In linear algebra, a basis is a set of vectors that, in a linear combination, can represent every vector in a given vector space. In other words, a basis is a linearly independent spanning set.

Definition

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A basis B of a vector space V is a linearly independent subset of V that spans (or generates) V.

In more detail, suppose that B = { v₁, …, v_n } is a finite subset of a vector space V over a field F. Then, B is a basis, if it satisfies the following conditions:

• the linear independence property,

for all a₁, …, a_n ∈ F, if a₁v₁ + … + a_nv_n = 0, then necessarily a₁ = … = a_n = 0; and

• the spanning property,

for every x in V it is possible to choose a₁, …, a_n ∈ F such that x = a₁v₁ + … + a_nv_n.

A vector space that admits 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 B₀ ⊆ B obeys the independence property shown above; and
• for every x in V it is possible to choose a₁, …, a_n ∈ F and v₁, …, v_n ∈ B such that x = a₁v₁ + … + a_nv_n.

The axioms of a vector space do not permit us to meaningfully speak about an infinite sum of vectors. That is why the sums in the above definition are all finite. Settings that permit infinite linear combinations allow alternative definitions of the basis concept: see ''Related notions below.

When we want to describe the matrix of a linear transformation and in some other situations, it is convenient to list the basis vectors in a specific order. 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. Here is another way to think about this:

every ordered basis of a finite-dimensional vector space V corresponds to a linear isomorphism from f:Rⁿ → V, and vice versa.

Proof. Given such an isomorphism, the sequence f(e₁), ... , f(e_n), where e₁, ... e_n is the standard basis of Rⁿ, is an ordered basis of V. Conversely, given an ordered basis, the mapping defined by

f(a) = a₁v₁ + … + a_nv_n,

where a= a₁e₁ + … + a_ne_n is an element of Rⁿ, is necessarily a linear isomorphism.

Properties

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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 but no proper subset of B is.
• 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.

The theorem that every vector space has a basis is implied by the well-ordering theorem, or any other equivalent of the axiom of choice. (Proof: Well-order the elements of the vector space. Create the subset of all elements not linearly dependent on their predecessors. This is easily shown to be a basis). The converse is also true. All bases of a vector space have the same cardinality (number of elements), called the dimension of the vector space. The latter result is known as the dimension theorem, and requires the ultrafilter lemma, a strictly weaker form of the axiom of choice.

Examples

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• Consider R², the vector space of all co-ordinates (a, b) where both a and b are real numbers. Then a very natural and simple basis is simply the vectors e₁ = (1,0) and e₂ = (0,1): suppose that v = (a, b) is a vector in R², 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 R² (see the section Proving that a set is a basis further down).

• More generally, the vectors e₁, e₂, ..., e_n are linearly independent and generate Rⁿ. Therefore, they form a basis for Rⁿ and the dimension of Rⁿ 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 Rdenote the vector space of real polynomials; then (1, x, x², ...) is a basis of R^* is therefore equal to aleph-0.

Basis extension

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Between any linearly independent set and any generating set there is a basis. More formally: if L is a linearly independent set in the vector space V and G is a generating set of V containing L, then there exists a basis of V that contains L and is contained in G. In particular (taking G = V), any linearly independent set L can be "extended" to form a basis of V. These extensions are not unique.

Proving that a set is a basis

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To prove that a set B is a basis for a (finite-dimensional) vector space V, it is sufficient to show that the number of elements in B equals the dimension of V, and one of the following:

• B is linearly independent, or
• span(B) = V.

Examples of proofs

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As an easy example, let us show that the vectors (1,1) and (-1,2) form a basis for R². The following proof methods require increasing amounts of sophistication and decreasing amounts of effort.

By brute force

We have to prove that these two vectors are linearly independent and that they generate R².

Part I: To prove that they are linearly independent, suppose that there are numbers a,b such that:

$a(1,1)+b(-1,2)=(0,0).\; \backslash ,$

Then:

$$

(a-b,a+2b)=(0,0) \,