Basis

1. Definition

Consider the vector equation

c1b1 + c2b2 + ... + cnbn = v

in a vector space V, where c1, c2, ..., cn are the unknown variables. The existence of solutions for all vV is equivalent to the vectors b1, b2, ..., bn spanning V. Moreover, the concept of linear independence among b1, b2, ..., bn was introduced to describe the uniqueness of the solution. Thus the best case for a collection of vectors is described in the following definition.

Vectors b1, b2, ..., bnV form a basis of V if

• b1, b2, ..., bn span V.
• b1, b2, ..., bn are linearly independent.

The discussion at the beginning immediately gives us the following meaning of basis.

B = {b1, b2, ..., bn} is a basis ⇔ c1b1 + c2b2 + ... + cnbn = v has a unique solution for any v.

Example The standard basis vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) span R3 and are linearly independent. Therefore the standard basis {e1, e2, e3} is indeed a basis of R3. The fact clearly extends to the standard basis of general Rn.

The monomials 1, t, t2, ..., tn span Pn and are linearly independent. Therefore the monomials form a basis of Pn.

Similarly, the equality

 A = [ a b ] = a[ 1 0 ] + b[ 0 1 ] + c[ 0 0 ] + d[ 0 0 ] = aE11 + bE12 + cE21 + dE22 c d 0 0 0 0 1 0 0 1

has the following two consequences

• Any 2 by 2 matrix A is a linear combination of E11, E12, E21, E22.
• If a linear combination A of E11, E12, E21, E22 (with a, b, c, d as coefficients) is the zero matrix, then all the entries of A (which are a, b, c, d) must be zero.

The first consequence means exactly that M(2, 2) is spanned by E11, E12, E21, E22. The second consequence means exactly that E11, E12, E21, E22 are linearly independent. Thus {E11, E12, E21, E22} is a basis of M(2, 2).

In general, let Eij be the m by n matrix with 1 as the ij-entry and 0 as all the other entries. Then {Eij: 1 ≤ im, 1 ≤ jn} is a basis of M(m, n).

Example In an earlier example, the system

 3x1 + x2 - x3 = 2 x1 - x2 + x3 = 2 2x1 + 2x2 + x3 = 6

was shown to have a unique solution. Since the number of rows is the same as the number of columns, by the basic principal of linear algebra, the general system

 3x1 + x2 - x3 = b1 x1 - x2 + x3 = b2 2x1 + 2x2 + x3 = b3

has a unique solution for any right side. Since the general system is the same as the vector equation

x1(3, 1, 2) + x2(1, -1, 2) + x3(-1, 1, 1) = (b1, b2, b3),

the existence for the general system means that (3, 1, 2), (1, -1, 2), (-1, 1, 1) span R3, and the uniqueness means the three vectors are linearly independent. Thus we conclude that the three vectors form a basis of R3.

The argument in the second example applies to general collections of euclidean vectors. After all, being a basis of Rn means always existence (span Rn) plus uniqueness (independence). According to the study on the basic principal of linear algebra, the exact condition is described as follows.

If the columns of a matrix A is a basis of Rn, then A has to be an n by n matrix. Moreover, for an n by n matrix A, the following are equivalent.

• Columns of A form a basis of Rn.
• Ax = b has a unique solution for any b.
• Ax = b has solutions for any b.
• The solution of Ax = b is unique.
• A is invertible.

For the equivalence to invertibility, see here. For the criterion of euclidean basis from the computational viewpoint, see here.