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A scalar is an element of the field over which a vector space is defined.

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.

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]

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.

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 $w\in W$.

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

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.

The nth tensor power is the iterated tensor product of modules $V^{\otimes n}=V\otimes \cdots \otimes V$.

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.

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.

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.

The tautological bundle over a projective space has as its fiber over a line precisely that line. The real tautological line bundle over ${\mathbb{RP}}^m$ has first Stiefel-Whitney class equal to the degree-one generator.

This is a fiberwise orientation compatible with local trivializations of a vector bundle. An $R$-orientation of a rank-$d$ vector bundle is a coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ for every fiber, equivalently a Thom class with the corresponding fiberwise restriction.

The tensor product of vector bundles is formed fiberwise. Tensor products of local trivializations have transition functions given by tensor products of the original transition matrices.

The signs of homogeneous linear forms $x\mapsto x^T\beta$ on ${\mathbb{R}}^p$ have VC dimension at most $p$. The Sauer-Shelah growth bound therefore gives

The map $x\mapsto (x_i{x}_j{)}_{i\le j}$ embeds homogeneous quadratic forms in a linear function space of dimension $d(d+1)/2$.

The indicators of closed Euclidean balls in ${\mathbb{R}}^d$ have VC dimension at most $d+2$. Indeed,wherebelongs to the span of $\Vert x{\Vert }_2^2,x_1,\dots ,x_d,1$.

For a complex field $\psi$, the canonical Hamiltonian evolution conventionis accompanied by its complex conjugate equation.

Several monochromatic arithmetic progressions are color-focused when they have different colors and extend by one further term to the same point. Such focused families give an elementary induction proof of the length-three case of the Van der Waerden theorem.

For all positive integers $k,m$, some $n$ has the following property: every $k$-coloring of $[n]$ contains $a,d>0$ for whichis monochromatic. Thus the common difference has the same color as the progression.