In mathematics, a set of elements (vectors) in a vector space V is called a basis, or a set of basis vectors, if the vectors are linearly independent and every vector in the vector space is a linear combination of this set.^{[1]} In more general terms, a basis is a linearly independent spanning set.
Given a basis of a vector space V, every element of V can be expressed uniquely as a linear combination of basis vectors, whose coefficients are referred to as vector coordinates or components. The computation of these components is sometimes called decomposition of a vector on a basis. A vector space can have several distinct sets of basis vectors; however each such set has the same number of elements, with this number being the dimension of the vector space.
A basis B of a vector space V over a field F is a linearly independent subset of V that spans V.
In more detail, suppose that B = { v_{1}, ..., v_{n} } is a finite subset of a vector space V over a field F (such as the real or complex numbers R or C). Then B is a basis if it satisfies the following conditions:
The numbers a_{i} are called the coordinates of the vector x with respect to the basis B, and by the first property they are uniquely determined.
A vector space that has a finite basis is called finitedimensional. To deal with infinitedimensional spaces, we must generalize the above definition to include infinite basis sets. We therefore say that a set (finite or infinite) B ? V is a basis, if
The sums in the above definition are all finite because without additional structure the axioms of a vector space do not permit us to meaningfully speak about an infinite sum of vectors. Settings that permit infinite linear combinations allow alternative definitions of the basis concept: see Related notions below.
It is often convenient to list the basis vectors in a specific order, for example, when considering the transformation matrix of a linear map with respect to a basis. We then speak of an ordered basis, which we define to be a sequence (rather than a set) of linearly independent vectors that span V: see Ordered bases and coordinates below.
Again, B denotes a subset of a vector space V. Then, B is a basis if and only if any of the following equivalent conditions are met:
Every vector space has a basis. The proof of this requires the axiom of choice. All bases of a vector space have the same cardinality (number of elements), called the dimension of the vector space. This result is known as the dimension theorem, and requires the ultrafilter lemma, a strictly weaker form of the axiom of choice.
Also many vector sets can be attributed a standard basis which comprises both spanning and linearly independent vectors.
Standard bases for example:
In R^{n}, where e_{i} is the ith column of the identity matrix.
In P_{2}, where P_{2} is the set of all polynomials of degree at most 2, is the standard basis.
In M_{22}, where M_{22} is the set of all 2×2 matrices and M_{m,n} is the 2×2 matrix with a 1 in the m,n position and zeros everywhere else.
Given a vector space V over a field F and suppose that and are two bases for V. By definition, if ? is a vector in V then for a unique choice of scalars in F called the coordinates of ? relative to the ordered basis The vector in F^{n} is called the coordinate tuple of ? (relative to this basis). The unique linear map with for is called the coordinate isomorphism for V and the basis Thus if and only if .
A set of vectors can be represented by a matrix of which each column consists of the components of the corresponding vector of the set. As a basis is a set of vectors, a basis can be given by a matrix of this kind. The change of basis of any object of the space is related to this matrix. For example, coordinate tuples change with its inverse.
Let S be a subset of a vector space V. To extend S to a basis of V means to find a basis B of V that contains S as a subset. This can be done if and only if S is linearly independent. Almost always, there is more than one such B, except in rather special circumstances (i.e. that S is already a basis, or S is empty and V has two elements).
A similar question is when does a subset S contain a basis. This occurs if and only if S spans V. In this case, S will usually contain several different bases.
Often, a mathematical result can be proven in more than one way. Here, using three different proofs, we show that the vectors (1,1) and (1,2) form a basis for R^{2}.
We have to prove that these two vectors are linearly independent and that they generate R^{2}.
Part I: If two vectors v and w are linearly independent, then (a and b scalars) implies .
To prove that they are linearly independent, suppose that there are numbers a, b such that:
(i.e., they are linearly dependent). Then:
Subtracting the first equation from the second, we obtain:
Adding this equation to the first equation then:
Hence we have linear independence.
Part II: To prove that these two vectors generate R^{2}, we have to let be an arbitrary element of R^{2}, and show that there exist numbers such that:
Then we have to solve the equations:
Subtracting the first equation from the second, we get:
Since (1,2) is clearly not a multiple of (1,1) and since (1,1) is not the zero vector, these two vectors are linearly independent. Since the dimension of R^{2} is 2, the two vectors already form a basis of R^{2} without needing any extension.
