Vector space ($V$)

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]

Addition makes $V$ an Abelian group: for all $u,v,w\in V$, $(u+v)+w=u+(v+w)$ and $u+v=v+u$, there is a zero vector $0$, and every $v$ has an additive inverse $-v$. Scalar multiplication satisfies, for $a,b\in \mathbb{F}$,Both operations take values in $V$. These requirements make manipulating linear combinations independent of how sums are grouped or expanded.

A sequence space is a vector space of scalar sequences, closed under coordinatewise addition and scalar multiplication. Examples include the l-p sequence space, the l-infinity sequence space and spaces of convergent sequences.

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.

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.

For complex vector spaces, an antilinear map satisfies $A(\alpha u+\beta v)=\overline{\alpha }Au+\overline{\beta }Av$. It is linear over the real numbers and conjugates complex scalar multiplication. This differs from two conjugate linear operators, which are related by similarity rather than conjugate-linearity.

An antilinear endomorphism $J$ with square $-1$ equips a complex vector space with a quaternionic structure: complex multiplication supplies the quaternionic generator $i$ and $J$ supplies $j$, satisfying $Ji=-iJ$. Such a space has even complex dimension. There is no invariant complex line, because $Jv=cv$ would imply $-v=|c{|}^{2}v$. In dimension four the invariant complex two-planes are quaternionic lines, forming ${\mathbb{HP}}^1\cong S^4$. This is the Euclidean reality structure on twistor space used in twistor theory.

An antiunitary operator between complex Hilbert spaces is a surjective antilinear map satisfying the displayed inner product identity. It preserves norms but conjugates scalars. In a chosen orthonormal basis it is $UK$, with $U$ a unitary operator and $K$ coefficientwise complex conjugation. Conversely any such $UK$ is antiunitary. Antilinearity by itself does not preserve norms.

A vector space over the field of complex numbers. A complex-linear map preserves addition and multiplication by complex scalars.

A real vector space is a vector space over the field $\mathbb{R}$ of real numbers.

For real vectors of equal length and equal sum, write $x\prec y$ when ${\sum }_{j=1}^k{x}_j^{\downarrow }\le {\sum }_{j=1}^k{y}_j^{\downarrow }$ for every proper initial segment; the downward arrow means decreasing rearrangement. Majorization compares how unevenly a fixed total is distributed. The uniform probability vector is majorized by every probability vector, while $(1,0,\dots ,0)$ majorizes every probability vector. The order controls exact pure-state entanglement conversion in Nielsen's pure-state conversion theorem.

For decreasing positive sequences in ${\ell }^1$, write $x\prec y$ if every initial partial sum of $x$ is at most the corresponding partial sum of $y$ and the finite total sums agree. The finiteness requirement matters: equality of two divergent totals interpreted as infinity is too weak to guarantee a doubly stochastic realization of summable-sequence majorization.

Decreasing strictly positive summable sequences with equal totals and $x\prec y$ admit an infinite doubly stochastic matrix realizing the displayed relation. Choose $k$ with $y_k\ge x_1>y_{k+1}$ and average the pair to obtain $x_1$ and $y_k+y_{k+1}-x_1$. The remaining decreasing sequence majorizes the remaining target sequence: short initial sums contain values at least $x_1$, and longer sums are the old sums minus $x_1$. Repeat on the tail, freezing one normalized finite-support row at each stage. Limit column sums are at most one; Tonelli's theorem, equal finite totals and strict positivity of every $y_j$ force every column sum to equal one.

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 subset of ${\mathbb{F}}_q^n$ is a finite-field Kakeya set when it contains an affine line in a vector space in every one-dimensional direction. The line position may depend on the direction. The polynomial method in combinatorics yields the finite-field Kakeya polynomial bound on its cardinality.

For odd characteristic, the union of the lines $y=mx-m^2/4$ contains a line of every finite slope. For fixed $x$, completing the square shows that it comprises the $(p+1)/2$ values of $y$ for which $x^2-y$ is a square, including zero. Adding the vertical line $x=0$ completes the directions and gives cardinality $(p^2+2p-1)/2=(1/2+o(1))p^2$. This is a finite-field Kakeya set whose density tends to one half.

