# Chapter 28. Dual Spaces # Chapter 28. Dual Spaces The dual space of a vector space is the vector space of all linear maps from that space to its field of scalars. If \(V\) is a vector space over \(F\), then its dual space is written $$ V^* $$ and is defined by $$ V^*=\{\varphi:V\to F:\varphi \text{ is linear}\}. $$ The elements of \(V^*\) are called linear functionals, covectors, or linear forms. A linear functional takes a vector as input and returns a scalar. The dual space itself is a vector space under pointwise addition and scalar multiplication. ## 28.1 Linear Functionals A linear functional on \(V\) is a linear map $$ \varphi:V\to F. $$ It satisfies $$ \varphi(u+v)=\varphi(u)+\varphi(v) $$ and $$ \varphi(cv)=c\varphi(v). $$ The output is a scalar, not a vector. For example, define $$ \varphi:\mathbb{R}^3\to\mathbb{R} $$ by $$ \varphi(x,y,z)=2x-y+5z. $$ Then \(\varphi\) is linear. It takes a vector in \(\mathbb{R}^3\) and returns one real number. ## 28.2 The Dual Space The dual space \(V^*\) is the set of all linear functionals on \(V\): $$ V^*=\operatorname{Hom}(V,F). $$ If \(\varphi,\psi\in V^*\), define $$ (\varphi+\psi)(v)=\varphi(v)+\psi(v). $$ If \(c\in F\), define $$ (c\varphi)(v)=c\varphi(v). $$ With these operations, \(V^*\) is a vector space. The zero vector of \(V^*\) is the zero functional: $$ 0(v)=0 $$ for every \(v\in V\). ## 28.3 Functionals on \(\mathbb{R}^n\) Every linear functional on \(\mathbb{R}^n\) has the form $$ \varphi(x)=a_1x_1+a_2x_2+\cdots+a_nx_n. $$ Equivalently, $$ \varphi(x)=a^Tx, $$ where $$ a= \begin{bmatrix} a_1\\ a_2\\ \vdots\\ a_n \end{bmatrix}. $$ Thus a linear functional on \(\mathbb{R}^n\) may be represented by a row vector: $$ \varphi= \begin{bmatrix} a_1&a_2&\cdots&a_n \end{bmatrix}. $$ This representation depends on the standard basis. ## 28.4 Covectors Vectors in \(V\) and vectors in \(V^*\) have different roles. A vector \(v\in V\) is an input object. A covector \(\varphi\in V^*\) is a scalar-valued linear measurement. The pairing is written $$ \varphi(v). $$ Some texts also write $$ \langle \varphi,v\rangle. $$ The pairing takes one covector and one vector and returns a scalar. It is linear in each argument. ## 28.5 The Dual Basis Let $$ B=(v_1,\ldots,v_n) $$ be a basis of \(V\). The dual basis is the basis $$ B^*=(v^1,\ldots,v^n) $$ of \(V^*\) defined by $$ v^i(v_j)=\delta^i_j. $$ Here \(\delta^i_j\) is the Kronecker delta: $$ \delta^i_j= \begin{cases} 1,& i=j,\\ 0,& i\neq j. \end{cases} $$ Thus \(v^i\) extracts the \(i\)-th coordinate relative to the basis \(B\). ## 28.6 Meaning of the Dual Basis If $$ v=c_1v_1+\cdots+c_nv_n, $$ then $$ v^i(v)=c_i. $$ So the dual basis functional \(v^i\) reads off the \(i\)-th coordinate of \(v\). For example, in \(\mathbb{R}^3\) with the standard basis, $$ e^1(x,y,z)=x, $$ $$ e^2(x,y,z)=y, $$ $$ e^3(x,y,z)=z. $$ The dual basis consists of the coordinate projection maps. ## 28.7 Basis of the Dual Space If \(V\) has basis $$ B=(v_1,\ldots,v_n), $$ then the dual basis $$ B^*=(v^1,\ldots,v^n) $$ is a basis of \(V^*\). Therefore $$ \dim V^*=\dim V $$ when \(V\) is finite-dimensional. Every functional \(\varphi\in V^*\) can be written uniquely as $$ \varphi=a_1v^1+\cdots+a_nv^n. $$ The coefficients are $$ a_i=\varphi(v_i). $$ Indeed, for $$ v=c_1v_1+\cdots+c_nv_n, $$ we have $$ \varphi(v)=c_1\varphi(v_1)+\cdots+c_n\varphi(v_n). $$ ## 28.8 Example in \(\mathbb{R}^2\) Let \(V=\mathbb{R}^2\) with standard basis $$ e_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad e_2= \begin{bmatrix} 0\\ 1 \end{bmatrix}. $$ The dual basis is $$ e^1(x,y)=x, \qquad e^2(x,y)=y. $$ Let $$ \varphi(x,y)=3x-5y. $$ Then $$ \varphi=3e^1-5e^2. $$ The coordinate vector of \(\varphi\) in the dual basis is $$ [\varphi]_{B^*} = \begin{bmatrix} 3\\ -5 \end{bmatrix}. $$ ## 28.9 Example with a Nonstandard Basis Let $$ B=(v_1,v_2) $$ where $$ v_1= \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 1\\ -1 \end{bmatrix}. $$ We seek the dual basis $$ B^*=(v^1,v^2). $$ Write $$ v^1(x,y)=ax+by. $$ The conditions are $$ v^1(v_1)=1, \qquad v^1(v_2)=0. $$ Thus $$ a+b=1, \qquad a-b=0. $$ Solving gives $$ a=b=\frac12. $$ So $$ v^1(x,y)=\frac12x+\frac12y. $$ Similarly, write $$ v^2(x,y)=cx+dy. $$ The conditions are $$ c+d=0, \qquad c-d=1. $$ Solving gives $$ c=\frac12, \qquad d=-\frac12. $$ Thus $$ v^2(x,y)=\frac12x-\frac12y. $$ ## 28.10 Dual Basis and Inverse Matrices Let \(B=(v_1,\ldots,v_n)\) be a basis of \(\mathbb{R}^n\), and let $$ P_B= \begin{bmatrix} |&|&&|\\ v_1&v_2&\cdots&v_n\\ |&|&&| \end{bmatrix}. $$ The dual basis functionals are the rows of $$ P_B^{-1}. $$ Indeed, the condition $$ v^i(v_j)=\delta^i_j $$ says that the \(i\)-th dual row applied to the \(j\)-th basis column gives the \((i,j)\)-entry of the identity matrix. Thus $$ P_B^{-1}P_B=I. $$ This is often the fastest way to compute a dual basis in coordinates. ## 28.11 The Natural Pairing There is a natural pairing $$ V^*\times V\to F $$ defined by $$ (\varphi,v)\mapsto \varphi(v). $$ This pairing is bilinear: $$ (\varphi+\psi)(v)=\varphi(v)+\psi(v), $$ $$ \varphi(u+v)=\varphi(u)+\varphi(v), $$ and scalar multiplication can be moved through either side. The pairing connects a vector space with its dual without requiring an inner product. ## 28.12 Dual Space Versus Inner Product Identification In \(\mathbb{R}^n\), every vector \(a\) defines a functional $$ \varphi_a(x)=a^Tx. $$ This uses the standard dot product. It is tempting to identify \(V\) and \(V^*\). In finite-dimensional inner product spaces, this can be done. But the identification depends on the inner product. Without an inner product, \(V\) and \(V^*\) are separate spaces. A vector is not automatically a covector. This distinction becomes important in geometry, tensor algebra, differential forms, and functional analysis. ## 28.13 The Double Dual The dual of the dual space is $$ V^{**}=(V^*)^*. $$ There is a natural map $$ J:V\to V^{**} $$ defined by $$ J(v)(\varphi)=\varphi(v). $$ Thus \(J(v)\) is a functional on \(V^*\): it takes a covector \(\varphi\) and evaluates it at \(v\). If \(V\) is finite-dimensional, then \(J\) is an isomorphism: $$ V\cong V^{**}. $$ This identification is canonical. It does not require choosing a basis. ## 28.14 Transpose of a Linear Map Let $$ T:V\to W $$ be a linear map. The dual map, also called the transpose or pullback, is $$ T^*:W^*\to V^* $$ defined by $$ T^*(\psi)=\psi\circ T. $$ That is, if \(\psi\) is a functional on \(W\), then \(T^*(\psi)\) is a functional on \(V\). For \(v\in V\), $$ (T^*\psi)(v)=\psi(Tv). $$ The direction reverses: \(T\) goes from \(V\) to \(W\), while \(T^*\) goes from \(W^*\) to \(V^*\). ## 28.15 Matrix of the Dual Map Suppose \(T:V\to W\) is represented by a matrix \(A\) with respect to bases of \(V\) and \(W\). Then \(T^*:W^*\to V^*\) is represented by the transpose matrix $$ A^T $$ with respect to the corresponding dual bases. This explains the name transpose. If $$ y=Ax $$ describes the original map on column vectors, then the dual map sends row functionals backward: $$ \psi \mapsto \psi A. $$ In column-coordinate notation for dual bases, this becomes multiplication by \(A^T\). ## 28.16 Annihilators Let \(S\subseteq V\). The annihilator of \(S\) is $$ S^0=\{\varphi\in V^*:\varphi(s)=0\text{ for all }s\in S\}. $$ It is a subspace of \(V^*\). If \(U\subseteq V\) is a subspace, then \(U^0\) consists of all linear functionals that vanish on \(U\). For example, in \(\mathbb{R}^2\), if $$ U=\operatorname{span} \left( \begin{bmatrix} 1\\ 0 \end{bmatrix} \right), $$ then a functional $$ \varphi(x,y)=ax+by $$ vanishes on \(U\) exactly when $$ a=0. $$ Thus $$ U^0=\{\varphi(x,y)=by:b\in\mathbb{R}\}. $$ ## 28.17 Dimension of an Annihilator If \(V\) is finite-dimensional and \(U\subseteq V\), then $$ \dim U^0=\dim V-\dim U. $$ This is analogous to the dimension of an orthogonal complement. Indeed, if $$ \dim V=n $$ and $$ \dim U=k, $$ then a basis of \(U\) can be extended to a basis of \(V\). The dual basis functionals that vanish on \(U\) correspond exactly to the basis vectors added outside \(U\). Thus the annihilator measures the number of independent linear conditions that ignore \(U\). ## 28.18 Quotients and Dual Spaces There is a natural relation between quotient spaces and annihilators: $$ (V/U)^*\cong U^0. $$ A functional on \(V/U\) corresponds to a functional on \(V\) that vanishes on \(U\). Indeed, if $$ \lambda:V/U\to F $$ is linear, then $$ \varphi(v)=\lambda(v+U) $$ defines a linear functional on \(V\), and \(\varphi\) vanishes on \(U\). Conversely, if \(\varphi\in U^0\), then $$ \lambda(v+U)=\varphi(v) $$ is well-defined. ## 28.19 Common Notation | Notation | Meaning | |---|---| | \(V^*\) | Dual space of \(V\) | | \(\varphi,\psi\) | Typical linear functionals | | \(B^*\) | Dual basis of \(B\) | | \(v^i\) | \(i\)-th dual basis functional | | \(\varphi(v)\) | Pairing of functional and vector | | \(T^*\) | Dual map or transpose of \(T\) | | \(S^0\) | Annihilator of \(S\) | | \(V^{**}\) | Double dual | ## 28.20 Summary The dual space \(V^*\) is the vector space of all linear functionals on \(V\). Its elements measure vectors by producing scalars. The key ideas are: | Concept | Meaning | |---|---| | Linear functional | Linear map \(V\to F\) | | Dual space | Space of all linear functionals | | Covector | Another name for element of \(V^*\) | | Dual basis | Functionals \(v^i\) satisfying \(v^i(v_j)=\delta^i_j\) | | Natural pairing | Evaluation \(\varphi(v)\) | | Double dual | \(V^{**}\), naturally containing \(V\) | | Dual map | \(T^*(\psi)=\psi\circ T\) | | Annihilator | Functionals that vanish on a subset | Dual spaces turn vectors into objects that can be measured linearly. They are the natural home of coordinates, linear equations, transpose maps, annihilators, and many constructions that appear later in geometry and analysis.