Simply compute the determinant
Since the above matrix has a nonzero determinant, its columns form a basis of R^{2}. See: invertible matrix.
A basis is a linearly independent set of vectors with or without a given ordering. For many purposes it is convenient to work with an ordered basis. For example, when working with a coordinate representation of a vector it is customary to speak of the "first" or "second" coordinate, which makes sense only if an ordering is specified for the basis. For finitedimensional vector spaces one typically indexes a basis {v_{i}} by the first n integers. An ordered basis is also called a frame.
Suppose V is an ndimensional vector space over a field F. A choice of an ordered basis for V is equivalent to a choice of a linear isomorphism ? from the coordinate space F^{n} to V.
Proof. The proof makes use of the fact that the standard basis of F^{n} is an ordered basis.
Suppose first that
is a linear isomorphism. Define an ordered basis {v_{i}} for V by
where {e_{i}} is the standard basis for F^{n}.
Conversely, given an ordered basis, consider the map defined by
where x = x_{1}e_{1} + x_{2}e_{2} + ... + x_{n}e_{n} is an element of F^{n}. It is not hard to check that ? is a linear isomorphism.
These two constructions are clearly inverse to each other. Thus ordered bases for V are in 11 correspondence with linear isomorphisms F^{n} > V.
The inverse of the linear isomorphism ? determined by an ordered basis {v_{i}} equips V with coordinates: if, for a vector v ? V, ?^{1}(v) = (a_{1}, a_{2},...,a_{n}) ? F^{n}, then the components a_{j} = a_{j}(v) are the coordinates of v in the sense that v = a_{1}(v) v_{1} + a_{2}(v) v_{2} + ... + a_{n}(v) v_{n}.
The maps sending a vector v to the components a_{j}(v) are linear maps from V to F, because of ?^{1} is linear. Hence they are linear functionals. They form a basis for the dual space of V, called the dual basis.
In the context of infinitedimensional vector spaces over the real or complex numbers, the term Hamel basis (named after Georg Hamel) or algebraic basis can be used to refer to a basis as defined in this article. This is to make a distinction with other notions of "basis" that exist when infinitedimensional vector spaces are endowed with extra structure. The most important alternatives are orthogonal bases on Hilbert spaces, Schauder bases, and Markushevich bases on normed linear spaces. In the case of the real numbers R viewed as a vector space over the field Q of rational numbers, Hamel bases are uncountable, and have specifically the cardinality of the continuum, which is the cardinal number where is the smallest infinite cardinal, the cardinal of the integers.
The common feature of the other notions is that they permit the taking of infinite linear combinations of the basis vectors in order to generate the space. This, of course, requires that infinite sums are meaningfully defined on these spaces, as is the case for topological vector spaces  a large class of vector spaces including e.g. Hilbert spaces, Banach spaces, or Fréchet spaces.
The preference of other types of bases for infinitedimensional spaces is justified by the fact that the Hamel basis becomes "too big" in Banach spaces: If X is an infinitedimensional normed vector space which is complete (i.e. X is a Banach space), then any Hamel basis of X is necessarily uncountable. This is a consequence of the Baire category theorem. The completeness as well as infinite dimension are crucial assumptions in the previous claim. Indeed, finitedimensional spaces have by definition finite bases and there are infinitedimensional (noncomplete) normed spaces which have countable Hamel bases. Consider , the space of the sequences of real numbers which have only finitely many nonzero elements, with the norm Its standard basis, consisting of the sequences having only one nonzero element, which is equal to 1, is a countable Hamel basis.
In the study of Fourier series, one learns that the functions {1} ? { sin(nx), cos(nx) : n = 1, 2, 3, ... } are an "orthogonal basis" of the (real or complex) vector space of all (real or complex valued) functions on the interval [0, 2?] that are squareintegrable on this interval, i.e., functions f satisfying
The functions {1} ? { sin(nx), cos(nx) : n = 1, 2, 3, ... } are linearly independent, and every function f that is squareintegrable on [0, 2?] is an "infinite linear combination" of them, in the sense that
for suitable (real or complex) coefficients a_{k}, b_{k}. But many^{[2]} squareintegrable functions cannot be represented as finite linear combinations of these basis functions, which therefore do not comprise a Hamel basis. Every Hamel basis of this space is much bigger than this merely countably infinite set of functions. Hamel bases of spaces of this kind are typically not useful, whereas orthonormal bases of these spaces are essential in Fourier analysis.