For a finite-field Kakeya set $A\subseteq {\mathbb{F}}_q^n$, a smaller cardinality would give a nonzero polynomial of total degree of a polynomial at most $q-1$ vanishing on $A$, by the dimension of a bounded-total-degree polynomial space. Its restriction to each complete affine line in a vector space has $q$ roots of a polynomial and degree less than $q$, hence vanishes identically. The top homogeneous polynomial part then vanishes in every direction. A polynomial of degree less than $q$ in each variable cannot vanish everywhere on ${\mathbb{F}}_q^n$ unless it is zero, giving a contradiction. For the prime field, this is the polynomial nonvanishing below the field size.

The linear span of a subset $S$ of a vector space is the set of all finite linear combinations of elements of $S$. It is the smallest vector subspace containing $S$: every vector subspace containing the original vectors must contain their sums and scalar multiples.

In ${\mathbb{R}}^3$, the linear span of $(1,0,0)$ is a line, the linear span of $(1,0,0),(0,1,0)$ is the plane $z=0$, and adding $(0,0,1)$ gives all of ${\mathbb{R}}^3$. Adding $(2,0,0)$ to the first set changes none of its attainable vectors, because it is already a scalar multiple of $(1,0,0)$. Thus a spanning set can contain redundant generators; linear independence identifies when this redundancy is absent. In an infinite-dimensional space, an algebraic linear span still uses finite sums; a closed linear span can also include limits in a specified norm.

In a normed vector space, the closed linear span of $S$ is the closure of its linear span in the norm topology. It includes limits of finite linear combinations. In a Hilbert space, an infinite orthonormal basis spans in this closed sense, allowing norm-convergent series in addition to finite sums.

A subset $S$ of a vector space is a spanning set when its linear span is the whole vector space.

For a vector subspace $W\le V$, the quotient vector space $V/W$ consists of cosets $v+W$, with vector operations induced from $V$.

For fixed $u\ne 0$, the parallel affine lines $l_x=x+\mathrm{span}(u)$ are precisely the cosets of the indicated quotient vector space. Changing representatives adds a multiple of $u$, so $l_x+l_y=l_{x+y}$ and $a{l}_x=l_{ax}$ are well-defined. The zero is the whole line $l_0$, not a single point. If $(u,b_1,b_2)$ is a basis, the two line classes $l_{b_1},l_{b_2}$ form a basis of the quotient by spanning and independence modulo $\mathrm{span}(u)$.

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 proper vector subspace of a vector space is one strictly smaller than the ambient space. In a finite-dimensional vector space, this is equivalent to having smaller dimension: a basis of the subspace can be extended to a basis of the ambient space, and strict inclusion requires at least one additional vector.

A flag is a strictly nested chain of vector subspaces of a fixed vector space. A complete flag contains a subspace of every possible dimension. For an action by linear maps, an invariant flag means that each of its vector subspaces is an invariant subspace. A basis adapted to an invariant complete flag makes all these maps upper triangular, connecting complete flags to simultaneous triangularization of a Lie algebra representation.

A complete flag in an $n$-dimensional vector space is a nested sequence of vector subspaces with $\dim V_i=i$. An action preserves such a flag exactly when its matrices are upper triangular in a basis adapted to the flag. The Lie theorem supplies an invariant complete flag for every complex finite-dimensional Lie algebra representation of a solvable Lie algebra.

The fixed coordinate subspace of dimension $k$ in ${\mathbb{R}}^p$ consists of vectors whose coordinates after the first $k$ vanish. Restricting a quadratic form to it uses only the leading $k$ by $k$ matrix block. It specifies one fixed support; allowing every support of size $k$ would introduce another combinatorial counting step.

The intersection of any family of vector subspaces of a common vector space is the set of vectors belonging to every member of the family. It is again a vector subspace.

The sum of vector subspaces $U,W\le V$ isIt is the smallest vector subspace containing both $U$ and $W$.

A hyperplane is a vector subspace of codimension one. A projective hyperplane in ${\mathbb{P}}^n$ is the zero set of one nonzero homogeneous linear form.

A closed vector subspace of a topological vector space is both a vector subspace and a closed set. The kernel of a continuous linear map is a closed vector subspace.

If $E$ is a nonzero closed linear subspace of a normed ambient space and $F$ is finite-dimensional with $E\cap F=\{0\}$, the unit sphere of $E$ has positive distance from $F$. Otherwise a sequence of nearly coincident points yields a bounded sequence in $F$, hence a convergent subsequence; closedness of $E$ puts its norm-one limit in both vector subspaces. In applications to a bidual space, completeness of the Banach space makes its canonical image closed.

