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A topological vector space is a vector space $E$ over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, equipped with a topology such that vector addition and scalar multiplication are jointly continuous:The domains carry the product topology, and the scalar field carries its usual topology. Joint continuity permits both inputs to vary at once. Here the definition does not impose the Hausdorff condition; we state it separately when needed. ^[ref].

Every normed vector space is a topological vector space: the triangle inequality gives $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$, whilecontrols scalar multiplication near any $({\lambda }_0,x_0)$. More generally, a seminorm topology supplies many measurements of smallness, rather than one norm. Translation by any vector is a homeomorphism, so a neighbourhood basis at zero determines all neighbourhoods by translation.

The compatibility matters: on a nonzero real vector space, the discrete topology makes vector addition continuous, but not scalar multiplication. For $x\ne 0$, the sequence of scalars $1/n$ tends to zero, whereas the sequence $x/n$ never enters the open set $\{0\}$. Thus choosing a topology arbitrarily need not respect the linear algebra.

This framework lets functional analysis choose the appropriate notion of limit of a sequence. Norm topology measures one global error, weak topology tests linear functionals, and the Fréchet space of smooth functions controls every order of derivative. In each case the operations forming linear combinations are continuous, making approximation compatible with the underlying equations.

Two norms $p,q$ on a vector space give the same topology if and only if $cp(x)\le q(x)\le Cp(x)$ for some positive constants. Indeed, continuity of the identity at zero gives $p(x)<\delta \Rightarrow q(x)<1$. For $x\ne 0$, apply this to $\delta x/(2p(x))$ to obtain $q(x)<2p(x)/\delta$. Reverse the roles of the norms for the lower bound. Conversely the two inequalities compare every translated open ball. This proof uses absolute homogeneity of a norm, not finite dimension.

A module $M$ over an integral domain $R$ is torsion-free when $rm=0$ with $0\ne r\in R$ implies $m=0$, equivalently when its torsion submodule is zero. Every free module over an integral domain is torsion-free, as one can check in each coefficient. Every vector space is torsion-free over its scalar field.

The converse requires hypotheses. A finitely generated module over a principal ideal domain is free if it is torsion-free, but the module $\mathbb{Q}$ over $\mathbb{Z}$ is torsion-free and not free. Any two nonzero rational numbers have a nontrivial integer relation, so a possible basis of a module could have at most one element; no single $q$ generates all rational numbers, since $q/2\notin \mathbb{Z}q$ for $q\ne 0$. The finite-generation hypothesis cannot be dropped. See Keith Conrad's Modules over a PID, Corollary 2.6.

A length-$n$, dimension-$k$ binary linear code has full support when no coordinate vanishes on every codeword, equivalently when the union of the supports of a vector of all codewords covers every coordinate. Every coordinate is then a nonzero linear functional on the message vector space ${\mathbb{F}}_2^k$. Translation by a vector on which it is one pairs the zero and one fibers, so exactly $2^{k-1}$ codewords contribute one in that coordinate. Summing these contributions proves the formula. If there are $r$ nonzero columns in a full-rank generator matrix, the same argument gives $r{2}^{k-1}$ instead.

On an affine space modeled on a vector space, the map $T_v(x)=x+v$ for a fixed vector $v$. Its derivative is the identity, and $T_v{T}_w=T_{v+w}$. Changes between flat coordinates on a translation surface have this form.

Choosing a positive system of a root system for a complex semisimple Lie algebra divides its root spaces into negative and positive ones. Their respective sums ${\mathfrak{n}}^{-}$ and ${\mathfrak{n}}^{+}$, together with the Cartan subalgebra $\mathfrak{h}$, give this direct sum of vector spaces. Both ${\mathfrak{n}}^{\pm }$ are nilpotent Lie algebras, and $\mathfrak{h}\oplus {\mathfrak{n}}^{+}$ is a Borel subalgebra. The Poincaré-Birkhoff-Witt theorem turns this into an ordered vector-space decomposition of the universal enveloping algebra, useful for building Verma modules.

The trivial representation sends every group element to the identity operator. A one-dimensional constant-coordinate line in a permutation representation is an example: relabelling positions leaves a uniform assignment unchanged. A trivial action on a higher-dimensional vector space has many invariant subspaces and is therefore not an irreducible representation.

Let $B_1,B_2$ be real square invertible matrices whose columns are lattice bases. They generate the same Euclidean lattice precisely when^[ref]

If $B_2=B_{1}U$, multiplication by $U$ merely permutes the integer coordinate vectors, so the two generated sets agree. Conversely, equality of the generated sets means every column of either basis is an integer linear combination of the other. Hence $B_2=B_{1}U$ and $B_1=B_{2}V$ for integer matrices $U,V$. Substitution followed by multiplication by $B_1^{-1}$ gives $UV=I$, so the integer inverse criterion for a unimodular matrix applies.

Equal real linear spans are a weaker condition: any two real square invertible matrices span the same ambient vector space. Even equal covolumes do not force the same Euclidean lattice. The requirement that the change matrix and its inverse matrix have integer entries provides the missing arithmetic information.

In ${\mathbb{F}}_2[u,v]/(u,v{)}^2$, the ideals $(u)$, $(v)$ and $(u+v)$ are pairwise incomparable and nonprime. Their union is the ideal $(u,v)$: the three lines cover its two-dimensional vector space over ${\mathbb{F}}_2$. This shows why a primality hypothesis matters in prime avoidance.

For a bilinear map $B:V\times W\to U$, there is exactly one linear map $L:V{\otimes }_{F}W\to U$ satisfying $L(v\otimes w)=B(v,w)$. Equivalently, $B=L\circ \tau$, where $\tau (v,w)=v\otimes w$. The universal property of a tensor product concerns the vector space together with this distinguished bilinear map.^[ref]

Here is a complete calculation over the real numbers. Define the bilinear mapIts values on pairs of basis vectors determine $L$ on the four tensor-product basis elements:Consequently, for $t={\sum }_{i,j=1}^2{c}_{ij}{e}_i\otimes f_j$,For $v=(2,1)$ and $w=(-1,3)$, the direct calculation gives $B(v,w)=(-2+12+3,-1-6)=(13,-7)$. Using the expansion of $v\otimes w$ instead gives $L(v\otimes w)=-2(1,0)+6(2,-1)-(0,1)+3(1,0)=(13,-7)$. The linear map also handles a nonseparable tensor such as $t=e_1\otimes f_1+e_2\otimes f_2$: $L(t)=(2,0)$, even though no single pair has $v\otimes w=t$.

A short proof uses a tensor-product basis: assign $L(e_i\otimes f_j)=B(e_i,f_j)$ and extend by linearity. Expanding both inputs verifies $L(v\otimes w)=B(v,w)$ by bilinearity; the assigned values on a basis force uniqueness. More generally, every multilinear map $F:V_1\times \cdots \times V_r\to U$ factors uniquely through a linear map from $V_1\otimes \cdots \otimes V_r$. This lets formulas involving several inputs define linear maps without choosing coordinates; tensor contraction uses the evaluation pairing between a vector and a covector.

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 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.