Dual space ($V^{\ast }={\mathrm{Hom}}_F(V,F)$)

The dual space of a vector space $V$ over a field $F$ is the vector space of all linear functionals $\phi :V\to F$. Think of a vector as an input and a linear functional as a scalar-valued measurement of that input. For $V={\mathbb{R}}^2$, the rules $(x,y)\mapsto x$, $(x,y)\mapsto y$, and $(x,y)\mapsto 2x-y$ are different elements of the dual space. None needs a notion of length or angle.

Its operations combine measurements pointwise:The dual basis explains the finite-dimensional structure: specifying a linear functional's values on a basis determines its value on every vector. Thus an $n$-dimensional vector space has an $n$-dimensional dual space. The spaces have the same dimension, but their elements play different roles: an input and a rule that measures an input. ^[ref].

This separation is useful when changing coordinates. The dual pairing $\phi (v)$ must stay the same while the component lists for $v$ and $\phi$ change in opposite ways. Covector transformation under a change of basis makes the compensation explicit. An inner product on a finite-dimensional real vector space gives one way to represent each linear functional by a vector, but that representation uses the chosen inner product; the dual space itself does not require one.

Here dual space means the algebraic dual: all linear functionals. For a topological vector space, the continuous dual space keeps only the continuous linear functionals. In infinite dimension, this can be a proper vector subspace; discontinuous functional detected by vanishing sequence tails gives an example. The distinction matters when a scalar-valued measurement must respond continuously to small input changes.

If a normed subspace $Z$ is dense in a normed vector space $X$ and carries the inherited norm, restriction gives an onto linear isometry from the dual space of $X$ to that of $Z$. A bounded functional on $Z$ extends uniquely by limits along sequences from $Z$; its norm is unchanged because its inequality passes to the limit. Thus completion does not change the continuous dual. For example, the subspace of sequences with $\sum j|a_j|<\infty$, equipped with the unweighted ${\ell }^1$ norm, has dual ${\ell }^{\infty }$, since it contains the dense subspace of finitely supported sequences. The inherited norm matters; the weighted norm would give a different dual norm.

The algebraic double dual of a vector space is the dual space of its algebraic dual. An element assigns a scalar linearly to each linear functional on the original space. The bidual evaluation map sends $v$ to the rule $\ell \mapsto \ell (v)$. It is a canonical linear isomorphism in finite dimension, unlike an identification of $V$ with $V^{\ast }$, which generally requires an additional choice. For spaces with topology, using the continuous dual space instead defines the topological bidual; that is a distinct construction in infinite dimension.

The dual pairing evaluates a linear functional on a vector:It is bilinear: it is linear in each argument separately. Even over the complex numbers, this algebraic evaluation has no complex conjugation. An inner product instead has two vectors as inputs and, over the complex numbers, is sesquilinear. For $\phi (x,y)=2x-y$ and $v=(3,1)$, the dual pairing equals five. Its value is independent of the basis used to describe its two inputs.

For a normed vector space $X$, a family $(x_i,f_i{)}_{i\in I}$ in $X\times X^{\ast }$ is biorthogonal when the displayed Kronecker-delta equations hold. Its vectors are linearly independent: applying $f_i$ to a finite linear dependence isolates its $i$th coefficient. No inner product is required, and the family need not span a dense subspace. If its linear span is dense, each prescribed coordinate functional is unique by continuity. For a countable total family, the finite sums $P_{n}x={\sum }_{i\le n}{f}_i(x)x_i$ are bounded projection operators, since biorthogonality gives $P_n^2=P_n$.

Restricting a linear functional to each summand gives a family of functionals, with no finite-support restriction. Conversely an arbitrary such family acts on a vector by the finite sum of its nonzero coordinates. These constructions are inverse and prove the displayed algebraic dual space identity. A group action on the summands induces the product action by their algebraic contragredients.

The bidual of a normed vector space $X$ is the continuous dual $X^{\ast \ast }=(X^{\ast }{)}^{\ast }$. Evaluation defines the canonical embedding into the bidual $J_X(x)(f)=f(x)$. The norm inequality gives $\Vert J_{X}x\Vert \le \Vert x\Vert$, and a norming functional from the Hahn-Banach theorem gives equality. Thus $J_X$ is isometric, and its image is dense in its closure, a completion of a normed space.

