Chapter 4: Vector Spaces - Part 4/Slides By Pearson

Find a Basis and Dimension of a Subspace
Find a basis and the dimension of the subspace
3 6
5 4
: , , , in
2
5
a b c
a d
H a b c d
b c d
d
   
      
   
    

H is the set of all linear combinations of the vectors
1
1
5
v
0
0
 
 
 
 
 
 
2
3
0
v
1
0
 
 
 
 
 
 
3
6
0
v
2
0
 
 
 
 
 
 
4
0
4
v
1
5
 
 
 
 
 
 
v3 = -2 v2
v1, v2, v4 are linearly independent

Therefore, {v1, v2, v4} is a set of linearly
independent vectors.
Also, H = span {v1, v2, v4}.
Hence {v1, v2, v4} is a basis for H.
Thus dim H =3.

4. Vector Spaces
ROW and COLUMN
SPACES of a MATRIX
and
RANK of a MATRIX

 Defn - Let A be an m x n matrix. The rows of A,
considered as vectors in Rn , span a subspace of Rn
called the row space of A.
Each row has n entries, so Row A is a subspace of Rn.
11 12 1
21 22 2
1 2
n
n
mnm m
a a a
a a a
a a a
 
 
 
 
 
  
A
Row Space of a Matrix : RowA

 Defn - Let A be an m x n matrix. The columns of
A, considered as vectors in Rm, span a subspace of
Rm called the column space of A
 The system Ax = b has a solution if and only if b
is in the column space of A.
 Each column has m entries, so Col A is a subspace
of Rm.
11 12 1
21 22 2
1 2
n
n
mnm m
a a a
a a a
a a a
 
 
 
 
 
  
A
Column Space of a Matrix : Col A

 Theorem - If A and B are mxn row (column)
equivalent matrices, then the row (column) spaces
of A and B are equal (the same).
 Recall: If A and B are row equivalent when the
rows of B are obtained from the rows of A by a
finite number of the three elementary row
operations.
Equality of Spaces

 Defn - The dimension of the row space of a matrix
A is called the row rank of A
 Defn - The dimension of the column space of a
matrix A is called the column rank of A
 Theorem - The row rank and the column rank of
an mxn matrix A = [ aij ] are equal.
Rank of A

 The rank of A corresponds to the number of linearly
independent rows (columns) of A.
 If A is an n x n matrix, then rank A = n if and only if A is
row equivalent to In
 Let A be an n x n matrix. A is nonsingular if and only if
rank A = n
 If A is an n x n matrix, then rank A = n if and only if
det(A) ≠ 0.
 The homogeneous system Ax = 0 where A is an n x n
matrix has a nontrivial solution if and only if rank A < n
 Let A be an n x n matrix. The linear system Ax = b has a
unique solution if and only if rank A = n.
Rank of A

Equivalent statements for nxn matrix A
 A is nonsingular
 AX  0 has the trivial solution only
 A is row equivalent to In
 The system AX  b has a unique solution
 A has rank n
 The rows (columns) of A form a linearly independent
set of vectors in Rn
Rank of A

Let A be an mxn matrix.
Then
rank A + nullity A = n.
Nullity A = dim Null(A)
Rank + Nullity Theorem

 Example:
a. If A is a matrix with a two-dimensional
null space, what is the rank of A?
b. Could a matrix have a two-dimensional
null space?
 Solution:
a. Since A has 9 columns, , and
hence rank .
b. No. If a matrix, call it B, has a two-
dimensional null space, it would have to have
rank 7, by the Rank Theorem.
7 9
6 9
(rank ) 2 9A  
7A 
6 9
Rank + Nullity Theorem

To identify the basis for Row A and Col A
1. Reduce A to echelon form E
2. The nonzero rows of E form a basis for Row A
3. The columns of A that correspond to the columns of E
containing the leading entries form a basis for Col A
Note: The same is true if the matrix E is in reduced
echelon form.
For a matrix A, the number of linearly independent rows
is equal to the number of linearly independent columns.
Bases for Row A and Col A

has m = n = 2 and rank r = 1
Column space contains all multiples of
Row space consists of all multiples of [ 1 2 ]
Null space consists of all multiples of
1 2
3 6
 
