# Basis (linear algebra)

*Basis vector redirects here. For basis vector in the context of crystals, see crystal structure. For a more general concept in physics, see frame of reference.*

A set of vectors in a vector space *V* is called a **basis**, or a set of **basis vectors**, if the vectors are linearly independent and every other vector in the vector space is linearly dependent on these vectors.^{[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**. A vector space can have many different sets of basis vectors, however each set has the same number of elements, which defines the dimension of the vector space.

## Contents

## Definition

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
*linear independence*property,

- for all
*a*_{1}, …,*a*_{n}∈**F**, if*a*_{1}*v*_{1}+ … +*a*_{n}*v*_{n}= 0, then necessarily*a*_{1}= … =*a*_{n}= 0; and

- for all

- the
*spanning*property,

- for every
*x*in*V*it is possible to choose*a*_{1}, …,*a*_{n}∈**F**such that*x*=*a*_{1}*v*_{1}+ … +*a*_{n}*v*_{n}.

- for every

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 finite-dimensional. To deal with infinite-dimensional 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

- every finite subset
*B*_{0}⊆*B*obeys the independence property shown above; and - for every
*x*in*V*it is possible to choose*a*_{1}, …,*a*_{n}∈**F**and*v*_{1}, …,*v*_{n}∈*B*such that*x*=*a*_{1}*v*_{1}+ … +*a*_{n}*v*_{n}.

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.

## Expression of a basis

There are several ways to describe a basis for the space. Some are made *ad hoc* for a specific dimension. For example, there are several ways to give a basis in dim 3, like Euler angles.

The general case is to give a matrix with the components of the new basis vectors in columns. This is also the more general method because it can express any possible set of vectors even if it is not a basis. This matrix can be seen as three things:

**Basis Matrix**: Is a matrix that represents the basis, because its columns are the components of vectors of the basis. This matrix represents any vector of the new basis as linear combination of the current basis.

**Rotation operator**: When orthonormal bases are used, any other orthonormal basis can be defined by a rotation matrix. This matrix represents the rotation operator that rotates the vectors of the basis to the new one. It is exactly the same matrix as before because the rotation matrix multiplied by the identity matrix I has to be the new basis matrix.

**Change of basis matrix**: This matrix can be used to change different objects of the space to the new basis. Therefore is called "change of basis" matrix. It is important to note that some objects change their components with this matrix and some others, like vectors, with its inverse.

## Properties

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:

*B*is a minimal generating set of*V*, i.e., it is a generating set and no proper subset of*B*is also a generating set.*B*is a maximal set of linearly independent vectors, i.e., it is a linearly independent set but no other linearly independent set contains it as a proper subset.- Every vector in
*V*can be expressed as a linear combination of vectors in*B*in a unique way. If the basis is ordered (see*Ordered bases and coordinates*below) then the coefficients in this linear combination provide*coordinates*of the vector relative to the basis.

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} {E1,...,En} where En is the n-th column of the identity matrix which consists of all ones in the main diagonal and zeros everywhere else. This is because the columns of the identity matrix are linearly independent can always span a vector set by expressing it as a linear combination.

In P_{2} where P_{2} is the set of all polynomials of degree at most 2 {1,x,x^{2}} is the standard basis.

In M_{22} {M_{1,1},M_{1,2},M_{2,1},M_{2,2}} 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. This again is a standard basis since it is linearly independent and spanning.

## Examples

- Consider
**R**^{2}, the vector space of all coordinates (*a*,*b*) where both*a*and*b*are real numbers. Then a very natural and simple basis is simply the vectors**e**_{1}= (1,0) and**e**_{2}= (0,1): suppose that*v*= (*a*,*b*) is a vector in**R**^{2}, then*v*=*a*(1,0) +*b*(0,1). But any two linearly independent vectors, like (1,1) and (−1,2), will also form a basis of**R**^{2}.

- More generally, the vectors
**e**_{1},**e**_{2}, ...,**e**_{n}are linearly independent and generate**R**^{n}. Therefore, they form a basis for**R**^{n}and the dimension of**R**^{n}is*n*. This basis is called the*standard basis*.

- Let
*V*be the real vector space generated by the functions*e*^{t}and*e*^{2t}. These two functions are linearly independent, so they form a basis for*V*.

- Let
**R**[x] denote the vector space of real polynomials; then (1, x, x^{2}, ...) is a basis of**R**[x]. The dimension of**R**[x] is therefore equal to aleph-0.

## Extending to a basis

Let *S* be a subset of a vector space *V*. To extend *S* to a basis means to find a basis *B* 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. *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.

## Example of alternative proofs

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}.

### From the definition of *basis*

We have to prove that these two vectors are linearly independent and that they generate **R**^{2}.

Part I: If two vectors v,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:

- and and

Subtracting the first equation from the second, we obtain:

- so

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 (a,b) be an arbitrary element of **R ^{2}**, and show that there exist numbers r,s ∈

**R**such that:

Then we have to solve the equations:

Subtracting the first equation from the second, we get:

### By the dimension theorem

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.

### By the invertible matrix theorem

Simply compute the determinant

Since the above matrix has a nonzero determinant, its columns form a basis of **R**^{2}. See: invertible matrix.

