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.
• OV ∈ 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 = {v1, v2, v3, …, vn} 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 0V is linearly dependent. The set {0V} or a 'trivial subspace,' of V. The null set, Ø, is considered to be the bases for the trivial subspace, {0V}, which is therefore a zero-dimensional subspace.

• A standard basis is the 'obvious' basis used to describe the vector space Rn, using the form:
e1 = (1, 0, 0, …, 0)
e2 = (0, 1, 0, …, 0)
e3 = (0, 0, 1, …, 0)
¦             ¦
en = (0, 0, 0, …, 1)

For example, the standard basis for R3 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 R3.

• The standard basis for polynomials of degree n or less is:
S = {1, x, x2, x3, …, xn}.
• 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 Rn, S must span Rn and must be linearly independent (however, note that Rn 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 R2.

Let x = (x1, x2) represent an arbitrary vector in R2. We must show that x can be written as a linear combination of elements in the vector space:
v1=(1,1) and v2=(1,-1).
Let λ ∈ R be a scalar. This is illustrated by:
x = λ1v1 + λ2v2
Then,
x = λ1(1,1) + λ2(1,-1) by substitution
x = (λ1+λ2, λ1-λ2)
This can be written as two equations:
x1= λ1+λ2
x2= λ1-λ2
whose coefficients can be written as an augmented matrix:
[1   1  x1]
[1  -1  x2]
This matrix is then row-reduced to reduced row-echelon form, or Hermite form, using elementary row operations. The Hermite matrix is:
[1  0  (x1 + x2)/2]
[0  1  (x1  - x2)/2]
which states that λ1 = (x1+ x2)2 , λ2 = (x1- x2)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 λ1v1 + λ2v2 = 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 v1 and v2, 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=R3. Because the subspace is a copy of the R2 plane within R3, 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.

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