Basis
Definition of basis:
The vectors
v1 , v2 ,, vk
in a vector space V are said to form a basis of V if
span(v1 , v2 ,, vk )  V
(a)
v1 , v2 ,, vk
span V (i.e.,
(b)
v1 , v2 ,, vk
are linearly independent.
).
Example:
1
0
0 




e1  0, e2  1, e3  0, and S  e1 , e2 , e3  . Are e1 , e2 and e3 a
0
0
1
basis in R 3 ?
[solution:]
e1 , e2 and e3 form a basis in R 3 since
(a)
span( S )  span (e1 , e2 , e3 )  R 3
(see the example in the previous
section).
(b) e1 , e2 and e3 are linearly independent (also see the example in the previous
section).
Example:
1
0 
 3
v1   , v2   , v3    . Are v1 ,v2 and v3 a basis in R 2 ?
0 
1
 4
[solution:]
v1 ,v2 and v3 are not a basis of R 2 since v1 ,v2 and v3 are linearly dependent,
1
3v1  4v2  v3  0 .
2
Note that span (v1 , v2 , v3 )  R .
Example:
1 
  2
8




v1  2, v2   1 , v3   6 . . Are v1 ,v2 and v3 a basis in R 3 ?
3
 1 
10
[solution:]
v1 ,v2 and v3 are not a basis in R 3 since v1 ,v2 and v3 are linearly independent,
8
1 
  2




v3   6   42  2 1   4v3  2v 2 .
10
3
 1 
Example:
Let
1 
1 
1




v1  2, v2  0, v3  1, and S  v1 , v2 , v3 .
1
2
0
Are S a basis in R 3 ?
[solution:]
(a)
span( S )  R
3
a 
 For any vector v  b   R 3 , there exist real numbers
 
 c 
c1 , c2 , c3 such that
2
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,
  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 and
span( S )  R 3 .
(b)
Since
c1  c2  c3  0
c1v1  c2 v2  c3 v3   2c1  c3   0  c1  c2  c3  0
,
 c1  2c2  0
v1 , v2 , v3
are linearly independent.
By (a) and (b),
v1 , v2 , v3
are a basis of R 3 .
Important result:
3
If
S  v1 , v2 ,, vk  is a basis for a vector space V, then every vector in V
can be written in an unique way as a linear combination of the vectors in
S.
Example:
1
0
0 
e1  0, e2  1, e3  0, and S  e1 , e2 , e3  . S is a basis of R 3 .
0
0
1
a 
Then, for any vector v  b  ,
 
 c 
a 
1
0
0
v  b   a 0  b 1  c 0  ae1  be2  ce3
 c 
0
0
1
is uniquely determined.
Important result:
Let
S  v1 , v2 ,, vk 
let
W  spanv1 , v2 ,, vk . Then, some subset of S is a basis of
be a set of nonzero vectors in a vector space V and
W.
How to find a basis (subset of S) of W:
There are two methods:
Method 1:
The procedure based on the proof of the above important result.
Method 2:
Step 1: Form equation
4
c1v1  c2 v2    ck vk  0 .
Step 2: Construct the augmented matrix associated with the equation in
step 1 and transform this augmented matrix to the reduced row
echelon form.
Step 3: The vectors corresponding to the columns containing the leading
1’s form a basis. For example, if
k 6
and the reduced row
echelon matrix is
1
0

0

0



0





0
1



0
0
0
0
1
0

0

0





0
0
0
0
0
0
0

0

0 ,


0

then the 1’st, the 3’nd, and the 4’th columns contain a leading 1 and thus
v1 , v3 , v 4
are a basis of
W  spanv1 , v2 ,, v6  .
Example:
Let
1 0 2 0 1 


S  e1 , e2 , a1 , e3 , a2   0, 1, 3, 0, 3 
0 0 0 1 2 
          
and
spanS   R3 . Please find subsets of S which form a basis of
[solution:]
Method1:
5
R3 .
We first check if e1 and e2 are linearly independent. Since they are linearly
independent, we continue to check if e1 , e2 and a1 are linearly independent.
Since
2e1  3e2  a1  0 ,
we delete a1 from S and form a new set S1 , S1  e1 , e2 , e3 , a2  . Then, we continue
to check if e1 , e2 and e3 are linearly independent. They are linearly independent.
Thus, we finally check if e1 , e2 e3 and a 2 are linearly independent. Since
e1  3e2  2e3  a 2  0 ,
we delete a1 from S1 and form a new set S 2 , S 2  e1 , e2 , e3 . Therefore,
S 2  e1 , e2 , e3 
is the subset of S which form a basis of form a basis of R 3 .
Method 2:
Step 1:
The equation is
1
0 
 2
0
1 
c1 0  c2 1  c3 3  c4 0  c5 2  0 .
0
0
0
1
3
Step 2:
The augmented matrix and its reduced row echelon matrix is
1 0 2 0 1 0
0 1 3 0 2 0

.
0 0 0 1 3 0
The 1’st, the 2’nd and 4’th columns contain the leading 1’s. Thus,
e1 , e2 , e3 
forms a basis.
6
Important result:
Let
S  v1 , v2 ,, vn 
be a basis for a vector space V and let
T  w1, w2 ,, wr 
r  n
is a linear independent set of vectors in V. Then,
.
Corollary:
Let
S  v1 , v2 ,, vn 
vector space V. Then,
and
T  w1 , w2 ,, wm 
be two bases for a
nm.
Note:
For a vector space V, there are infinite bases. But the number of
vectors in two different bases are the same.
Example:
For the vector space R 3 ,
1 
1 
1
v1  2, v2  0, v3  1, S  v1 , v2 , v3  is a basis for R 3 (see the
1
2
0
previous example). Also,
1
0 
0 




e1  0, e2  1, e3  0, T  e1 , e2 , e3  is basis for R 3 .
0
0
1
 There are 3 vectors in both S and T.
7