4.5 The Dimension of a Vector Space
We saw in section 4.4 that a vector space V
with a basis β containing n vectors is
isomorphic to Rn. The number n is called the
dimension of V.
Theorem 4.9: If a vector space V has a basis
  b1 , b 2 ,, b n  then any set in V containing
more than n vectors must be linearly
dependent.
Proof: Suppose u1 , u2 ,..., u p  is a set of
vectors in V with p > n.
Then the coordinate vectors
u1  , u2  ,..., u p  form a set in Rn. The set


has more vectors than entries in each vector
(since p > n), so by Theorem 1.8, it is linearly
dependent. Therefore the set u1 , u 2 ,..., u p  is
a set of vectors in V is linearly dependent
since the coordinate mapping x  x  is an
isomorphism, that is, it preserves every vector
space calculation.
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Theorem 4.10: If a vector space V has a basis
of n vectors, then every basis of V has n
vectors.
Proof: Let β1 be a basis for V consisting of
exactly n vectors, and β2 be any other basis
for V.
We must show that β2 has the same number
of vectors as β1.
Strategy: show that β1 can’t have more
vectors than β2, and that β2 can’t have more
vectors than β 1.
By Th. 4.9, if β 1 has more vectors than β 2,
then β1 is a linearly dependent set. But by the
definition of basis, β 1 is a linearly
independent set, so β 1 can’t have more
vectors than β 2.
Similarly, By Th. 4.9, if β 2 has more vectors
than β 1, then β 2 is a linearly dependent set.
But by the definition of basis, β 2 is a linearly
independent set, so β 2 can’t have more
vectors than β 1.
Since neither can have more vectors than the
other, they must both have the same number
of vectors, and β 2 has n vectors also. QED
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Definition: If V is spanned by a finite set, then
V is said to be finite dimensional, the
dimension of V, written dim V, is the number
of vectors in a basis of V.
The dimension of the zero vector space {0} is
defined to be zero.
If V is not spanned by a finite set, then V is
said to be infinite dimensional.
Example: The standard basis for P3 is
{1, t, t2, t3}, so dim P3 = 4.
►In general, dim Pn = n + 1
Example: The standard basis of Rn is
e1, e2 ,..., en  where e1 , e 2 ,..., e n are the
columns of In. So, for example, dim R3 = 3
Example: Find dim H if
 a  b  2c 




2a  2b  4c  d 

H 
: a, b, c, d  R 

 b  c  d



 3a  3c  d 

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Note: dim H = the number of vectors in the
basis of H. Find the basis of H. Let h ÎH.
 a   b  2c   0 
 2 a   2b   4 c   d 
h    
 0   b   c  d 
       
 3a   0   3c  d 
1  1   2 
0 
 2  2  4
1
a   b   c   d  
 0  1  1 
1
     
 
 3 0   3
1
Now,
 1  1   2   0  
        
2 2 4 1 
H  Span  , , , 
0 1 1 1 
3 0 3 1 
A basis is a linearly independent set, and
h3 = h1 + h2, so by the Spanning set
Theorem, we can throw out h3 to get a basis:
Since  3 vectors in the basis of H, dim H = 3.
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Dimensions of Subspaces of R3:
Zero-Dimensional Subspace: contains only 0,
the origin
One-Dimensional Subspaces: span{v} where
v ≠ 0 in R3, a line through the origin.
Two-Dimensional Subspaces: span{u, v}
where u and v are in R3, and are linearly
independent, a plane through the origin.
Three-Dimensional Subspaces: span{u,v,w}
where u, v, and w are linearly independent
vectors in R3, this subspace is R3 because the
columns of A = [u v w] (a 3 x 3 matrix) span
R3 by the IMT.
Theorem 4.11: Let H be a subspace of a
finite-dimensional vector space V. Any
linearly independent set in H can be
expanded, if necessary, to a basis of H. Also,
H is finite-dimensional and dim H ≤ dim V
Example: Let
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 1  1  
    
H  Span 0, 1 
0 0  , then H is a subspace of
    
R3 and dim H ≤ dim R3.
We could expand the spanning set
 1  1   0  
1 1 
      
    
  0  , 1  ,  0  
0, 1 
0 0  to 0 0 1 
      
    
to form a basis for R3.
Theorem 4.12: (The Basis Theorem) Let V
be a p–dimensional vector space p ≥ 1. Any
linearly independent set of exactly p vectors
in V is automatically a basis for V.
And any set of exactly p vectors that spans V
is automatically a basis for V.
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Example: show that {t, 1 – t, 1 + t – t2} is a
basis for P2.
Let v1 = t, v2 = 1 – t, v3 = 1 + t – t2.
v1 and v2 are linearly independent since they
are not multiples of each other. And v3 is not
a linear combination of v1 and v2 since no
combination of these could produce t2, so
{t, 1 – t, 1 + t – t2} is a linearly independent
set of three vectors. Since dim P2 = 3,
{t, 1 – t, 1 + t – t2} is a basis of P2 by the The
Basis Theorem.
Dimension of Col A and Nul A
Example: Let
1 2 3 4 5 
A

2
4
7
8
6


Find dim Col A and Nul A.
Col A: To find linearly independent columns of
A, row reduce A.
1 2 3 4 5 1 2 0 4 17 
 2 4 7 8 6  ~ 0 0 1 0  4 

 

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1 3 
Col A  Span  ,   
2 7 
So, dim Col A = 2
Nul A: Since Nul A is the set of all solutions to
the homogeneous vector equation Ax = 0,
solve Ax = 0.
The corresponding augmented matrix is:
1 2 3 4 5 0 1 2 0 4 17
2 4 7 8 6 0 ~ 0 0 1 0  4

 
0

0
The solution is
x1  2 x2  4 x4  17 x5
x2 is free
x 3  4 x5
x4 is free
x5 is free
or
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  2
  4
 17
1
0
 0 
 
 


x  x 2  0   x 4  0   x5  4 
 
 


0
1
0
 
 


 0 
 0 
 1 
So,
   2  4  17 
    

1
0
0
     
 
  0 ,  0 ,  4  
 0   1   0  
  is a basis for Nul A, and
    
  0   0   1  
dim Nul A = 3
Note: dim Col A is the number of pivot
columns in A.
dim Nul A is the number of free variables in
the system associated to Ax = 0.
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