*A basis is a minimal subset (of vectors) that spans a particular vector space. (Plural: “bases.”)*

__BACKGROUND CONCEPTS__

A vector space (V) is *any nonempty* set (of vectors, matrices, polynomials, or functions) that satisfies *specific properties* of addition and scalar multiplication. A vector space consists of four entities: a set of vectors, a field of scalars, and two operations. Most commonly, these elements are in the field of real numbers, *R*.

A subspace W is a nonempty subset of a vector space V that satisfies the following conditions:

- If the vectors u, v ∈ W, then u + v ∈ W
- If u ∈ W, λ ∈ F (any scalar in a field F), then λu ∈ W.
- O
_{V} ∈ W

A subset is said to span a vector space if every vector in V can be written as a linear combination of the vectors in the subset. The span(S) is a subspace of V.

__DEFINITION__

A basis S is a linearly independent set of vectors S = {v_{1}, v_{2}, v_{3}, …, v_{n}} that spans a vector space V.

This states that the basis must have enough vectors to span V but not so many vectors that there is redundancy within the set, i.e. every element of V can be written as a linear combination of the basis S in a unique way.

__CONTINUATION__

A basis can be of two forms: A finite dimensional basis consists of a finite number of vectors. Otherwise, the set is an infinite dimensional basis. A basis cannot contain a zero element (however, a subspace must) because any set with 0_{V} is linearly dependent. The set {0_{V}} or a 'trivial subspace,' of V. The null set, Ø, is considered to be the bases for the trivial subspace, {0_{V}}, which is therefore a zero-dimensional subspace.

- A standard basis is the 'obvious' basis used to describe the vector space
*R*^{n}, using the form:
e_{1} = (1, 0, 0, …, 0)

e_{2} = (0, 1, 0, …, 0)

e_{3} = (0, 0, 1, …, 0)

¦ ¦

e_{n} = (0, 0, 0, …, 1)

For example, the standard basis for *R*^{3} is S = {(1,0,0), (0,1,0), (0,0,1)}.

Other bases can be created that span the same vector space by replacing any element of the bases by a scalar multiple of itself, or adding scalar multiples of other basis elements to it.

Example: The set S = {(-7,0,0), (0, ¼, 0), (0,0,666)} and {(1,2,3), (0,1,0), (0,0,1)} are also bases for **R**^{3}.

The standard basis for polynomials of degree *n* or less is:

S = {1, x, x^{2}, x^{3}, …, x^{n}}.

The standard basis for 2x2 matrices is:

[1 0] [0 1] [0 0] [0 0]

S = { [0 0], [0 0], [1 0], [0 1] }

Bases for mxn matrices are formed similarly with '1' as a single entry with '0' everywhere else (There are mn such matrices in this basis.)

__MORE PROPERTIES__

- Every linearly independent set can be extended to a basis of a vector space (if the set is not already a basis).
- Similarly, every spanning set of a vector space can be reduced to a basis of V (again, if the set is not already a basis).
- Every basis of the same vector space V has the same number of elements.

__VERIFYING BASES__

Recall that according to the definition of a basis for *R*^{n}, S must span *R*^{n} and must be linearly independent (however, note that *R*^{n} is only one type of vector space for which a basis can be created).

Example: Show that S = {(1,1), (1,-1)} is a basis for *R*^{2}.

Let **x** = (x_{1}, x_{2}) represent an arbitrary vector in *R*^{2}. We must show that **x** can be written as a linear combination of elements in the vector space:
v_{1}=(1,1) and v_{2}=(1,-1).

Let λ ∈ *R* be a scalar. This is illustrated by:
**x** = λ1v_{1} + λ2v_{2}

Then,
**x** = λ1(1,1) + λ2(1,-1) by substitution

**x** = (λ1+λ2, λ1-λ2)

This can be written as two equations:
x_{1}= λ1+λ2

x_{2}= λ1-λ2

whose coefficients can be written as an augmented matrix:
[1 1 x_{1}]

[1 -1 x_{2}]

This matrix is then row-reduced to reduced row-echelon form, or Hermite form, using elementary row operations. The Hermite matrix is:
[1 0 (x_{1} + x_{2})/2]

[0 1 (x_{1} - x_{2})/2]

which states that λ1 = (x_{1}+ x_{2})*2 , λ2 = (x*_{1}- x_{2})2.

This matrix is of full rank, meaning that all rows are linearly independent. This shows that the system of equations has a unique solution. Therefore, we conclude that S spans *R*.

To check for linear independence, we examine the equation λ1v_{1} + λ2v_{2} = 0. If S is linearly independent, then only one solution (for the values of λ) exits, the trivial solution, λ1 = λ2 = 0. Plugging in values for v_{1} and v_{2}, we have:
λ1(1,1) + λ2(1,-1) = (0,0)

(λ1+λ2 , λ1-λ2) = (0,0)

λ1+λ2 = 0

λ1 - λ2 = 0

By inspection (or by setting up another row-reduction of the augmented matrix) we see that the system has a unique solution, the trivial solution. Therefore, S is linearly independent.
[1 1 0] [1 0 0]

[1 -1 0] → [0 1 0] , which states λ1 = 0 and λ2 = 0.

__FINDING A BASIS__

- Use the fact that any linearly independent set can be extended to a basis.
- Limitations on the vector space will also reduce the size of the basis.
- Example: The equation y=z defines a subspace of the vector space V=
*R*^{3}. Because the subspace is a copy of the *R*^{2} plane within *R*^{3}, the basis will only contain two elements.
- When building a basis, choose linearly independent vectors that are contained in the subspace until the entire subspace is spanned by the set.
- S={(0,1,1), (5,9,9)} satisfies the y=z condition (above) and has linear independence.

__DIMENSION OF A BASIS__

If a basis of a vector space V contains *n* elements (finitely many), then the dimension of V, dim(V), is *n*.

Example: The dimension of the plane y=z (with basis S={(0,1,1), (5,9,9)}) is 2. Note: a subspace has dimension, a basis does not.

- If W is a subspace of V, then dim(W) ≤ dim(V).
- If dim(W) = dim(V), then W=V (only when W is a subspace of V).

Sources:

Blyth, T.S., and E.F. Robertson. __Basic Linear Algebra.__ 2nd ed. London: Springer-Verlag London Limited, 2002.

Edwards, Bruce H., and Ron Larson. __Elementary Linear Algebra.__ Boston: Houghton Mifflin Company, 2000.