An affine subspace of a vector space is a translate $v+W$ of a vector subspace $W$. Its dimension is the dimension of $W$.

The smallest affine subspace containing a nonempty set $S$. For any chosen $a\in S$, the displayed formula proves existence and minimality: every containing affine subspace must contain all the indicated differences in its direction space. Equivalently it consists of finite affine combinations ${\sum }_i{\lambda }_i{s}_i$ with ${\sum }_i{\lambda }_i=1$. Its dimension can be smaller than the dimension of an affine hyperplane in which the set has been embedded.

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.

In a vector space over a field $k$, an affine line is a translate of a one-dimensional vector subspace. Its direction is that vector subspace, so multiplying $v$ by a nonzero scalar preserves the direction. Over a finite field of size $q$ it contains exactly $q$ distinct points. This usage concerns affine subsets of vector spaces, rather than the affine line as an affine scheme.

Let $V$ be a vector space over a specified field $F$, and let $v_1,\dots ,v_n\in V$. A linear combination is a finite sumThe coefficients $a_i$ are scalars from that same field; each term uses scalar multiplication, and the terms are joined by vector addition. For coefficients in the real numbers, negative values reverse arrows and fractional values scale them by fractional amounts. The collection of attainable vectors is the linear span. The column picture in Strang's MIT notes interprets solving a linear system as choosing coefficients to reach a target vector.^[ref]

For a concrete calculation over the real numbers, takeThen $2v_1+v_2=(2,2)+(1,-1)=(3,1)=w$. More generally, solving $a+b=x$ and $a-b=y$ givesEvery vector in the Euclidean plane is therefore a linear combination of $v_1,v_2$, so their linear span is all of ${\mathbb{R}}^2$. The target $(3,1)$ has coefficients $(2,1)$; the target $(0,2)$ has coefficients $(1,-1)$, since $(1,1)-(1,-1)=(0,2)$. The displayed formula also makes the coefficients unique. Consequently $(v_1,v_2)$ is an ordered basis, with coordinate vector $[(x,y){]}_{(v_1,v_2)}=((x+y)/2,(x-y)/2{)}^T$.

The payoff is a reusable language for synthesis: linear combinations build vectors from chosen ingredients, linear span decides which targets are reachable, and linear independence decides whether the coefficients are unique. The ingredients can also be polynomials or functions, as long as they belong to a vector space over the specified field.

An integer linear combination is a linear combination whose coefficients are integers. The lattice basis definition uses unique integer linear combinations, rather than arbitrary real coefficients. For example, $(1,0)$ is a real linear combination of $(2,0)$ but is not an integer linear combination of that single vector.

A scalar multiplying a specified term in a linear combination, polynomial or power series. When the term family has linear independence, the coefficients are unique; uniqueness of finite Laurent polynomial coefficients follows after multiplying by a sufficiently large power to obtain an ordinary polynomial.

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.

The vector sum of two vectors $v,w$ is the vector $v+w$ obtained by vector addition.

Scalar multiplication is the vector space operation $(a,v)\mapsto av$ taking a scalar and a vector to a vector.

A scalar multiple of a vector $v$ is a vector of the form $av$ for a scalar $a$.

The dimension of a vector space is the cardinality of any of its bases. In finite dimension this counts the scalars needed to specify an arbitrary vector once an ordered basis is chosen. The exchange argument behind the basis extension theorem shows that any two finite bases have the same size: an independent family cannot be larger than a spanning family.

Thus ${\mathbb{R}}^2$ has dimension two even with an oblique basis, $\mathbb{R}[x{]}_{\le 2}$ has dimension three, and the homogeneous plane in vector subspace has dimension two. This count concerns independent freedom, so an embedded plane remains two-dimensional while sitting in a three-dimensional ambient vector space. Dimension also depends on the field, as the two interpretations of $\mathbb{C}$ in vector space illustrate.

A vector space is finite-dimensional when it has a finite set as a basis.

An infinite-dimensional vector space has no finite basis, so its dimension is an infinite cardinal.

The direct sum $U\oplus W$ of two vector spaces consists of pairs $(u,w)$ with componentwise operations. A vector space is an internal direct sum of subspaces $U$ and $W$ exactly when every vector has a unique expression $u+w$ with $u\in U$ and $w\in W$.