For finite-dimensional vector spaces $V,W$ over a field $F$, a bilinear pairing $B:V\times W\to F$ is perfect when the induced linear maps $V\to W^{\ast }$ and $W\to V^{\ast }$ are isomorphisms. Equivalently, both dimensions agree and the matrix of the pairing in any two bases is invertible. The evaluation pairing between a finite-dimensional vector space and its dual space is perfect. This notion allows two different spaces; a nondegenerate bilinear form is the case $V=W$.

A linear subspace $W\subseteq X^{\ast }$ is separating if every nonzero $x\in X$ has $T(x)\ne 0$ for some $T\in W$. For finite-dimensional $X$, an evaluation map $X\to F^{\dim W}$ is injective, so the rank-nullity theorem forces $\dim W\ge \dim X=\dim X^{\ast }$ and $W=X^{\ast }$. For the infinite-dimensional polynomial space, the span of its coefficient functionals is a proper separating linear subspace of the dual space, which contains all sequences of coefficients, not just finitely supported ones.

For an algebraic dual space $X^{\ast }={\mathrm{Hom}}_F(X,F)$, the linear map $J:X\to X^{\ast \ast }$ sends $x$ to evaluation at $x$. It is injective because linear functionals separate nonzero vectors: extend any nonzero vector to a basis and use its coordinate functional. If $X$ has a finite basis $e_i$ with dual basis $e_i^{\ast }$, every $B\in X^{\ast \ast }$ equals $J({\sum }_{i}B(e_i^{\ast })e_i)$, so $J$ is an isomorphism. For infinite-dimensional spaces, surjection does not follow from this finite sum argument.

A covector is an element of the dual space, hence a linear functional on a vector space. In finite-dimensional coordinates, it acts as a row multiplied by a coordinate vector column. The row $\left(\begin{array}{cc}2& -1\end{array}\right)$ acts on $\left(\begin{array}{c}3\\ 1\end{array}\right)$ to give five. The terminology emphasizes its role as a scalar-valued measurement and its behavior under a change of basis.

For example, the total derivative of $f(x,y)=x^2+y^2$ at $(1,2)$ sends a small displacement $h$ to the linear change $df(h)=2h_x+4h_y$. This covector describes the first-order response. The gradient $\nabla f=(2,4)$ represents the same rule through the chosen Euclidean inner product. In differential geometry, the cotangent space collects these scalar-valued measurements of tangent vectors.

Let the columns of an invertible matrix $B$ be a new basis expressed in the old basis. The same vector has old coordinate vector $x$ and new coordinate vector $x^{\prime }=B^{-1}x$. If a covector has old row $\ell$, its new row is ${\ell }^{\prime }=\ell B$, sinceThus the scalar dual pairing stays unchanged. If covector coefficients are stored as a column $c={\ell }^T$, their new column is $c^{\prime }=B^{T}c$; this is the inverse transpose of the transformation $x\mapsto B^{-1}x$ on vector coordinates. These conventions express the covariance and contravariance of vectors. ^[ref].

For the basis $b_1=(1,0)$, $b_2=(1,1)$,The linear functional $\phi (x,y)=2x-y$ has rows $\ell =\left(\begin{array}{cc}2& -1\end{array}\right)$ and ${\ell }^{\prime }=\left(\begin{array}{cc}2& 1\end{array}\right)$. Both calculations give five: $2\cdot 3-1\cdot 1=2\cdot 2+1\cdot 1$. Changing the basis changes the descriptions, not the measured value.

A linear functional is a linear map from a vector space $V$ to its scalar field $F$:Its output is one scalar, rather than an arbitrary vector in another space. In ${\mathbb{R}}^2$, $\phi (x,y)=2x-y$ gives $\phi (3,1)=5$, $\phi (1,2)=0$, and $\phi (4,3)=5$: adding the two inputs adds their measured values.

The level sets $\phi (v)=c$ visualize what the measurement retains and discards. For this example they are parallel straight lines $y=2x-c$. Moving parallel to $(1,2)$ changes the input but leaves $\phi$ unchanged. The zero level set is its kernel of a linear map, while the other level sets are translates of that vector subspace.

Inputs need not be geometric arrows. On the vector space of polynomials of polynomial degree at most two, evaluating at zero is a linear functional, as is taking the definite integral from zero to one. For $p(t)=a+bt+c{t}^2$, they give $p(0)=a$ and ${\int }_0^{1}p(t)dt=a+b/2+c/3$. Thus polynomial evaluation and integration are two different measurements of the same polynomial. Coordinate extraction is another measurement, developed under dual basis.

A linear functional is defined by linearity, not by continuity. In finite-dimensional real or complex normed vector spaces it is automatically continuous; in infinite dimension it need not be. The continuous dual space adds this requirement when small changes in the input should give small changes in the output.