The geometric notions of an affine space, projective space, convex set, and cone have related notions of basis.^{[3]} An affine basis for an ndimensional affine space is points in general linear position. A projective basis is points in general position, in a projective space of dimension n. A convex basis of a polytope is the set of the vertices of its convex hull. A cone basis^{[4]} consists of one point by edge of a polygonal cone. See also a Hilbert basis (linear programming).
For a probability distribution in R^{n} with a probability density function, such as the equidistribution in a ndimensional ball with respect to Lebesgue measure, it can be shown that n randomly and independently chosen vectors will form a basis with probability one, which is due to the fact that n linearly dependent vectors x_{1}, ..., x_{n} in R^{n} should satisfy the equation (zero determinant of the matrix with columns x_{i}), and the set of zeros of a nontrivial polynomial has zero measure. This observation has led to techniques for approximating random bases.^{[5]}^{[6]}
It is difficult to check numerically the linear dependence or exact orthogonality. Therefore, the notion of ?orthogonality is used. For spaces with inner product, x is ?orthogonal to y if (that is, cosine of the angle between x and y is less than ?).
In high dimensions, two independent random vectors are with high probability almost orthogonal, and the number of independent random vectors, which all are with given high probability pairwise almost orthogonal, grows exponentially with dimension. More precisely, consider equidistribution in ndimensional ball. Choose N independent random vectors from a ball (they are independent and identically distributed). Let ? be a small positive number. Then for

N random vectors are all pairwise ?orthogonal with probability .^{[6]} This N growth exponentially with dimension n and for sufficiently big n. This property of random bases is a manifestation of the socalled measure concentration phenomenon.^{[7]}
The figure (right) illustrates distribution of lengths N of pairwise almost orthogonal chains of vectors that are independently randomly sampled from the ndimensional cube as a function of dimension, n. A point is first randomly selected in the cube. The second point is randomly chosen in the same cube. If the angle between the vectors was within then the vector was retained. At the next step a new vector is generated in the same hypercube, and its angles with the previously generated vectors are evaluated. If these angles are within then the vector is retained. The process is repeated until the chain of almost orthogonality breaks, and the number of such pairwise almost orthogonal vectors (length of the chain) is recorded. For each n, 20 pairwise almost orthogonal chains where constructed numerically for each dimension. Distribution of the length of these chains is presented.
Let V be any vector space over some field F. Let X be the set of all linearly independent subsets of V.
The set X is nonempty since the empty set is an independent subset of V, and it is partially ordered by inclusion, which is denoted, as usual, by ?.
Let Y be a subset of X that is totally ordered by ?, and let L_{Y} be the union of all the elements of Y (which are themselves certain subsets of V).
Since (Y, ?) is totally ordered, every finite subset of L_{Y} is a subset of an element of Y, which is a linearly independent subset of V, and hence every finite subset of L_{Y} is linearly independent. Thus L_{Y} is linearly independent, so L_{Y} is an element of X. Therefore, L_{Y} is an upper bound for Y in (X, ?): it is an element of X, that contains every element Y.
As X is nonempty, and every totally ordered subset of (X, ?) has an upper bound in X, Zorn's lemma asserts that X has a maximal element. In other words, there exists some element L_{max} of X satisfying the condition that whenever L_{max} ? L for some element L of X, then L = L_{max}.
It remains to prove that L_{max} is a basis of V. Since L_{max} belongs to X, we already know that L_{max} is a linearly independent subset of V.
If L_{max} would not span V, there would exist some vector w of V that cannot be expressed as a linear combination of elements of L_{max} (with coefficients in the field F). In particular, w cannot be an element of L_{max}. Let L_{w} = L_{max} ? {w}. This set is an element of X, that is, it is a linearly independent subset of V (because w is not in the span of L_{max}, and L_{max} is independent). As L_{max} ? L_{w}, and L_{max} ? L_{w} (because L_{w} contains the vector w that is not contained in L_{max}), this contradicts the maximality of L_{max}. Thus this shows that L_{max} spans V.
Hence L_{max} is linearly independent and spans V. It is thus a basis of V, and this proves that every vector space has a basis.
This proof relies on Zorn's lemma, which is equivalent to the axiom of choice. Conversely, it may be proved that if every vector space has a basis, then the axiom of choice is true; thus the two assertions are equivalent.