If $V=A_1\oplus A_2=A_2\oplus A_3=A_1\oplus A_3$ with nonzero subspaces, choose $0\ne v\in A_3$ and write $v=a_1+a_2$ using the first direct sum of vector spaces. Both components are nonzero. The plane spanned by them meets $A_1$ and $A_2$ in their component lines and meets $A_3$ in the line through $v$. Its intersection with $A_3$ cannot be the whole plane, since $a_1\notin A_3$.

An orthogonal direct sum is a direct sum whose distinct summands are mutually orthogonal. For $v\in V$ and $w\in W$, the Pythagorean identity gives $\Vert v+w{\Vert }^2=\Vert v{\Vert }^2+\Vert w{\Vert }^2$. In a Hilbert space, a closed subspace of a Hilbert space and its orthogonal complement give an orthogonal direct sum of the entire space. The component maps are orthogonal projections and have operator norm one when their ranges are nonzero.

A submodule $N$ of a module $M$ is a direct summand if some submodule $L$ gives $M=N\oplus L$. Equivalently, the inclusion has a left inverse R-module homomorphism, or $N$ is the image of an idempotent endomorphism of $M$.

Integer row vectors forming a square unimodular matrix are a basis of the corresponding free abelian group: the adjugate formula makes the inverse integral. The subgroup spanned by any subset of that basis is therefore a direct summand, complemented by the remaining basis vectors. In contrast, a proper finite-index subgroup of a free abelian group cannot be a direct summand: its nonzero finite quotient would have to embed as a subgroup of a torsion-free group.

For finite-dimensional subspaces $U,W$ of a common vector space,

For every subspace $U$ of a finite-dimensional vector space $V$, extending a basis of $U$ to a basis of $V$ produces a subspace $W$ such that $V=U\oplus W$.

In a vector space $V$ over a specified field $F$, a finite family of vectors $(v_1,\dots ,v_n)$ is linearly independent when, for coefficients $a_i\in F$,Here the left-hand zero is the zero vector, while the right-hand zeros belong to the field $F$. Thus linear independence excludes every nontrivial linear relation. For an infinite family, require this condition for every finite subfamily.^[ref]

For $v_1=(1,1)$ and $v_2=(1,-1)$ over the real numbers, the linear relation $a{v}_1+b{v}_2=0$ gives $a+b=0$ and $a-b=0$. Adding and subtracting yields $2a=2b=0$, hence $a=b=0$. This verifies linear independence directly and explains why the linear combination giving $(3,1)$ has exactly the coefficients $(2,1)$.

A linearly independent family cannot contain the zero vector: a coefficient of $1$ on that entry and $0$ on every other entry would be a nontrivial linear relation. The empty family is linearly independent, because its empty sum is the zero vector and there are no coefficients that could be nonzero. Linear independence identifies a collection without redundant vectors; combined with a spanning set, it gives a basis. The uniqueness of linear combination coefficients makes its representations unambiguous.

A family is linearly independent if and only if every vector in its linear span has unique coefficients as a linear combination of that family. For a short proof, subtract two representations to get ${\sum }_i(a_i-b_i)v_i=0$; linear independence forces each difference to vanish. Conversely, uniqueness applied to the zero vector forces every linear relation to use only zero coefficients. This explains the practical role of linear independence: a coordinate vector relative to an ordered basis is unambiguous.

A finite vector family is minimally dependent if it is linearly dependent and every proper subfamily is linearly independent. For $n\ge 2$, the family $e_1,\dots ,e_{n-1},{\sum }_{j=1}^{n-1}{e}_j$ in ${\mathbb{R}}^n$ is an example. The sum supplies a nonzero dependence. If the sum vector is omitted, the remaining vectors are a coordinate basis; if $e_j$ is omitted, its coordinate in any proposed relation forces the coefficient of the sum vector to vanish, then all other coefficients vanish. Every smaller proper subfamily lies in one of these independent $(n-1)$-element subfamilies.

Let the first $k$ columns of a matrix be the first dependent prefix, with no zero column. Their relation space has dimension one, and every nonzero relation $b$ has at least two nonzero coordinates. If both $Ab=0$ and $A\mathrm{diag}({\lambda }_j)b=0$, one-dimensionality gives ${\lambda }_j{b}_j=c{b}_j$ for all $j$. Two nonzero coordinates would force two of the distinct weights ${\lambda }_j$ to agree, a contradiction.