## Ordered bases and coordinates

A basis is just 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 finite-dimensional 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 *n*-dimensional vector space over a field **F**. A choice of an ordered basis for *V* is equivalent to a choice of a linear isomorphism *φ* 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

*φ*:**F**^{n}→*V*

is a linear isomorphism. Define an ordered basis {*v*_{i}} for *V* by

*v*_{i}=*φ*(**e**_{i}) for 1 ≤*i*≤*n*

where {**e**_{i}} is the standard basis for **F**^{n}.

Conversely, given an ordered basis, consider the map defined by

*φ*(*x*) =*x*_{1}*v*_{1}+*x*_{2}*v*_{2}+ ... +*x*_{n}*v*_{n},

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 *φ* is a linear isomorphism.

These two constructions are clearly inverse to each other. Thus ordered bases for *V* are in 1-1 correspondence with linear isomorphisms **F**^{n} → *V*.

The inverse of the linear isomorphism *φ* determined by an ordered basis {*v*_{i}} equips *V* with *coordinates*: if, for a vector *v* ∈ *V*, *φ*^{−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 *φ*^{−1} is linear. Hence they are linear functionals. They form a basis for the **dual space** of *V*, called the **dual basis**.

## Related notions

### Analysis

In the context of infinite-dimensional vector spaces over the real or complex numbers, the term * Hamel basis* (named after Georg Hamel) or

*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 infinite-dimensional 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. The term*

**algebraic basis***is also commonly used to mean a basis for the real numbers*

**Hamel basis****R**as a vector space over the field

**Q**of rational numbers. (In this case, the dimension of

**R**over

**Q**is uncountable, specifically the continuum, the cardinal number

**2**

^{ℵ0}.)

The common feature of the other notions is that they permit the taking of infinite linear combinations of the basic 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 infinite-dimensional spaces is justified by the fact that the Hamel basis becomes "too big" in Banach spaces: If *X* is an infinite-dimensional 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, finite-dimensional spaces have by definition finite bases and there are infinite-dimensional (*non-complete*) normed spaces which have countable Hamel bases. Consider , the space of the sequences of real numbers which have only finitely many non-zero elements, with the norm Its standard basis, consisting of the sequences having only one non-zero element, which is equal to 1, is a countable Hamel basis.

#### Example

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, 2π] that are square-integrable 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 square-integrable on [0, 2π] is an "infinite linear combination" of them, in the sense that

for suitable (real or complex) coefficients *a*_{k}, *b*_{k}. But most 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.

### Affine geometry

The related notions of an affine space, projective space, convex set, and cone have related notions of **Template:Visible anchor**^{[2]} (a basis for an *n*-dimensional affine space is points in general linear position), **Template:Visible anchor** (essentially the same as an affine basis, this is points in general linear position, here in projective space), **Template:Visible anchor** (the vertices of a polytope), and **Template:Visible anchor**^{[3]} (points on the edges of a polygonal cone); see also a Hilbert basis (linear programming).

## Proof that every vector space has a basis

Let **V** be any vector space over some field **F**. Every vector space must contain at least one element: the zero vector **0**.

Note that if **V** = {**0**}, then the empty set is a basis for **V**. Now we consider the case where **V** contains at least one nonzero element, say **v**.

Define the set **X** as all linear independent subsets of **V**. Note that since **V** contains the nonzero element **v**, the singleton subset L = {**v**} of **V** is necessarily linearly independent.

Hence the set **X** contains at least the subset L = {**v**}, and so **X** is nonempty.

We let **X** be partially ordered by inclusion: If L_{1} and L_{2} belong to **X**, we say that L_{1} ≤ L_{2} when L_{1} ⊂ L_{2}. It is easy to check that (**X**, ≤) satisifies the definition of a partially ordered set.

We now note that if **Y** is a subset of **X** that is totally ordered by ≤, then the union L_{Y} of all the elements of **Y** (which are themselves certain subsets of **V**) is an upper bound for **Y**. To show this, it is necessary to verify both that a) L_{Y} belongs to **X**, and that b) every element L of **Y** satisfies L ≤ L_{Y}. Both a) and b) are easy to check.

Now we apply Zorn's lemma, which asserts that because **X** is nonempty, and every totally ordered subset of the partially ordered set (**X**, ≤) has an upper bound, it follows 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}.)

Finally we claim that L_{max} is a basis for **V**. Since L_{max} belongs to **X**, we already know that L_{max} is a linearly independent subset of **V**.

Now **suppose** L_{max} does **not** span **V**. Then there exists some vector **w** of **V** that cannot be expressed as a linearly combination of elements of L_{max} (with coefficients in the field **F**). Note that such a vector **w** cannot be an element of L_{max}.

Now consider the subset L_{w} of **V** defined by L_{w} = L_{max} ∪ {**w**}. It is easy to see that **a**) L_{max} ≤ L_{w} (since L_{max} is a subset of L_{w}), and that **b**) L_{max} ≠ L_{w} (because L_{w} contains the vector **w** that is not contained in L_{max}).

But the combination of **a**) and **b**) above contradict the fact that L_{max} is a maximal element of **X**, which we have already proved. This contradiction shows that the assumption that L_{max} does not span **V** was not true.

Hence L_{max} **does** span **V**. Since we also know that L_{max} is linearly independent over the field **F**, this verifies that L_{max} is a basis for **V**. Which proves that the arbitrary vector space **V** has a basis.

**Note**: This proof relies on Zorn's lemma, which is logically equivalent to the Axiom of Choice. It turns out that, conversely, the assumption that every vector space has a basis can be used to prove the Axiom of Choice. Thus the two assertions are logically equivalent.

## See also

## Notes

## References

### General references

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## External links

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