Lecture Notes Friday, October 7, 2011
Subspaces and Spans, Basis and Dimension
I. Subspaces
Recall
A vector space is a set …
 with two operations (vector addition and scalar multiplication) satisfying 4
properties each
 that is closed under linear combinations (unlimited application of the two
operations)
A set that is a vector space may contain a subset of vectors that is also in itself a vector
space. This subset is called a subspace.
Definition
A subset W of a vector space V is a subspace of V if W is itself a vector space.
Theorem: A nonempty subset W of a vector space V is a subspace of V if it closed under
linear combinations.
Note:
This is all that is required because the subspace W of V inherits the other properties of V!
Examples of Subspaces of R3
Vector Space R3: the set of vectors (a,b,c) where a,b,c ℝ equipped with the usual
operations of vector addition and scalar multiplication is a vector space.
Two examples of subspaces of this vector space:

The subset of vectors (a,b,0) where a,b ℝ (it is equivalent to R 2)

The subset of vectors (a,0,0) where aℝ (it is equivalent to R)
II. Spans
Generating Subspaces
A way of generating subspaces of V to consider any set of vectors in V and then consider
the closure of those vectors under linear combinations.
Definition
The span of a set of vectors v1 , v 2 ,..., v n is the set of all linear combinations of
v1 , v 2 ,..., v n .
(The vectors v1 , v 2 ,..., v n are called the spanning set that generate the set, and we say the
resulting set is spanned by the vectors.)
Let V be a vector space.
Let R  v1 , v 2 ,..., v n be any subset of V.
Let Span(R) be the set of all linear combinations
of the vectors in R.
Span(R) is a subspace of V because it is closed
under linear combinations.
Common Subspaces
The span of any set of vectors generates a subspace. Here are some common examples of
subspaces generated by the span of a set of vectors:



Row Space = the set spanned by the row vectors of a matrix
Column Space = the set spanned by the column vectors of a matrix
Lines, planes (and hyperplanes) are generated by the spans of 1, 2 (and n) linearly
independent spatial vectors.
Example: Span of a single vector is a line.
The span of u=[1,3] is the set of all linear combinations of u. Span(u)= au a  .
Example: Span of two linearly independent vectors is a plane.
The span of u=[1,3], v=[0,1] is the set of all linear combinations of u,v.
Span(u,v)= au  bv a, b    a,3a  b a, b  .
III. Relationship Between Linear Dependence of Vectors and
their Span
Consider a set of vectors R  v1 , v 2 ,..., v n .
Each linearly dependent vector does not contribute to the span of R.
A linearly independent vector increases the size of the span of R.
(If a vector is linearly dependent on the other, it is a linear combination of the other
vectors and will have already been included in the span. If a vector is linearly
independent of the others, it is not a linear combination of the other vectors and all new
linear combinations must now be included in the span.)
1 0 0 1 
Example A: What is the span of S= 
, 
 ?
0 0 0 0 
 1 0 0 1
  a b 

 b
such that a, b  R  
such that a, b  R



 0 0 0 0
   0 0

Span of S = a 
1 0 0 1 0 2 
Example B: What is the span of S= 
, 0 0, 0 0  ?
0
0

 
 


Span of S = a 1 0  b0 1  c 0 2 such that a, b, c  R  a b  2c such that a, b, c  R
 0 0
0 0
0 0

 0
0 

(still the same set of 22 matrices with zeros in the second row)
1 0 0 1 0 0 
Example C: What is the span of S= 
, 
, 
 ?
0 0 0 0 0 1 
 1 0 0 1 0 0
 a b 

