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The Banach space ${\ell }^1$ consists of absolutely summable scalar sequences, with norm $\Vert x{\Vert }_1={\sum }_n|x_n|$. Its continuous dual is ${\ell }^{\infty }$, under the pairing ${\sum }_n{x}_n{y}_n$. Finitely supported sequences are norm dense, which identifies a linear functional from its values on coordinate vectors. The space of sequences converging to zero inside this dual is a norming subspace of infinite codimension.

For example, $x=(2^{-1},2^{-2},\dots )$ belongs to the absolutely summable sequence space and has norm one, by the geometric series formula. Merely tending to zero is insufficient: $(1/n)$ tends to zero but its series of absolute values diverges. Thus this sequence space records total absolute size, rather than just the eventual size of individual coordinates. Its completeness ensures that a Cauchy sequence in this total-error norm has an absolutely summable limit.^[ref]

For a bounded linear operator $T:H\to K$ between real or complex Hilbert spaces, its adjoint is the unique bounded linear operator $T^{\ast }:K\to H$ satisfyingThe inner product here is linear in its first argument. The adjoint operator transfers an output measurement back to the input: measuring $Tx$ against the vector $y$ gives exactly the same scalar as measuring $x$ against $T^{\ast }y$. It depends on both inner products.

In this real example,Both inner products equal $3$. The dashed lines collect all vectors giving that same measurement: the output line $u_1=3$ becomes the input line $2x_1+x_2=3$. The adjoint operator therefore moves the direction that describes the measurement. It generally differs from the inverse matrix: here $A^{-1}y=(1/2,0{)}^{\mathrm{T}}$.

For an explicitly complex example, use the standard inner product $⟨u,v⟩={\sum }_j{u}_j\overline{v_j}$ on ${\mathbb{C}}^2$. In the standard orthonormal basis, the adjoint operator is represented by the conjugate transpose:For $x=(1+i,2{)}^{\mathrm{T}}$ and $y=(1,i{)}^{\mathrm{T}}$, direct matrix multiplication givesand henceThe complex conjugate is essential to this agreement.

Existence follows from the Riesz representation theorem: for fixed $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ has a unique representing vector $T^{\ast }y$. The resulting adjoint operator has the same operator norm as $T$, andMIT's lecture 18 develops the representation argument and the conjugate transpose example.^[ref]

This transfer explains the normal equation for a linear inverse problem: a least-squares solution has residual $Tx-b$ orthogonal to every attainable direction $Th$, so $⟨h,T^{\ast }(Tx-b)⟩=0$ for every $h$, equivalently $T^{\ast }Tx=T^{\ast }b$. An output residual becomes an input condition through the same adjoint operator. All identities in this section concern bounded operators defined on the whole source Hilbert space; for a densely defined operator that is not bounded, use the adjoint of a densely defined operator and check its operator domain.

In a finite-dimensional real vector space, a level set of a nonzero linear functional is an affine hyperplane. Choosing $v_0$ in that level set gives $H=v_0+\ker \ell$, so its dimension is one less than that of the ambient space. With a Euclidean inner product, translation followed by an orthonormal basis of $\ker \ell$ identifies it isometrically with the lower-dimensional Euclidean space. The coordinate-sum constraint on balanced cut vectors is an example.

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.

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 $U$.

Every linearly independent subset of a finite-dimensional vector space extends to a basis. Starting with its vectors, add a vector outside their current span until the span is the whole space. Each addition preserves linear independence, and the process stops because the dimension is finite. This also lets a linear functional on a vector subspace extend to the ambient space by assigning arbitrary values on the additional basis vectors.

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 bilinear form takes two vectors in a vector space $V$ over a field $F$ and returns a scalar in $F$. Its defining property is separate linearity: holding either input fixed gives a linear functional of the other. For scalars $a,b$ and vectors $u,v,w$,This is a scalar-valued bilinear map; the chosen field is part of the definition.^[ref]

Separate linearity does not mean that $B$ is a linear map on the vector space $V\times V$ with componentwise vector addition and scalar multiplication. Over the real numbers, $B(2v,2w)=4B(v,w)$, whereas a linear map on the pair would give $2B(v,w)$. The simplest example is $B(x,y)=xy$ on the one-dimensional vector space $\mathbb{R}$: fixing $y=3$ gives $x\mapsto 3x$, but doubling both $x$ and $y$ quadruples their product.

A bilinear form measures how two inputs interact. The dot product measures alignment; a symmetric bilinear form can describe a quadratic form; an alternating bilinear form can measure signed area. A nondegenerate bilinear form turns each vector into a unique linear functional and so connects geometry with the dual space. The same short definition therefore supports lengths, bilinear orthogonal complements, and symplectic vector spaces, depending on the additional properties imposed.