For a nonzero Euclidean vector $w$, the Cauchy-Schwarz inequality gives $z\cdot w\ge -\Vert w\Vert$ when $\Vert z\Vert =1$. Equality is attained uniquely at $z=-w/\Vert w\Vert$. On a sphere of radius $r>0$ the minimizer is $-rw/\Vert w\Vert$, and the minimum is $-r\Vert w\Vert$. This turns many constrained vector problems into a normalization of one fixed vector. If $w=0$, every point minimizes the functional.

If a linear functional on a normed vector space has $f(v_N)=1$ for a sequence tending to zero in norm, it is discontinuous and unbounded. An explicit example is the algebraic span in ${\ell }^2$ of the coordinate vectors and $w=(2^{-n}{)}_{n\ge 1}$. Define $f(w)=1$ and $f(e_n)=0$. The tails $v_N=w-{\sum }_{n=1}^N{2}^{-n}{e}_n$ have norm $2^{-N}/\sqrt{3}$ but image one. Thus the functional vanishes on the dense finite-support subspace while being nonzero elsewhere.

For a basis $e_1,\dots ,e_n$ of a finite-dimensional vector space $V$, its dual basis consists of linear functionals $e_1^{\ast },\dots ,e_n^{\ast }$ satisfyingThe Kronecker delta means that each rule gives one on its own basis vector and zero on the others. These rules extract coordinates:Applying $e_i^{\ast }$ to a linear combination removes every term except the $i$th. Consequently the dual basis is a basis of the dual space, and every linear functional has expansion $\phi ={\sum }_i\phi (e_i)e_i^{\ast }$. ^[ref].

Coordinate extraction works for a nonorthogonal basis too. With $b_1=(1,0)$ and $b_2=(1,1)$, the dual basis isThe checks $b_1^{\ast }(b_1)=1$, $b_1^{\ast }(b_2)=0$, $b_2^{\ast }(b_1)=0$, and $b_2^{\ast }(b_2)=1$ fix the two rules. For $v=(3,1)$ they give coordinates $(2,1)$, so $v=2b_1+b_2$. The usual dot products with $b_1,b_2$ instead give $3,4$ and do not extract these coordinates: simple projection formulas depend on an orthonormal basis.

The same idea explains polynomial interpolation. On polynomials of polynomial degree at most $n$, evaluation at distinct interpolation nodes $x_0,\dots ,x_n$ gives the dual basis to the Lagrange cardinal polynomials, because ${\ell }_i(x_j)={\delta }_{ij}$. Samples are coordinates in this basis, so the reconstruction $p={\sum }_{i}p(x_i){\ell }_i$ is a dual-basis expansion. Differentiation weights from nodal evaluations and factorial monomial basis for polynomial differentiation apply the same principle to different measurements.

For $p(X)=a+bX+c{X}^2$, the three point evaluation functionals at $0,1,-1$ form a basis of the dual space of real polynomials of degree at most two. The displayed recovery formulas express the coordinate functionals, hence the dual basis to $1,X,X^2$, as linear combinations of these evaluations. More generally distinct-node polynomial interpolation identifies values with coefficient data through an invertible Vandermonde matrix.

For the basis $v_j=e_1+\cdots +e_j$ obtained from a basis $e_1,\dots ,e_n$, the dual basis is obtained by successive differences of the original dual basis. Indeed, if $x={\sum }_j{c}_j{v}_j$, its original coordinates are $x_i={\sum }_{j\ge i}{c}_j$. Hence $c_i=x_i-x_{i+1}$ for $i<n$, and $c_n=x_n$. These are precisely the new coordinate linear functionals. This gives an elementary example of the inverse relation between change of basis and covector transformation under a change of basis.

A coordinate functional extracts one coefficient in a specified finite basis. It is an element of that basis's dual basis, rather than a coordinate rule independent of the basis. For the standard basis of $F^n$, the $i$th coordinate functional sends $(a_1,\dots ,a_n)$ to $a_i$.

On polynomials of degree at most $n$, the derivative-evaluation functionals $e_j(\varphi )={\varphi }^{(j)}(0)$ are dual to $p_j=x^j/j!$, since $e_i(p_j)={\delta }_{ij}$. Differentiation maps $p_j$ to $p_{j-1}$, with $p_0$ mapped to zero. Its matrix is a single nilpotent Jordan block with ones above the diagonal, and its dual has the transpose matrix.

