Spanning Sets
Definition of linear combination:
Let
v1 , v2 ,, vk
be vectors in a real vector space V. A vector
a linear combination of
v1 , v2 ,, vk
v in V is called
if
v  c1v1  c2 v2    ck vk ,
where
c1 , c2 ,, ck .
Example:
Let
1
0 
0 
1 
e1  0, e2  1, e3  0, v  2 .
0
0
1
3
Since
1 
1
0
0






v  2  1  0  2  1  3  0  1e1  2e2  3e3 ,
3
0
0
1
v
is a linear combination of
e1 , e2 , e3 .
Example:
Let
0 2 
  1 3
  2 0
0 8
v1  
,
v

,
v

,
v

 2  1 2 3  1 3
2 1
1 0






be vectors in the vector space consisting of all 2  2 matrices. Then,
0 8
0 2 
  1 3
  2 0
v

1


2


(

1
)


1 0
 1 2
 1 3  v1  2v2  v3 .
2 1






1
That is,
v
is a linear combination of
v1 , v2 , v3 .
Example:
For linear system
1 0  1  x1  1
Ax  2 1 0   x2   1  b .
3 2 1   x3  1
2
 x   3 is a solution for the above linear system. Thus,
 1 
 2  1 0  1  2 
1 
0 
 1
A 3  2 1 0   3  2  2  3  1   1   0 
 1  3 2 1   1 
3
2
 1 
1
 2col1 ( A)  3col2 ( A)  col1 ( A)  1  b
1
That is,
b
is a linear combination of the column vectors of A,
col1 ( A), col2 ( A), col3 ( A) .
Note:
For a linear system
Amn xn1  bm1 ,
solution or solutions 
b
the linear system has
is a linear combination of the column
vectors of A,
col1 ( A), col2 ( A),, coln ( A) .
2
 c1 
c 
c   2
   is a solution of Amn xn1  bm1 ,
For example, if
 
c n 
then
c1col1 ( A)  c2 col 2 ( A)    cn col n ( A)  b .
On the other hand, the linear system has no solution 
b
linear combination of the column vectors of A
Example:
4
Is the vector v  5  a linear combination of the vectors
 
5 
1 
 1
 3
v1  2, v2   1 , v3  3
.
3
 4 
2
[solution:]
We need to find the constants c1 , c2 , c3 such that
 4
1 
 1
 3
v  5  c1 2  c 2  1   c3 3  c1v1  c2 v2  c3 v3 .
5
3
 4 
2
 we need to solve for the linear system
 c1  1  1 3  c1  4
Ac2   2 1 3 c 2   5 .
c3  3 4 2 c3  5
The solutions are
3
is not a
c1  2t  3, c2  t  1, c3  t , t  R .
Thus,
v
v   2t  3v1  t  1v2  tv3 , t  R
is a linear combination of
v1 , v2 , v3
with infinite number of expressions.
Example:
3
Is the vector v   4 a linear combination of the vectors
 
  6
1
  1
1
v1  2, v2    1, v3  4
.
3
 2
5
[solution:]
We need to find the constants c1 , c2 , c3 such that
3
1
  1
1






v   4  c1 2  c2   1  c3 4  c1v1  c2 v2  c3v3 .
 6
3
 2
5
 we need to solve for the linear system
 c1  1  1 1  c1   3 
Ac2   2  1 4 c2    4 .
c3  3  2 5 c3   6
 The linear system has no solution.

v
is not a linear combination of
v1 , v2 , v3
4
Note:
v1 , v2 ,, vm
Let
column vectors
Ax  v
and
v
be vectors in R n and let A be the matrix with
col j ( A)  v j , j  1,2,, m . Thus,
has solution or solutions 
v is a linear combination of
v is
not a linear combination of
v1 , v2 ,, vm .
Ax  v
has no solution 
v1 , v2 ,, vm .
Definition of spanning set:
S  v1 , v2 ,, vk 
be a set of vectors in a real vector space V. Then, the
span of S, denoted by
span(S ) , is the set consisting of all the vectors that
Let
are linear combinations of
v1 , v2 ,, vk . That is,
span(S )  c1v1  c2v2    ck vk | c1 , c2 ,, ck  R.
If
span ( S )  V
, it is said that V is spanned by S or S spans V.
Example:
Let
1
0
0 
e1  0, e2  1, e3  0, and S  e1 , e2 , e3  .
0
0
1
Then,
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

 c1 


span ( S )  c1e1  c2e2  c3e3  c2  | c1 , c2 , c3  R   R 3
,




c3 


Example:
Let
1 
1 
1




v1  2, v2  0, v3  1, and S  v1 , v2 , v3 .
1
2
0
Does span( S )  R 3 ?
[solution:]
span( S )  R
3
a 
 For any vector v  b   R 3 , there exist real numbers
 
 c 
c1 , c2 , c3 such that
a 
1
1
1
v  b   c1 2  c2 0  c3 1  c1v1  c2 v2  c3 v3 .
 c 
1
2
0
 we need to solve for the linear system
1 1 1  c1  a 
 2 0 1  c    b 

 2    .
1 2 0 c3   c 
The solution is
c1 
 2a  2b  c
abc
4a  b  2c
, c2 
, c3 
.
3
3
3
Thus,
6
  2a  2b  c 
 a bc
 4a  b  2c 
v
v1  
v 2  
 v3 .
3
3
3






That is, every vector in R 3 can be a linear combination of v1 , v2 , v3
Important result:
Let
S  v1 , v2 ,, vk 
span(S )
be a set of vectors in a real vector space V. Then,
is a subspace of V.
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