Over the complex numbers, a bilinear form is linear in both inputs without conjugating either scalar. A Hermitian form instead is a sesquilinear form and conjugates when its inputs are exchanged. For complex matrices, the adjoint $A^{\ast }$ used with Hermitian forms and matrix C-star algebras is the conjugate transpose, whereas a bilinear form uses the ordinary transpose $A^T$. For example, $B(z,w)=zw$ has $B(i,i)=-1$, while the Hermitian form $H(z,w)=z\overline{w}$ has $H(i,i)=1$.^[ref]

A bilinear form $B:X\times X\to \mathbb{R}$ on a real normed vector space is bounded when one constant $0\le C<\infty$ satisfiesThis is an upper bound on the magnitude of every pairing. It says nothing about the sign of $B(v,v)$ and does not require a symmetric bilinear form. The chosen norm is part of the hypothesis. Boundedness is equivalent to joint continuity, as shown by the boundedness criterion for a bilinear form.^[ref]

For example, on ${\mathbb{R}}^2$ with the Euclidean norm, the bilinear formsboth satisfy the upper bound with $C=1$, by the Cauchy-Schwarz inequality. Yet their quadratic forms are $B_{-}(v,v)=x^2-y^2$ and $B_0(v,v)=x^2$: the former takes negative values and the latter vanishes at the nonzero vector $(0,1)$. Thus neither is a coercive bilinear form. The latter is also a degenerate bilinear form, whereas $B_{-}$ is nondegenerate. Boundedness controls size, not positivity or loss of directions.

Every bilinear form on a real or complex finite-dimensional vector space equipped with any norm is bounded. In coordinates, write $B(u,v)=u^{T}Av$ using its matrix of a bilinear form. The Cauchy-Schwarz inequality gives $|u^{T}Av|\le \Vert A{\Vert }_2\Vert u{\Vert }_2\Vert v{\Vert }_2$, with $\Vert A{\Vert }_2$ the spectral norm; finite-dimensional equivalence of norms transfers this to the chosen norm. The conclusion concerns each fixed norm and fixed bilinear form: it does not supply a single constant for an arbitrary family of matrices or changing dimensions.

On an infinite-dimensional vector space, the upper bound is an additional condition. Equip the finitely supported sequences $c_{00}$ with the norm inherited from the l2 sequence space and setEach pairing is a finite sum, so this is a well-defined bilinear form. For the basis vector $e_n$, however, $\Vert e_n{\Vert }_2=1$ and $B(e_n,e_n)=n$; no finite $C$ works. It is even a coercive bilinear form with $\alpha =1$, since $\sum n{v}_n^2\ge \sum v_n^2$. Consequently coercivity alone does not imply boundedness on a normed vector space. This example uses $c_{00}$, which is not a Hilbert space; it does not define this pairing on all of the l2 sequence space.

A bounded bilinear form defines a bounded linear operator $T:X\to X^{\prime }$ by $(Tu)(v)=B(u,v)$, where $X^{\prime }$ is the continuous dual space. Its operator norm is the smallest admissible $C$:This converts an upper bound for pairings into an upper bound for a linear map taking an unknown vector to a linear functional.

Suppose $\Vert s_{\ast }{\Vert }_{\infty }=1$ and $s_{\ast }(x_j)=(-1{)}^j$ at increasing sites satisfying the Schoenberg–Whitney theorem conditions. For the B-spline collocation matrix $A$, the coefficient linear functional in the dual basis is ${\mu }_i(s)={\sum }_j(A^{-1}{)}_{ij}s(x_j)$. Therefore $\Vert {\mu }_i\Vert \le {\sum }_j|(A^{-1}{)}_{ij}|$. The checkerboard inverse of a totally nonnegative matrix makes ${\mu }_i(s_{\ast })=(-1{)}^i{\sum }_j|(A^{-1}{)}_{ij}|$, attaining that bound. Since ${\mu }_i(s_{\ast })=a_i^{\ast }$,The conclusion is an extremal property of an alternating unit-norm spline, not a claim that every bounded spline has these extremal coefficients.

For a vector subspace $W$ of $X$, its codimension is the dimension of the quotient vector space $X/W$. This definition also applies when $X$ is infinite-dimensional. If $X=W\oplus E$ is a direct sum, the map $e\mapsto e+W$ is a linear isomorphism from $E$ to $X/W$, so ${\mathrm{codim}}_{X}W=\dim E$. In particular, the kernel of a nonzero linear functional has codimension one: normalize $Te=1$ and write $x=(x-(Tx)e)+(Tx)e$.

For a compact group $G$, positive normalized linear functionals on $C(G)$ form a nonempty weak-star topology compact convex set. For finitely many translations, apply the Schauder-Tychonoff fixed-point theorem to the average of their induced affine dual maps. A fixed functional is stationary under this average. For each real continuous $f$, its orbit function $h(g)=\varphi (l_{g}f)$ is continuous on the compact subgroup generated by those translations and satisfies a finite averaging identity. At a maximum, every summand must attain that same maximum; iteration and density of the positive-word semigroup force $h$ to be constant. Thus the functional is fixed by each of the finitely many translations. Compactness and the finite intersection property now give a functional invariant under every translation. The Riesz-Markov-Kakutani representation theorem turns it into a normalized Haar measure, without assuming commutativity of the group.