For a vector subspace $U\le V$,In finite dimension, $\dim U^{\circ }=\dim V-\dim U$.

An annihilator of a vector subspace collects the linear functionals that ignore every input from that vector subspace. For $U=\mathrm{span}\{(1,2)\}\subseteq {\mathbb{R}}^2$, a linear functional $\phi (x,y)=ax+by$ vanishes on $U$ exactly when $a+2b=0$. Hence $U^{\circ }$ is the one-dimensional span of $(x,y)\mapsto 2x-y$. The zero level set in the linear functional diagram is precisely $U$.

For a vector subspace $U$ of a finite-dimensional vector space $V$, its annihilator of a vector subspace $U^0\subseteq V^{\ast }$ determines $U$ exactly: $U={\bigcap }_{f\in U^0}\ker f$. Indeed, a basis of $U$ can be extended to one of $V$. If $v\notin U$, include $v$ among the added basis vectors and define a linear functional taking value one at $v$ and zero on all the other basis vectors. This functional vanishes on $U$ but not at $v$, excluding every outside vector from the intersection.

It suffices to intersect the kernels of any basis $g_1,\dots ,g_k$ of $U^0$: every other annihilating functional is their linear combination. The dual basis for a basis of $V$ extending a basis of $U$ shows $k=\dim V-\dim U$. If $U=V$, the empty intersection is $V$. This translates a subspace description by spanning vectors into an exact description by independent linear equations.

For a linear map $T:V\to V$, an invariant subspace $U$ has an annihilator of a vector subspace $U^0$ invariant under the dual map $T^{\prime }$. Indeed, for $\phi \in U^0$ and $u\in U$,For a finite-dimensional vector space, the converse also holds. If $T^{\prime }(U^0)\subseteq U^0$, every $\phi \in U^0$ vanishes on $Tu$ for each $u\in U$. The dual basis description of $U^0$ shows that the common zero set of its linear functionals is exactly $U$, so $Tu\in U$. ThusThe dimension identity $\dim U^0=\dim V-\dim U$ exchanges a one-dimensional invariant subspace with an invariant hyperplane in the dual space.

For a linear map $T:V\to W$, the dual map $T^{\ast }:W^{\ast }\to V^{\ast }$ sends a linear functional $\phi$ on the output space to the input measurement $T^{\ast }(\phi )=\phi \circ T$. It reverses direction because a measurement of the output can be evaluated on an input by first applying $T$:For $T(x,y)=(x+y,2y)$ and $\phi (u,v)=2u-v$, the input measurement is $T^{\ast }\phi (x,y)=2(x+y)-2y=2x$. The output measurement ignores the part of this transformation contributed by $y$.

If $T$ has matrix $A$ in specified bases, a row representing $\phi$ changes to $\ell A$. Storing the coefficients of a linear functional as a column in the corresponding dual basis gives $A^T{\ell }^T$, explaining why the transpose of a linear map is represented by the matrix transpose. This algebraic operation uses no inner product; it agrees with a matrix transpose even over the complex numbers, rather than automatically introducing complex conjugation. The adjoint operator is defined using an inner product and is a different construction. ^[ref].

In finite dimension, the kernel and image of the dual map are described by annihilators of vector subspaces:

For a linear map between finite-dimensional vector spaces, the image of its dual map is the annihilator of a vector subspace given by its kernel. Inclusion follows by composition. Conversely a linear functional vanishing on the kernel descends to the image and extends to the target by basis extension. The identity and the rank-nullity theorem show that a linear map and its dual have equal rank.

For the map represented by a matrix $A$, column rank is the rank of the map and row rank is the rank of its dual. Since the dual map has kernel equal to the annihilator of the original image, rank-nullity gives equal ranks.

If the columns $v_1,\dots ,v_m$ of an $n$ by $m$ matrix are independent, some $m$ by $m$ row minor is nonzero. A nonzero term of that minor's determinant selects distinct nonzero coordinates $f(i)$, and replacing the corresponding standard basis vectors $e_{f(i)}$ by the $v_i$ leaves a basis.

For a finite-dimensional vector space and $U\le V$, dualizing the quotient and inclusion maps gives

If $q_1,\dots ,q_n\in V^{\ast }$ are linearly independent, thenmaps $V$ onto $F^n$. Otherwise a nonzero functional annihilating its proper image would give a nontrivial linear relation among the $q_j$.

A polynomial moment functional evaluates a polynomial through a weighted integral, such as $p\mapsto \int p(t)w(t)dt$; finitely many moments determine it on a bounded-degree polynomial space.