 b
 c
such that a, b, c  R  
such that a, b, c  R




 0 0 0 0 0 1
  0 c 

Span of S = a 
(this is a larger set of 22 matrices)
How many vectors to span the entire vector space of 22 matrices?
Since linearly dependent vectors don’t contribute to the span, we can consider the
minimal spanning set of a vector space – what is the smallest number of vectors that are
needed to span a vector space?
Answer: we need exactly one linearly independent vector for each degree of freedom.
IV. Basis and Dimension of a Vector Space
A set of linearly independent vectors that span a vector space is called a basis for that
vector space (subspace). The number of vectors that are needed to span a space is called
the dimension of that space.
The basis of a vector space is not unique but always has the same number of vectors. It
may be thought of as a minimal spanning set for a vector space, because it contains the
fewest number of vectors exactly needed to span that space.
Example:
ℝ3 is spanned by (1,0,0),(0,1,0),(0,0,1) but also by any other set of 3 linearly independent
vectors.
(for example ℝ3 is spanned by (1,1,0),(0,1,1),(1,0,1).)
For homework:
A common problem in linear algebra is to find or complete a basis for a vector space.
A set of vectors is a basis for a vector space if the following two conditions are met:
1. The vectors are linearly independent.
2. The vectors exactly span the set. (They do not span a larger set and do not miss any
vectors in the set.)
Homework Problems
Problems 13 and 14: Verifying if a set of vectors are a spanning set.
(You’ve already done 4.14, let’s look at 4.13.)
Homework Problem 4.13
1
1
1




Show that the vectors u1  1 , u 2  2 , u 3  5 do span R3.

 
 
1
3
8
Certainly the span will be a subset of R3… we need to show they span ALL of R3. That
a 
is, any vector v  b  can be expressed as a linear combination of u1, u2, u3. Thus we
 
 c 
require the following equation always has a solution:
1
1 
1   a 




k1u1  k 2 u 2  k3u3  v yields the system k1 1  k 2 2  k 3 2  b  .
1
3
8  c 
1 1 1 a 


This corresponds to the augmented matrix 1 2 5 b  .
1 3 8 c 
Step 2: Solve the linear system by row reducing.

1 1 1 a  1 1 1 a  1 1 1
a
 


 
ba 
1 2 5 b  ~ 0 1 4 b  a  ~ 0 1 4
1 3 8 c  0 1 3 c  b  0 0 1 2b  a  c 
R 2  R 2  R1
R3  R3  R 2
R 3   R 3  R 2 
… As soon as we identify there are 3 pivots, we know there will always a solution. (The
solution will also be unique since there are no free parameters).
Note that it would have been enough just to determine that there are 3 pivots, or
equivalently that the three vectors are linearly independent.
Homework Problem 4.14
a 
Find conditions on a,b and c so that v  b  belongs to the span of u1, u2, u3.
 
 c 
Step 1: Reorganize this question as a problem for solving a linear system.
What is the definition of span?
The span of a set of vectors u1, u2, …, um in a vector spave V is the set of all linear
combinations of u1, u2, …, um.
Applying the definition:
a 
Find conditions on a,b and c so that v  b  is a linear combination of u1, u2, u3.
 c 
That is, find conditions on a,b and c so that
k1u1  k 2 u 2  k3u3  v is consistent.
1
 1
 3  a 
k1u1  k 2 u 2  k3u3  v yields the system k1 2  k 2  1   k3  0   b  .
0
 2 
 4  c 
1  1 3 a 


0 b .
This corresponds to the augmented matrix 2 1
0 2  4 c 
Step 2: Solve the linear system by row reducing.
 1 0 1  a  b  c 
 1  1 3
1  1 3 a  1  1 3
a
a
 
 


 
0 b  ~ 0 1  2 b  2 a  c  ~ 0 1  2  2 a  b  c  ~ 0 1  2  2 a  b  c 
2 1
 0 0
0 2  4 c  0 2  4
0 4a  2b  3c  0 0 0 4a  2b  3c 
c
R3  R3  2 R 2
R 2  R 2  2 R1
R1  R1  R 2
… For the system to be consistent, we must have 4a  2b  3c  0.
This condition means that the vectors do not span all of R3. What is the subspace spanned
by these vectors?
The subspace spanned by these vectors:
= row space of
1  1 3 
2 1
0 

0 2  4
 
= row space of



0
1 0  4    1 



3

= a 0

b
1




 for
2
0 1

3
2 

   4 
0   
3  3
0 0
 
Two free parameters  a 2-dimensional subspace.


a, b  