A finite family of vectors in a vector space over a specified field $F$ is linearly dependent if it has a nontrivial linear relation: ${\sum }_i{a}_i{v}_i=0$ with $a_i\in F$ and at least one coefficient nonzero. One can then solve for a vector whose coefficient is nonzero as a linear combination of the others; removing that entry preserves the linear span. Repeated vectors in a list give linear dependence: for example $(v,v)$ has the linear relation $1v+(-1)v=0$. A list keeps both entries even though the set $\{v,v\}$ has only one element. Strang's notes illustrate redundant columns by expressing one as the sum of two others.^[ref]

Keep $v_1=(1,1)$ and $v_2=(1,-1)$, and add $v_3=(2,0)=v_1+v_2$. The three vectors still have linear span ${\mathbb{R}}^2$, but nowThe two coefficient triples $(2,1,0)$ and $(1,0,1)$ differ by the nontrivial linear relation $(-1,-1,1)$. In fact all linear combination representations of this target areTo check both completeness and correctness, use $c_1{v}_1+c_2{v}_2+c_3{v}_3=(c_1+c_3)v_1+(c_2+c_3)v_2$. Linear independence of $v_1,v_2$ forces $c_1+c_3=2$ and $c_2+c_3=1$, giving exactly the displayed family. Redundancy therefore preserves attainable targets while allowing their coefficients to move. For solving a linear system, these alternatives matter; for assigning unique coordinate vectors, one instead chooses a basis.

The following example shows why linear independence depends on the scalar field.

Consider the vectors $1,i$ in the complex numbers, where $i$ is the imaginary unit. With scalar field $\mathbb{R}$, a linear relation $a\cdot 1+b\cdot i=0$ has $a,b\in \mathbb{R}$. Taking its real part and imaginary part gives $a=b=0$, so the pair is linearly independent. Every complex number $x+iy$ also has coefficients $(x,y)$ in the real numbers, so $1,i$ form a basis and the dimension of a vector space is two.

With scalar field $\mathbb{C}$, the same pair is linearly dependent, sinceuses the nonzero coefficients $(i,-1)$ from the complex numbers. The single vector $1$ is now a basis, because $z=z\cdot 1$ for every complex number $z$, giving dimension of a vector space one. For example $3+i=3\cdot 1+1\cdot i=(3+i)\cdot 1+0\cdot i$: the second representation uses a coefficient unavailable over the real numbers. Specifying the field is part of the definition of linear independence.

A linear relation among vectors $v_1,\dots ,v_n$ is a coefficient tuple $(a_1,\dots ,a_n)$ satisfying ${\sum }_i{a}_i{v}_i=0$. It is nontrivial if at least one coefficient is nonzero. For the matrix with these vectors as columns, the space of relations is its kernel. The vectors are linearly independent exactly when this kernel is zero. An invertible change of coordinates preserves every relation, since $S{\sum }_i{a}_i{v}_i=0$ if and only if ${\sum }_i{a}_i{v}_i=0$.

A basis is a linearly independent spanning set of vectors in a vector space. Every vector is a unique finite linear combination of basis vectors. Independence supplies uniqueness, and spanning supplies existence. An ordered basis turns the coefficients into a coordinate vector.

For $\mathcal{B}=(b_1,b_2)=((2,1),(1,2))$, the matrix with these vectors as columns isHence the coordinate vector of $v=(5,4)$ is $[v{]}_{\mathcal{B}}=B^{-1}v=(2,1)$, and $B[v{]}_{\mathcal{B}}=v$ recovers the usual coordinates. The determinant $3$ is nonzero, so this recovery works for every vector in ${\mathbb{R}}^2$. A basis need not have perpendicular or unit-length vectors; an orthonormal basis adds those geometric conveniences when an inner product is available.

The point of changing basis is to simplify the operation being studied. Diagonalization uses an eigenbasis to turn a coupled linear map into independent scalings. In a space of polynomials, the basis $1,x,x^2$ makes differentiation easy to encode, while orthogonal polynomials can make approximation errors easier to measure. Gilbert Strang's MIT lecture on independence, basis and dimension develops this connection between attainable columns, redundant columns and coordinates.^[ref]

An ordered basis is a basis equipped with an ordering of its basis vectors. The order fixes which position of a coordinate vector corresponds to each basis vector.