A bounded complex-linear linear functional on a complex linear subspace of a normed space extends to the whole space with the same operator norm. To deduce this from the real Hahn-Banach theorem, extend the real part $u$ under the bound $\Vert \ell \Vert \Vert x\Vert$ on the underlying real vector space, then put $L(x)=u(x)-iu(ix)$. This is complex-linear and agrees with $\ell$ on its original domain. Given $x$, choose a complex number $a$ of modulus one with $aL(x)=|L(x)|$. Then $|L(x)|=\mathrm{Re}L(ax)=u(ax)\le \Vert \ell \Vert \Vert x\Vert$. Restriction gives the reverse operator norm inequality.

When the columns of $H$ are every nonzero vector in ${\mathbb{F}}_2^l$, a nonzero row vector $a$ defines a nonzero linear functional $v\mapsto a\cdot v$. Exactly half the $2^l$ input vectors take value $1$: translation by any vector on which the functional is $1$ pairs its zero and one fibres. Removing the zero input removes only a zero value. Thus every nonzero word in the binary simplex code has Hamming weight $2^{l-1}$.

A linear functional on $X^{\ast }$ is continuous for the weak-star topology exactly when it is evaluation at an element of $X$. Continuity makes it vanish on the common kernel of finitely many evaluations, so it is a linear combination of those evaluations. The canonical embedding into the bidual identifies this dual with $X$.

In a Hausdorff locally convex space, a separating family of continuous seminorms has $p(x)>0$ for any chosen nonzero $x$ and some $p$. Define a linear functional on its one-dimensional span attaining $p(x)$ and extend it under that seminorm by the real Hahn-Banach theorem. The extension is continuous because it is seminorm-bounded, and distinguishes $x$ from zero. The Hausdorff condition is essential.

The continuous dual space $X^{\ast }$ of a topological vector space is the vector space of its continuous linear functionals. Unlike the algebraic dual, it requires that a small change in an input produce a small change in its scalar measurement. For a normed vector space, this is equivalent to a finite boundfor all vectors $x$ and some constant $C\ge 0$. The least such constant is the operator norm of $\phi$.

This is the continuity-restricted subspace of the algebraic dual space; in infinite dimension the two can differ.

For the linear functional $\phi (x,y)=2x-y$ on ${\mathbb{R}}^2$ with its Euclidean norm, the Cauchy-Schwarz inequality givesEquality holds on the unit vector $(2,-1)/\sqrt{5}$, so its operator norm is exactly $\sqrt{5}$. The same estimate controls measurement error: $|\phi (v+h)-\phi (v)|\le \sqrt{5}\Vert h{\Vert }_2$. The linear functional diagram shows which changes are entirely invisible to this measurement; its kernel of a linear map consists of the changes with zero response. All linear functionals on a finite-dimensional real or complex normed vector space are continuous, but discontinuous functional detected by vanishing sequence tails shows why the restriction matters in infinite dimension.

A real convex function locally bounded on an open convex set has a continuous supporting linear functional at every interior point. Apply the dominated Hahn-Banach theorem to its sublinear directional derivative $p_x$ to obtain $l\le p_x$. Increasing secant slopes imply $p_x(z)\le f(x+z)-f(x)$ whenever the endpoint is in the domain. A local Lipschitz continuity bound gives both $l(y)\le L\Vert y\Vert$ and $-l(y)=l(-y)\le L\Vert y\Vert$, proving continuity.

A linear functional $f$ dominated by a finite real convex function $\varphi$ on a vector subspace $W$ extends with the same domination to the whole real vector space. To add $v\notin W$, choose $g(v)$ between the supremum of $(f(w)-\varphi (w-tv))/t$ and the infimum of $(\varphi (w+sv)-f(w))/s$, over $w\in W$ and $s,t>0$. Convexity, applied to $(s(w_1-tv)+t(w_2+sv))/(s+t)$, makes every lower candidate at most every upper candidate. The candidates at $w=0,s=t=1$ bound the interval by finite numbers. Chain unions preserve domination, so Zorn's lemma gives a full extension. No homogeneity of $\varphi$ is assumed.

A functional on a real vector space is convex when it is a convex function: $J(\theta v+(1-\theta )w)\le \theta J(v)+(1-\theta )J(w)$ for $0\le \theta \le 1$. For a symmetric nonnegative bilinear form $a$ and a linear functional $\ell$, the quadratic functional $J(v)=\frac{1}{2}a(v,v)-\ell (v)$ is convex, since its convexity difference is $\frac{1}{2}\theta (1-\theta )a(v-w,v-w)\ge 0$.