 
 
A
1
3
 
 
 
1
2
 
 
 
Example

Example - Find a basis for the subspace V of R3
spanned by the rows of the matrix A
1 2 1
1 1 1
1 3 3
3 5 1
1 4 5
 
 
 
 
 
 
 
 


 


A

 Example (continued) - Apply elementary row
operations to A to reduce it to B, in echelon form
1 0 3
0 1 2
0 0 0
0 0 0
0 0 0
 
 
 
 
 
 
 
 


B
A basis for the row space of B
consists of the vectors
w1 = [ 1 0 3 ] and w2 = [ 0 1 2 ].
So { w1, w2 } is a basis for V
v1 = w1  2w2, v2 = w1 + w2,
v3 = w1  3w2, v4 = 3w1  5w2,
v5 = w1  4w2

Example (continued) - Since a matrix A is row
equivalent to a matrix B that is in row echelon form,
then
the nonzero rows of B form a basis for the row space
of A and the row space of B.
w1 = [ 1 0 3 ] and w2 = [ 0 1 2 ].

DIMENSIONS OF NUL A AND COL A
 Let A be an matrix, and suppose the equation
has k free variables.
 A spanning set for Nul A will produce exactly k
linearly independent vectors—say, —one for
each free variable.
 So is a basis for Nul A, and the number of
free variables determines the size of the basis.
m n
x 0A 
1
u ,...,uk
1
{u ,...,u }k

 The dimension of Nul A is the number of free
variables in the equation , and the dimension
of Col A is the number of pivot columns in A.
 Example: Find the dimensions of the null space and
the column space of
x 0A 
3 6 1 1 7
1 2 2 3 1
2 4 5 8 4
A
   
   
 
   

 Solution: Row reduce the matrix to echelon form:
 There are three free variable—x2, x4 and x5.
 Hence the dimension of Nul A is 3.
 Also dim Col A = 2 because A has two pivot
columns (column 1 of A and column 3 of A).
1 2 2 3 1
0 0 1 2 2
0 0 0 0 0
  
 
 
  

Find bases for the row space, the column space, and the
null space of the matrix
2 5 8 0 17
1 3 5 1 5
3 11 19 7 1
1 7 13 5 3
   
 
 
 
   
Example

To find bases for the row space and the column space,
row reduce A to an echelon form:
 The first three rows of B form a basis for the row
space of A (as well as for the row space of B).
 Basis for Row A:
Example (cont) : Row space
1 3 5 1 5
0 1 2 2 7
0 0 0 4 20
0 0 0 0 0
A B
 
  
 
 
 
 
{(1,3, 5,1,5),(0,1, 2,2, 7),(0,0,0, 4,20)}   

 For the column space, observe from B that the pivots
are in columns 1, 2, and 4.
 Hence columns 1, 2, and 4 of A (not B) form a basis
for Col A:
Basis for Col
 Notice that any echelon form of A provides (in its
nonzero rows) a basis for Row A and also identifies
the pivot columns of A for Col A.
Example (cont) : Column space
2 5 0
1 3 1
: , ,
3 11 7
1 7 5
A
       
      
       
      
            

 However, for Nul A, we need the reduced echelon
form.
 Further row operations on B yield
Example (cont) : Null space
1 0 1 0 1
0 1 2 0 3
0 0 0 1 5
0 0 0 0 0
A B C
 
 
 
 
 
 

 The equation is equivalent to , that is,
 So , , , with x3
and x5 free variables.
x 0A  x 0C 
1 3 5
2 3 5
4 5
0
2 3 0
5 0
x x x
x x x
x x
  
  
 
1 3 5
x x x   2 3 5
2 3x x x  4 5
5x x

 The calculations show that
Basis for Nul A
 Observe that, unlike the basis for Col A, the bases for
Row A and Nul A have no simple connection with the
entries in A itself.
1 1
2 3
: ,1 0
0 5
0 1
     
         
    
    
    
        

Section 4.5
p.268, p.269
1 to 16
Problems

Chapter 4: Vector Spaces - Part 4/Slides By Pearson

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Chapter 4: Vector Spaces - Part 4/Slides By Pearson