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The sign class of a $d$-dimensional real vector space of functions has VC dimension at most $d$. If $n>d$ points were shattered, the evaluation map from the function space to ${\mathbb{R}}^n$ would have a nonzero vector orthogonal to its image. Choosing labels according to the signs of that vector contradicts orthogonality.

A vector is an element of a vector space. Its defining operations are vector addition and scalar multiplication. An arrow in a Euclidean space is one example; a polynomial or a function can also be a vector when the chosen vector space uses the corresponding addition and scalar multiplication.

For instance, the polynomials $p(x)=1+x$ and $q(x)=x-x^2$ belong to the real vector space of polynomials of degree at most two. Their vector addition gives $(p+q)(x)=1+2x-x^2$, and scalar multiplication gives $(2p)(x)=2+2x$. In the monomial basis $(1,x,x^2)$ these are the coordinate vectors $(1,1,0)$, $(0,1,-1)$, $(1,2,-1)$ and $(2,2,0)$ respectively. The vectors are the polynomials; these tuples are their representations in a chosen basis.

A general vector space does not itself prescribe lengths or angles. A norm supplies a size, while an inner product also supplies angle information and orthogonality. The familiar arrow geometry below belongs to Euclidean vectors, where this extra structure is available.

Vector addition is the vector space operation combining two vectors into their sum. It is commutative and associative, has the zero vector as identity, and distributes with scalar multiplication as specified by the vector space axioms. Coordinatewise addition describes it after choosing a basis.

A rank-$r$ vector bundle over a topological space $B$ is a fiber bundle $\pi :E\to B$ whose fibers are vector spaces over a fixed field $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. Each point has an open neighborhood $U$ and a vector bundle trivialization ${\pi }^{-1}(U)\cong U\times {\mathbb{F}}^r$ that is a linear isomorphism on each fiber.

On overlaps the vector bundle trivializations change by $(x,v)\mapsto (x,g_{ij}(x)v)$, where the invertible matrices $g_{ij}(x)$ vary continuously and satisfy the Čech cocycle condition. For a smooth vector bundle, the base and total space are smooth manifolds and these functions are smooth. The fiberwise vector addition and scalar multiplication are therefore globally well-defined.

The tangent bundle collects all tangent spaces of a smooth manifold. Its transition functions of a vector bundle are Jacobian matrices, so the chain rule supplies the cocycle identity. The Möbius line bundle supplies another example: every small patch is a product, but no global frame of a vector bundle exists. A global fiber-linear product isomorphism is the extra condition defining a trivial vector bundle.

The vector representation is the defining action of a matrix group on the vector space on which its matrices act. For an orthogonal or Lorentz group, it preserves the corresponding symmetric bilinear form.

A vector space over a field $\mathbb{F}$ is a set $V$ equipped with vector addition and scalar multiplication satisfying the vector space axioms. Its vectors can represent directed displacements, lists of measurements, polynomials, or entire functions. The common structure is that adding two elements and multiplying an element by a scalar remain meaningful operations inside the same space.

For example, ${\mathbb{R}}^2$ uses coordinatewise addition and scaling. Withwe have $v=2b_1+b_2$. Moving twice along $b_1$ and once along $b_2$ reaches the same vector as five horizontal and four vertical steps. The two descriptions refer to different bases of the same space. A basis makes such descriptions unique; a spanning set makes them possible.

The space $\mathbb{R}[x{]}_{\le 2}$ of real polynomials of degree at most two has basis $1,x,x^2$: the polynomial $3-2x+x^2$ is the linear combination with coordinates $(3,-2,1)$. The zero vector here is the zero polynomial, and addition combines coefficients. A derivative is a linear map on this space; its matrix records how it acts on the three basis vectors. The same coordinate machinery therefore handles both geometric motion and operations on functions.

The choice of field matters. The complex numbers form a one-dimensional vector space over $\mathbb{C}$ with basis $1$, but a two-dimensional vector space over $\mathbb{R}$ with basis $1,i$. The question "how many independent directions?" has an answer only after the allowed scalars are specified. Likewise, polynomials of degree exactly two do not form a vector space: subtraction can remove the leading term, and the zero polynomial is absent.

The usefulness of the abstraction is that a linear map is determined by its values on a basis. Once those images are known, linear combinations give its action on every vector. The rank-nullity theorem relates the number of attainable outputs to the number of invisible input directions, while change of basis lets us choose coordinates suited to the problem. Eigenvectors, for example, choose directions on which repeated application of a linear operator becomes scalar multiplication.

A historical source is Hermann Grassmann's Die lineale Ausdehnungslehre (1844). Its full title advertises applications to other branches of mathematics, statics, mechanics, magnetism and crystallography; the original edition is digitized by ETH-Bibliothek. This gives an early setting for treating geometric quantities algebraically across different applications, rather than attributing the modern axiom list to that title alone.^[ref]

Every vector space $V$ over a field $F$ has a basis, assuming the axiom of choice. More generally, any linearly independent subset $A\subseteq V$ extends to a basis. This is an algebraic Hamel basis: each representation uses a finite linear combination, even when the basis is infinite.^[ref]

Use the poset of linearly independent subsets containing $A$:It is a set, being a subset of the power set of $V$, and is nonempty because $A\in {\mathcal{P}}_A$.

A short proof outline supplies both essential steps:

Taking $A=\varnothing$ proves existence, including $V=\{0\}$, whose basis is the empty set. The chain in a partial order need not be indexed by the natural numbers: only the finitely many members needed for a particular linear relation are compared. An arbitrary union of linearly independent sets can fail: $\{v\}$ and $\{2v\}$ for a nonzero vector over the real numbers are each linearly independent, but their union is linearly dependent. The nesting of a chain in a partial order is what prevents this failure.

