190 CHAPTER 7. VECTOR SPACES AND FIELDS

7.2 Subspaces and Bases

7.2.1 Basic Definitions

Definition 7.2.1 If {v1, · · · ,vn} ⊆ V, a vector space, then

span (v1, · · · ,vn) ≡

{n∑

i=1

αivi : αi ∈ F

}.

A subset, W ⊆ V is said to be a subspace if it is also a vector space with the same field ofscalars. Thus W ⊆ V for W nonempty is a subspace if ax+ by ∈W whenever a, b ∈ F andx, y ∈W. The span of a set of vectors as just described is an example of a subspace.

Example 7.2.2 Consider the real valued functions defined on an interval [a, b]. A subspaceis the set of continuous real valued functions defined on the interval. Another subspace isthe set of polynomials of degree no more than 4.

Definition 7.2.3 If {v1, · · · ,vn} ⊆ V, the set of vectors is linearly independent if

n∑i=1

αivi = 0

impliesα1 = · · · = αn = 0

and {v1, · · · ,vn} is called a basis for V if

span (v1, · · · ,vn) = V

and {v1, · · · ,vn} is linearly independent. The set of vectors is linearly dependent if it is notlinearly independent.

7.2.2 A Fundamental Theorem

The next theorem is called the exchange theorem. It is very important that you understandthis theorem. It is so important that I have given several proofs of it. Some amount to thesame thing, just worded differently.

Theorem 7.2.4 Let {x1, · · · ,xr} be a linearly independent set of vectors such that each xi

is in the span{y1, · · · ,ys} . Then r ≤ s.

Proof 1: Define span{y1, · · · ,ys} ≡ V, it follows there exist scalars c1, · · · , cs suchthat

x1 =

s∑i=1

ciyi. (7.5)

Not all of these scalars can equal zero because if this were the case, it would follow thatx1 = 0 and so {x1, · · · ,xr} would not be linearly independent. Indeed, if x1 = 0, 1x1 +∑r

i=2 0xi = x1 = 0 and so there would exist a nontrivial linear combination of the vectors{x1, · · · ,xr} which equals zero.

Say ck ̸= 0. Then solve 7.5 for yk and obtain

yk ∈ span

x1,

s-1 vectors here︷ ︸︸ ︷y1, · · · ,yk−1,yk+1, · · · ,ys

 .