A First Course in Linear Algebra, 2017a

486 Vector Spaces
Let
B =
{[ 1 0
0 0
] [
,
0 − 1 2
− 1 2
1
Then span(B) =U, and it is routine to verify that B is an independent subset of M 22 . Therefore B is a
basis of U, and dim(U)=2.
♠
The following theorem claims that a spanning set of a vector space V can be shrunk down to a basis of
V . Similarly, a linearly independent set within V can be enlarged to create a basis of V .
Theorem 9.43: Basis of V
If V = span{⃗u 1 ,···,⃗u n } is a vector space, then some subset of {⃗u 1 ,···,⃗u n } is a basis for V . Also,
if {⃗u 1 ,···,⃗u k }⊆V is linearly independent and the vector space is finite dimensional, then the set
{⃗u 1 ,···,⃗u k }, can be enlarged to obtain a basis of V .
]}
.
Proof. Let
S = {E ⊆{⃗u 1 ,···,⃗u n } such that span{E} = V}.
For E ∈ S,let|E| denote the number of elements of E. Let
Thus there exist vectors
such that
m = min{|E| such that E ∈ S}.
{⃗v 1 ,···,⃗v m }⊆{⃗u 1 ,···,⃗u n }
span{⃗v 1 ,···,⃗v m } = V
and m is as small as possible for this to happen. If this set is linearly independent, it follows it is a basis
for V and the theorem is proved. On the other hand, if the set is not linearly independent, then there exist
scalars, c 1 ,···,c m such that
⃗0 =
m
∑
i=1
and not all the c i are equal to zero. Suppose c k ≠ 0. Then solve for the vector ⃗v k in terms of the other
vectors. Consequently,
V = span{⃗v 1 ,···,⃗v k−1 ,⃗v k+1 ,···,⃗v m }
contradicting the definition of m. This proves the first part of the theorem.
To obtain the second part, begin with {⃗u 1 ,···,⃗u k } and suppose a basis for V is
c i ⃗v i
{⃗v 1 ,···,⃗v n }
If
then k = n. If not, there exists a vector
span{⃗u 1 ,···,⃗u k } = V ,
⃗u k+1 /∈ span{⃗u 1 ,···,⃗u k }