Order the linearly independent subsets of a vector space by inclusion. A chain's union is independent, since every finite relation is contained in one chain member. Zorn's lemma gives a maximal independent subset. A vector outside its span could be adjoined while preserving independence, so it spans. Sending finite-support coefficient families to their linear combinations gives an isomorphism from the free module on that subset to $V$. This supplies an arbitrary, possibly infinite-dimensional basis using the axiom of choice; no pre-existing basis theorem is invoked.

The real univariate polynomials form a vector space with the countable Hamel basisThe subspace of polynomials of degree at most $n$ has dimension $n+1$.

A module over a division ring has the usual basis and linear independence theory of a vector space, with the side of scalar multiplication specified. For a right module, a finite relation has the form ${\sum }_i{v}_i{a}_i=0$. Every nonzero endomorphism of a one-dimensional right module is invertible. If $V$ is a simple module on the left over $R$, evaluation makes it a right vector space over the opposite ring ${\mathrm{End}}_R(V{)}^{\mathrm{op}}$, a division ring by Schur's lemma. This is the scalar structure used by the Jacobson density theorem.

If ${\mathbb{F}}_q$ is a finite field, a vector space of finite dimension $d$ over ${\mathbb{F}}_q$ has $q^d$ elements. Every infinite-dimensional vector space over ${\mathbb{F}}_p$ is an infinite elementary abelian $p$-group under addition.

A vector space with scalar field $\mathbb{Q}$ is equivalently a torsion-free divisible Abelian group. An embedding of such groups automatically respects rational scalar multiplication because solutions to $nb=ma$ are unique.

A vector subspace is a nonempty subset of a vector space closed under vector addition and scalar multiplication, and is therefore itself a vector space. Equivalently, whenever $u,v$ belong to the subset, so does $au+bv$ for all allowed scalars $a,b$. In particular it contains the zero vector.

For example, the plane $W=\{(x,y,z):x+y+z=0\}\subseteq {\mathbb{R}}^3$ is a vector subspace, with basis $(1,-1,0),(1,0,-1)$. Indeed every element is $(-y)(1,-1,0)+(-z)(1,0,-1)$. The parallel plane $x+y+z=1$ misses the zero vector and is an affine subspace instead. Homogeneous constraints produce vector subspaces: $W$ is the kernel of the linear map $(x,y,z)\mapsto x+y+z$. This lets us describe all solutions of a homogeneous linear equation through a basis of its kernel.

A vector-valued function is a function $F:X\to V$ taking values in a vector space. For example, $F(t)=(\cos t,\sin t)$ describes the unit circle in ${\mathbb{R}}^2$: one real number controls a two-component vector.

In a real or complex finite-dimensional vector space equipped with a norm, choosing a basis writes $F={\sum }_j{f}_j{e}_j$ in scalar-valued functions. Continuity and differentiability can then be checked in these coordinates, and $F^{\prime }={\sum }_j{f}_j^{\prime }{e}_j$ when the derivatives exist. Finitely many coordinates matter: coordinatewise convergence in an infinite-dimensional vector space need not imply norm convergence.

A vertex operator algebra is an integer-graded vector space equipped with a state-field correspondence $a\mapsto Y(a,z)$, a vacuum, translation operator and conformal vector satisfying vacuum, translation, locality and compatibility axioms. It packages the local operator product expansions of a chiral conformal field theory. A string vertex operator is an individual state insertion, rather than this complete algebraic structure.

In characteristic $p>0$, multiplication by $x_i$ and formal differentiation by $x_i$ act on the displayed vector space, because differentiation preserves the ideal generated by the $p$th powers. They satisfy the Weyl algebra relations and yield a nonzero module of dimension $p^n$. The central elements $x_i^p$ and ${\partial }_i^p$ also show why the characteristic-zero simplicity argument cannot apply. In particular the characteristic-zero Bernstein inequality for Weyl algebra modules fails for this finite-dimensional module.

// Destination: module-theory.bigb

The Witt index of a nondegenerate quadratic form $q$ on a finite-dimensional vector space $V$ is the largest dimension of a totally isotropic subspace: a linear subspace $U$ for which $q(u)=0$ for every $u\in U$. In characteristic different from two, the Witt decomposition writeswhere $H$ is a hyperbolic plane of a quadratic form, $q_{\mathrm{an}}$ is an anisotropic quadratic form, and $r$ is the Witt index. The anisotropic part has no nonzero vector on which the form vanishes.^[ref]

For the real form $x_1^2+x_2^2-x_3^2-x_4^2$, the subspace spanned by $(1,0,1,0)$ and $(0,1,0,1)$ is totally isotropic. Projection onto the two negative coordinates is injective on any totally isotropic subspace, giving the matching upper bound two. More generally, a real form of signature $(a,b)$ has Witt index $\min (a,b)$.

The base field matters: $x^2+y^2$ has index zero over $\mathbb{R}$ but index one over $\mathbb{C}$, using the line spanned by $(1,i)$. In characteristic two, vanishing of the associated polar form on $U$ alone is insufficient; the condition remains $q{|}_U=0$. Over a finite field, the two nondegenerate types in dimension $2m$ have Witt indices $m$ and $m-1$.^[ref]

The zero polynomial has every coefficient zero. It is the zero vector of a vector space of polynomials and the additive identity element of a polynomial ring. Its degree of a polynomial is often set to $-\infty$ when a formula requires one.

The zero vector of a vector space is its additive identity: $v+0=0+v=v$ for every vector $v$. It is unique, because if $0^{\prime }$ is another such identity then $0=0+0^{\prime }=0^{\prime }$. Distributivity gives $\alpha 0=\alpha (0+0)=\alpha 0+\alpha 0$, so $\alpha 0=0$ after cancellation. In a coordinate vector space it has every coordinate zero.