For an ordered basis $\mathcal{B}=(b_1,\dots ,b_n)$, the coordinate vector $[v{]}_{\mathcal{B}}=(a_1,\dots ,a_n{)}^T$ records the unique coefficients satisfying $v={\sum }_i{a}_i{b}_i$. Change of basis changes this representation while preserving the underlying vector.

For $b_1=(1,0{)}^T$, $b_2=(1,1{)}^T$, the vector $v=(3,1{)}^T$ has coordinate vector $(2,1{)}^T$, since $2b_1+b_2=v$. These are the same coefficients extracted by the dual basis in its worked example. The vector is independent of the chosen basis; its list of coordinates is not. In an orthonormal eigenbasis, the coordinates additionally separate the action of a self-adjoint operator into independent scalar multiplications.

A coordinate of a vector in a chosen basis is one entry of its coordinate vector. The corresponding coordinate functional in the dual basis extracts it. For an orthonormal basis, it can instead be read as an inner product with the basis vector, as in orthonormal eigenbasis. For a nonorthogonal basis, ordinary dot products need not give the coordinates.

A basis vector is one of the vectors in a specified basis of a vector space. Its coordinate column in that basis has one entry equal to one and all other entries zero. Columns of a matrix of a linear map are the images of domain basis vectors, expressed in the chosen codomain basis.

For a basis $e_1,\dots ,e_n$ with cyclic indexing, the vectors $e_j+e_{j+1}$ have coefficient matrix $I+S$, where $S$ cyclically shifts coordinates. Its determinant is $1-(-1{)}^n$. They therefore form a basis exactly when $n$ is odd over a field of characteristic different from two. For even $n$, alternating coefficients produce a nontrivial linear dependence.

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.

The standard basis of the vector space ${\mathbb{R}}^n$ consists of the vectors $e_i$ whose $i$th coordinate is one and whose other coordinates are zero. Every vector $x$ has the unique expression $x={\sum }_i{x}_i{e}_i$ in this basis.

If a vector space is spanned by $n$ vectors, every linearly independent family in it has at most $n$ members. Consequently, any two finite bases have the same number of vectors.

For polynomials in fixed variables and of a fixed degree, the monomials of that degree form a basis of the corresponding homogeneous component.

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.

An axial vector transforms as ${\mathbf{a}}^{\prime }=(\det Q)Q\mathbf{a}$ under an orthogonal matrix $Q$. It agrees with a polar vector under a proper rotation matrix and has the additional determinant sign under an orientation-reversing transformation. Angular momentum and angular velocity are axial vectors.

A polar vector transforms as ${\mathbf{v}}^{\prime }=Q\mathbf{v}$ under an orthogonal matrix $Q$, including an orientation-reversing transformation. Displacement, velocity and force are polar vectors. An axial vector instead transforms as ${\mathbf{a}}^{\prime }=(\det Q)Q\mathbf{a}$.

A Euclidean vector is a vector in a finite-dimensional real inner product space. Choosing an orthonormal basis identifies it with its real vector components in ${\mathbb{R}}^n$. A geometric arrow represents a Euclidean vector by a length and a direction; translating the arrow without changing either represents the same Euclidean vector. The zero vector has length zero and no distinguished direction.^[ref]

For points $P=(1,1)$ and $Q=(4,5)$ in a Cartesian coordinate system, the displacement is $Q-P=(3,4)$. The arrow from $(-2,0)$ to $(1,4)$ represents the same Euclidean vector, even though its endpoints differ. An endpoint is a position; their difference is a vector. In an orthonormal basis $({\mathbf{e}}_1,{\mathbf{e}}_2)$,The Euclidean norm comes from the inner product: $\Vert \mathbf{v}{\Vert }_2^2=⟨\mathbf{v},\mathbf{v}⟩$. The sum of squared vector components is valid in an orthonormal basis; an arbitrary basis requires the appropriate Gram matrix. This distinction lets the geometric vector stay the same while its coordinate vector changes.^[ref]

Vector addition is componentwise. With $\mathbf{u}=(2,1)$ and $\mathbf{w}=(1,3)$,Geometrically, translate the second arrow so its tail starts at the first arrow's head. The vector sum joins the starting point to the final endpoint. Reversing the order gives the other route around the same parallelogram, so vector addition is commutative. This is the useful bookkeeping behind adding successive displacements, velocities, or forces: direction information survives the calculation. For a force, the Euclidean vector alone does not specify its point of application, which also affects the torque.^[ref]