A First Course in Linear Algebra, 2017a

200 R n
Definition 4.80: Basis of a Subspace
Let V be a subspace of R n .Then{⃗u 1 ,···,⃗u k } is a basis for V if the following two conditions hold.
1. span{⃗u 1 ,···,⃗u k } = V
2. {⃗u 1 ,···,⃗u k } is linearly independent
Note the plural of basis is bases.
The following is a simple but very useful example of a basis, called the standard basis.
Definition 4.81: Standard Basis of R n
Let⃗e i be the vector in R n which has a 1 in the i th entry and zeros elsewhere, that is the i th column of
the identity matrix. Then the collection {⃗e 1 ,⃗e 2 ,···,⃗e n } is a basis for R n and is called the standard
basis of R n .
The main theorem about bases is not only they exist, but that they must be of the same size. To show
this, we will need the the following fundamental result, called the Exchange Theorem.
Theorem 4.82: Exchange Theorem
Suppose {⃗u 1 ,···,⃗u r } is a linearly independent set of vectors in R n , and each ⃗u k is contained in
span{⃗v 1 ,···,⃗v s } Then s ≥ r.
In words, spanning sets have at least as many vectors as linearly independent sets.
Proof. Since each ⃗u j is in span{⃗v 1 ,···,⃗v s }, there exist scalars a ij such that
⃗u j =
s
∑
i=1
Suppose for a contradiction that s < r. Then the matrix A = [ a ij
]
has fewer rows, s than columns, r. Then
the system AX = 0 has a non trivial solution ⃗ d, that is there is a ⃗ d ≠⃗0 such that A ⃗ d =⃗0. In other words,
Therefore,
r
∑ d j ⃗u j =
j=1
a ij ⃗v i
r
∑ a ij d j = 0, i = 1,2,···,s
j=1
=
r s
∑ d j ∑
j=1 i=1
s
∑
i=1
a ij ⃗v i
(
r
∑ a ij d j
)⃗v i =
j=1
s
∑
i=1
0⃗v i = 0
which contradicts the assumption that {⃗u 1 ,···,⃗u r } is linearly independent, because not all the d j are zero.
Thus this contradiction indicates that s ≥ r.
♠