Download Supplement: Basis, Dimension and Rank

Supplement: Basis, Dimension and Rank
1. Basis
Definition of basis:
The vectors
v1 , v2 ,, vk
in a vector space V are said to form a basis of V
if
(a)
v1 , v2 ,, vk
span V (i.e., span(v1 , v2 ,, vk )  V ).
(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,
3v1  4v2  v3  0 .
2
Note that span (v1 , v2 , v3 )  R .
1
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   4v1  2v2 .
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
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
2
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:
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 
3
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  be a set of nonzero vectors in a vector space V and
let
W  spanv1 , v2 ,, vk . Then, some subset of S is a basis of
W.
Important result:
Let
S  v1 , v2 ,, vn 
be a basis for a vector space V and let
T  w1, w2 ,, wr 
is a linear independent set of vectors in V. Then,
r  n.
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,
4
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.
2. Dimension
Definition of dimension:
The dimension of a vector space V is the number of vectors in a basis for
V.
Example:
1
0
0
e1  0, e2  1, e3  0, T  e1 , e2 , e3  is basis for R 3 .
0
0
1
 The dimension of
R 3 is 3.
Important result:
Let V be an n-dimensional vector space, and let
S  v1 , v2 ,, vn 
be a
set of n vectors in V.
(a) If S is linearly independent, then S is a basis for V.
(b) If S spans V, then S is a basis for V.
Example:
1
1
1
 
 
 
Is v1  2, v2  0, v3  1, S  v1 , v2 , v3  a basis for R 3 ?
1
2
0
[solutions:]
Since R 3 is a 3-dimensional vector space, not like in the previous example, we only
need to examine whether S is linearly independent or S spans R 3 . We don’t need to
examine S being both linearly independent and spans V.
Example:
1
 2
 4




 
Is v1   3 , v2  1, v3  2, S  v1 , v2 , v3  a basis for R 3 ?
 1
0
1
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[solutions:]
Since R 3 is a 3-dimensional vector space, we only need to examine whether S is
linearly independent or S spans R 3 . Because
c1  2c2  4c3  0
c1v1  c2 v2  c3v3  3c1  c2  2c3   0  c1  c2  c3  0
,
  c1  c3  0
v1 , v2 , v3
are linearly independent. Therefore,
v1 , v2 , v3
are a basis of R 3
3. Rank of a Matrix:
Recall: et
 a11
a
A   21
 

a m1
a12
a 22

am2




a1n 
a 2 n 
 .

a mn 
The i’th row of A is
rowi ( A)  ai1 ai 2  ain , i  1,2,, m, ,
and the j’th column of A is
 a1 j 
a 
2j
col j ( A)   , j  1,2,, n.
  
 
amj 
Definition of row space and column space:
spanrow1 ( A), row2 ( A), , rowm ( A),
which is a vector space under standard matrix addition and scalar
multiplication, is referred to as the row space. Similarly,
spancol1 ( A), col2 ( A),, coln ( A),
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which is also a vector space under standard matrix addition and scalar
multiplication, is referred to as the column space.
Definition of row equivalence:
A matrix B is row equivalent to a matrix A if B result from A via
elementary row operations.
Example:
Let
1 2 3
 4 5 6
 2 4 6
1 2 3
A  4 5 6, B1  1 2 3, B2  4 5 6, B3  3 3 3
7 8 9
7 8 9
7 8 9
7 8 9
Since
(1) 1 2 3
 4 5 6
1) ( 2 )
A  (2) 4 5 6 (

 B1  1 2 3 ,
7 8 9
(3) 7 8 9
(1) 1 2 3
2 4 6
(1) 2*(1)
A  (2) 4 5 6 

 B2  4 5 6
,
7 8 9
(3) 7 8 9
(1) 1 2 3
2 ) ( 2 ) (1)
A  (2) 4 5 6 (
 B3
(3) 7 8 9
B1 , B2 , B3
are all row equivalent to
1 2 3
 3 3 3 ,
7 8 9
A.
Important Result:
If A and B are two m n row equivalent matrices, then the row spaces
of A and B are equal.
Definition of row rank and column rank:
The dimension of the row space of A is called the row rank of A and the
7
dimension of the column space of A is called the column rank of A.
Example (continue):
Let
 1 2
3 2
A
2 3

 1 2
0
8
7
0
3  4
1 4 
2 3

4  3
Since the basis of the row space of A is
1
0 2 0 1, 0 1 1 0 1, 0 0 0 1  1,
the dimension of the row space is 3 and the row rank of A is 3. Similarly,
  1    2   3 

 
  
 3   2  1  
,
,






 2 
2
3

 
  

 1  2  4 
is the basis of the column space of A. Thus, the dimension of the column space is 3
and the column rank of A is 3.
Important Result:
The row rank and column rank of the
m n
matrix A are equal.
Definition of the rank of a matrix:
Since the row rank and the column rank of a m  n matrix A are equal,
we only refer to the rank of A and write rank  A .
Important Result:
Let A be an
nn
matrix.
 A is nonsingular if and only if rank  A  n .
rank  A  n  A is nonsingula r  det  A  0

 Ax  b has a unique solution.
 rank  A  n  Ax  0 has a nontrivial solution.
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