A First Course in Linear Algebra, 2017a

4.10. Spanning, Linear Independence and Basis in R n 201
We are now ready to show that any two bases are of the same size.
Theorem 4.83: Bases of R n are of the Same Size
Let V be a subspace of R n with two bases B 1 and B 2 . Suppose B 1 contains s vectors and B 2 contains
r vectors. Then s = r.
Proof. This follows right away from Theorem 9.35. Indeed observe that B 1 = {⃗u 1 ,···,⃗u s } is a spanning
set for V while B 2 = {⃗v 1 ,···,⃗v r } is linearly independent, so s ≥ r. Similarly B 2 = {⃗v 1 ,···,⃗v r } is a spanning
set for V while B 1 = {⃗u 1 ,···,⃗u s } is linearly independent, so r ≥ s.
♠
The following definition can now be stated.
Definition 4.84: Dimension of a Subspace
Let V be a subspace of R n . Then the dimension of V, written dim(V) is defined to be the number
of vectors in a basis.
Thenextresultfollows.
Corollary 4.85: Dimension of R n
The dimension of R n is n.
Proof. You only need to exhibit a basis for R n which has n vectors. Such a basis is the standard basis
{⃗e 1 ,···,⃗e n }.
♠
Consider the following example.
Example 4.86: Basis of Subspace
Let
⎧⎡
⎪⎨
V = ⎢
⎣
⎪⎩
a
b
c
d
⎤
⎥
⎦ ∈ R4 : a − b = d − c
Show that V is a subspace of R 4 , find a basis of V ,andfinddim(V).
⎫
⎪⎬
.
⎪⎭
Solution. The condition a − b = d − c is equivalent to the condition a = b − c + d, sowemaywrite
⎧⎡
⎪⎨
V = ⎢
⎣
⎪⎩
b − c + d
b
c
d
⎤
⎧ ⎫⎪ ⎬ ⎪⎨
⎥
⎦ : b,c,d ∈ R = b
⎪ ⎭ ⎪⎩
⎡
⎢
⎣
1
1
0
0
⎤
⎡
⎥
⎦ + c ⎢
⎣
−1
0
1
0
⎤ ⎡
⎥
⎦ + d ⎢
⎣
1
0
0
1
⎤
⎥
⎦ : b,c,d ∈ R ⎫⎪ ⎬
⎪ ⎭
This shows that V is a subspace of R 4 ,sinceV = span{⃗u 1 ,⃗u 2 ,⃗u 3 } where