x}) = \dotprod{\vec{a} }{\vec{x}}$ , is a linear function.

The linear map that assigns to $\vec{a}$ the element $L_{\vec{a}}$ of $V^\star$ , is injective. In particular, it is an isomorphism $V\to V^\star$ if $V$ is finite-dimensional.

This follows from the linearity in the second argument of the inner product: $\dotprod{\vec{a} }{\left( \lambda\, \vec{x} + \mu\, \vec{y}\right)}= \lambda\, (\dotprod{\vec{a}}{ \vec{x}}) + \mu \,(\dotprod{\vec{a}}{ \vec{y}})$ .

To prove the injectivity we assume that $\vec{a}$ satisfies $L_{\vec{a}} =0$ , the zero map $V\to \mathbb{R}$ . Then $\dotprod{\vec{a} }{\vec{x}}=0$ all $\vec{x}$ , and so also for $\vec{x}=\vec{a}$ . This means that $\dotprod{\vec{a} }{\vec{a}}=0$ . The positivity condition of the inner product then yields $\vec{a} = \vec{0}$ . Therefore, the kernel of the linear map that assigns to $\vec{a}$ the map $L_{\vec{a}}$ , consists only of $\vec{0}$ . From Criteria for injectivity and surjectivity it follows that the map is injective.

If $\dim{V}=n\lt\infty$ , then $\dim{V^\star} = n$ by the above theorem. Consequenly, the first Invertability criteria for a linear mapping implies that the map is invertible, and thus is an isomorphism.

The null space of the map $L_{\vec{a}}$ is exactly the orthoplement $\{\vec{a}\}^{\perp}$ .

The image is $\mathbb{R}$ if $\vec{a} \neq \vec{0}$ and $\{\vec{0}\}$ if $\vec{a} =\vec{0}$ .

Let $V$ be the vector space $\mathbb{R}^3$ with standard inner product. Furthermore, let $\varepsilon =\basis{\vec{e_1},\vec{e_2},\vec{e_3}}$ be the standard basis of $\mathbb{R}^3$ , and $\beta = \basis{e_1^\star,e_2^\star,e_3^\star}$ the basis of $V^\star$ where $e_1^\star$ is given by $e_i^\star(\vec{x}) = \dotprod{\vec{e_i}}{\vec{x}}$ .

What is the matrix of the image $L:V\to V^\star$ with respect to the bases $\alpha$ of $V$ and $\beta$ of $V^\star$ that assigns to $\vec{a}$ in $V$ the dual vector $L_{\vec{a}}$ , given by $L_{\vec{a}}(\vec{v}) = \dotprod{\vec{a}}{\vec{v}}$ ?

If $\vec{a} = \sum_{i=1}^3a_i\vec{e_i}$ , then
$L_{\vec{a}}(\vec{v}) = \dotprod{\vec{a}}{\vec{v}}=\sum_{i=1}^3a_i\dotprod{\vec{e_i}}{\vec{v}}= \sum_{i=1}^3a_i{e_i}^\star(\vec{v})$
so $L_{\vec{a}} = \sum_{i=1}^3a_i{e_i}^\star$
Therefore, the image of the basis vector $\vec{e_j}$ under $L$ is ${e_j}^\star$ , which implies that the requested matrix is equal to ${}_{\beta}L_{\varepsilon} = I_3$

New example