# Basis (linear algebra)

In mathematics, a set B of vectors in a vector space *V* is called a **basis** if every element of *V* may be written in a unique way as a finite linear combination of elements of B. The coefficients of this linear combination are referred to as **components** or **coordinates** of the vector with respect to B. The elements of a basis are called **basis vectors**.

Equivalently, a set B is a basis if its elements are linearly independent and every element of V is a linear combination of elements of B.^{[1]} In other words, a basis is a linearly independent spanning set.

A vector space can have several bases; however all the bases have the same number of elements, called the *dimension* of the vector space.

This article deals mainly with finite-dimensional vector spaces. However, many of the principles are also valid for infinite-dimensional vector spaces.

A **basis** *B* of a vector space *V* over a field *F* (such as the real numbers **R** or the complex numbers **C**) is a linearly independent subset of *V* that spans *V*. This means that a subset B of *V* is a basis if it satisfies the two following conditions:

A vector space that has a finite basis is called finite-dimensional. In this case, the finite subset can be taken as *B* itself to check for linear independence in the above definition.

It is often convenient or even necessary to have an ordering on the basis vectors, for example, when discussing orientation, or when one considers the scalar coefficients of a vector with respect to a basis without referring explicitly to the basis elements. In this case, the ordering is necessary for associating each coefficient to the corresponding basis element. This ordering can be done by numbering the basis elements. In order to emphasize that an order has been chosen, one speaks of an **ordered basis**, which is therefore not simply an unstructured set, but a sequence, an indexed family, or similar; see § Ordered bases and coordinates below.

Many properties of finite bases result from the Steinitz exchange lemma, which states that, for any vector space V, given a finite spanning set S and a linearly independent set L of n elements of V, one may replace n well-chosen elements of S by the elements of L to get a spanning set containing L, having its other elements in S, and having the same number of elements as S.

Most properties resulting from the Steinitz exchange lemma remain true when there is no finite spanning set, but their proofs in the infinite case generally require the axiom of choice or a weaker form of it, such as the ultrafilter lemma.

be a basis of V. By definition of a basis, every **v** in V may be written, in a unique way, as

Typically, the new basis vectors are given by their coordinates over the old basis, that is,

be the column vectors of the coordinates of **v** in the old and the new basis respectively, then the formula for changing coordinates is

The formula can be proven by considering the decomposition of the vector **x** on the two bases: one has

If one replaces the field occurring in the definition of a vector space by a ring, one gets the definition of a module. For modules, linear independence and spanning sets are defined exactly as for vector spaces, although "generating set" is more commonly used than that of "spanning set".

Like for vector spaces, a *basis* of a module is a linearly independent subset that is also a generating set. A major difference with the theory of vector spaces is that not every module has a basis. A module that has a basis is called a *free module*. Free modules play a fundamental role in module theory, as they may be used for describing the structure of non-free modules through free resolutions.

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.

for suitable (real or complex) coefficients *a*_{k}, *b*_{k}. But many^{[3]} square-integrable 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.

For a probability distribution in **R**^{n} with a probability density function, such as the equidistribution in an *n*-dimensional 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 det[**x**_{1} ⋯ **x**_{n}] = 0 (zero determinant of the matrix with columns **x**_{i}), and the set of zeros of a non-trivial polynomial has zero measure. This observation has led to techniques for approximating random bases.^{[6]}^{[7]}

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 *n*-dimensional ball. Choose *N* independent random vectors from a ball (they are independent and identically distributed). Let *θ* be a small positive number. Then for

The figure (right) illustrates distribution of lengths N of pairwise almost orthogonal chains of vectors that are independently randomly sampled from the *n*-dimensional cube [−1, 1]^{n} 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 π/2 ± 0.037π/2 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 π/2 ± 0.037π/2 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 were 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 L_{Y} is linearly independent. Thus 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 of **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**.

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 has been proved that if every vector space has a basis, then the axiom of choice is true.^{[9]} Thus the two assertions are equivalent.