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Mathematical Expression Editor

We discuss existence of bases of and subspaces of , and define dimension.

VSP-0035: Bases and Dimension

Recall that a basis of a subspace of is a subset of that is linearly independent and
spans . A basis allows us to uniquely express every element of as a linear
combination of the elements of the basis. Several questions may come to mind at this
time. Does every subspace of have a basis? We know that bases are not
unique. If there is more than one basis, what, if anything, do they have in
common?

Exploring Dimension

How would you describe
If you answered that is a line in , you are correct. While the two vectors span the
line, it is not necessary to have both of them in the spanning set to describe the
line.

What is the minimum number of vectors needed to span a line?

Answer: .

Observe also that the vectors in the given spanning set are not linearly independent,
so they do not form a basis for . How many vectors would a basis for have?

Answer: .

Now consider another subspace of :
Geometrically, is a plane in . Note that the vectors in the spanning set are linearly
independent. Can we remove one of the vectors and have the remaining vector span
the plane?

What is the minimum number of vectors needed to span a plane?

Answer: .

How many vectors would a basis for a plane have?

Answer: .

Our observations in Exploration init:dimensionintro hint at the idea of dimension. We know that a line
is a one-dimensional object, a plane is a two-dimensional object, and the space we
reside in is three-dimensional.

Based on our observations in Exploration init:dimensionintro, it makes sense for us to define dimension
of a vector space (or a subspace) as the minimum number of vectors required to span
the space (subspace). In other words, we would like to define dimension as the
number of elements in a basis.

We have to proceed carefully because we don’t want the dimension to depend on our
choice of a basis. So, before we state our definition, we need to make sure that
every basis for a given vector space (or subspace) has the same number of
elements.

Suppose and be two bases of (or a subspace of ). Then .

Proof

Suppose . Without loss of generality, assume that . Because spans ,
every of can be written as a linear combination of elements of :

Consider the vector equation

By substitution, we have:

Because ’s are linearly independent, we must have
For all . This gives us a system of equations and unknowns. We can write the
system as a matrix equation.

Recall our assumption that . By Theorem th:rankandsolutions of SYS-0030 we know that the system has
infinitely many solutions. This shows that equation (eq:sizeofbases) has a nontrivial solution. But
this shows that is linearly dependent and contradicts our assumption that is a basis
of . We conclude that .

Let be a subspace of . The dimension of is the number, , of elements in any basis of
. We write

We know that vectors form a basis of . Therefore .

The following section will guarantee that dimension is defined for every subspace of
.

Every Subspace of has a Basis

If a linearly independent subset of contains vectors, then .

Let be a linearly independent subset of . If is not in , then is linearly
independent.

Proof

Consider the equation

We need to show that . Suppose , then . But this contradicts the assumption that is
not in the span of . So, . But because are linearly independent.

This means that (eq:expandinglinindset) has only the trivial solution, and is linearly independent.

Let be a subspace of . Any linearly independent subset of can be expanded to a
basis of .

Proof

Suppose that is a linearly independent subset of . If then is already
a basis of . If , choose in such that is not in . The set is linearly independent
by Lemma lemma:expandinglinindset.

If we are done; otherwise choose such that is not in . Then is linearly
independent, and the process continues. We claim that a basis of will be reached
eventually. If no basis of is ever reached, the process creates arbitrarily large
independent sets in . But this is impossible by Lemma lemma:atmostnlinindinrn.