11.3 Subspaces Spans And Bases
As shown earlier, Fn is an example of a vector space with field of scalars F. Here is a short review of the
major exchange theorem. Here and elsewhere, when it is desired to emphasize that certain things are
vectors, bold face will be used. However, sometimes the context makes this sufficiently clear and bold face
is not used.
Theorem 11.3.1 If
,ur} are linearly independent, then r ≤ s.
Proof: Suppose r > s. Let Ep denote a finite list of vectors of
number of vectors in the list. Let
denote the first p
. In case
denote the empty set. For 0 ≤ p ≤ s,
have the property
is as small as possible for this to happen. I claim
≤ s − p
Here is why. For p = 0, it is obvious. Suppose true for some p < s. Then
and so there are constants, c1,
where m ≤ s − p
Then not all the di can equal zero because this would violate the linear independence of the
Therefore, you can solve for one of the zk
as a linear combination of
and the other
you can change Fp
and include one fewer vector in Ep.
≤ m −
1 ≤ s − p −
proves the claim.
Therefore, Es is empty and span
However, this gives a contradiction because it would
which violates the linear independence of these vectors. ■
Also recall the following.
Definition 11.3.2 A finite set of vectors,
is a basis for a vector space V if
is linearly independent. Thus if v ∈ V there exist unique scalars, v1,
,vr such that
i=1rvixi. These scalars are called the components of v with respect to the basis
Corollary 11.3.3 Let
be two bases
of Fn. Then r
Lemma 11.3.4 Let
be a set of vectors. Then V ≡ span
is a subspace.
Definition 11.3.5 Let V be a vector space. Then dim
read as the dimension of V is the number
of vectors in a basis.
Of course you should wonder right now whether an arbitrary subspace of a finite dimensional vector
space even has a basis. In fact it does and this is in the next theorem. First, here is an interesting lemma
which was also presented earlier.
Lemma 11.3.6 Suppose v
is linearly independent. Then
is also linearly independent.
Recall that this implies the following theorems also presented earlier.
Theorem 11.3.7 Let V be a nonzero subspace of Y a finite dimensional vector space having
dimension n. Then V has a basis.
In words the following corollary states that any linearly independent set of vectors can be enlarged to
form a basis.
Corollary 11.3.8 Let V be a subspace of Y, a finite dimensional vector space of dimension n and let
be a linearly independent set of vectors in V . Then either it is a basis for V or there exist
,vs such that
is a basis for V.
Theorem 11.3.9 Let V be a subspace of Y, a finite dimensional vector space of dimension n
and suppose span
V where the ui are nonzero vectors. Then there exist vectors,
is a basis for V .