5.5 Row Space, Column Space, and
Nullspace
Row Space, Column Space, and
Nullspace
 Definition:
For an mxn matrix
r1  a11
r2  a21
the vectors

rm  am1
a12  a1n 
a22  a2 n 
 a11 a12  a1n 
a

a

a
2n 
A   21 22
 

 


a
a

a
mn 
 m1 m 2

am 2  amn 
in Rn formed from the rows of A are called the row vectors
 a11 
 a21 
 am1 
of A, and the vectors
 




a
a
a
c1   12 , c2   22 ,  , cn   m 2 
  
  
  
 




a1n 
 a2 n 
 amn 
in Rm formed from the columns of A are called the column
vectors of A
2
Row Space, Column Space, and
Nullspace
 Definition:
 If A is an mxn matrix, then the subspace of Rn spanned
by the row vectors of A is called the row space of A,
 and the subspace of Rm spanned by the column
vectors is called the column space of A.
 The solution space of the homogeneous system of
equations Ax=0, which is a subspace of Rn, is called
the nullspace of A.
 Theorem 5.5.1: A system of linar equations Ax=b is
consistent iff b is in the column space of A
3
example
 1 3 2   x1 
 1 2 3   

  x2 
 2 1 2   x3 
1
  9 
 3 
Show that b in column space of A!
The solution by G.E. X1 = 2, X2 = -1, X3 = 3, the system is
consistent, b is in the column space of A
 1  3 
2 1
2  1    2   3  3    9 
 1   1 
 2   3
4
Row Space, Column Space, and
Nullspace
 Theorem 5.5.2: If x0 denotes any single solution
of a consistent linear system Ax=b, and if
v1,v2,...,vk form a basis for the nullspace of A,
that is, the solution space of the homogeneous
system Ax=0, then every solution of Ax=b can be
expressed in the form
x  x0  c1v1  c2 v 2   ck vk
and, conversely, for all choices of scalars
c1,c2,...,ck, the vector x in this formula is a
solution of Ax=b.
5
General and Particular Solutions
 Terminology:
 Vector x0 is called a particularly solution of Ax=b.
 The expression x0+c1v1+c2v2+...+ckvk is called the
general solution of Ax=b.
 The expression c1v1+c2v2+...+ckvk is called the general
solution of Ax=0.
6
Bases for Row Spaces, Column
Spaces, and Nullspaces



Theorem 5.5.3: Elementary row operations do not
change the nullspace of a matrix
Theorem 5.5.4: Elementary row operations do not
change the row space of a matrix
Theorem 5.5.5: If A and B are row equivalent
matrices, then:
a)
b)
A given set of column vectors of A is linearly
independent iff the corresponding column vectors of
B are linearly independent.
A given set of column vectors of A forms a basis for
the column space of A iff the corresponding column
vectors of B form a basis for the column space of B.
7
Bases for Row Spaces, Column
Spaces, and Nullspaces
 Theorem: If a matrix R is in row-echelon form, then the row
vectors with the leading 1’s (i.e., the nonzero row vectors) form a
basis for the row space of R, and the column vectors with the
leading 1’s of the row vectors form a basis for the column space
of R.
 Example: [Bases for Row and Column Spaces]
The matrix R is in row-echelon form, while the vectors r
1
0
R
0

0
2
1
0
0
5
3
0
0
0
0
1
0
3
0
0

0
r1  1  2 5 0 3
r2  0 1 3 0 0
r3  0 0 0 1 0
form a basis for the row space of R
8
Bases for Row Spaces, Column
Spaces, and Nullspaces
and the vectors
1
  2
0 
0 
 1 
0 
c1    , c 2    , c 4   
0 
 0 
1
 
 
 
0 
 0 
0 
form a basis for the column space of R
 Example: [Bases for Row and Column Spaces]
Find bases for the row and column spaces of
1  3 4  2 5 4 
1  3
2  6 9  1 8 2  elementary
0 0

A
R
2  6 9  1 9 7  row operations 0 0



1
3

4
2

5

4


0 0
4 2 5 4 
1 3  2  6
0 0 1 5

0 0 0 0
9
Bases for Row Spaces, Column
Spaces, and Nullspaces
The basis vectors are
r1  1  3
r2  0
r3  0
2
4
0 1 3
0
0
2
5
4
 6
0 1 5
The first, third, and fifth columns of R contain the
leading 1’s of the row vectors that form a basis for the
1
 4
 5 
column space of R.
 
 


0
1
 2
c1   , c3   , c5  
0 
0 
 1 
 
 


0
0
0
 
 


Thus the corresponding column vectors of A, form a
 4 
 5
basis for the column space of A  1 

2
 9 
 8 
, c5  

c1   , c3  
2
 9 
 9 
 




 1
  4
 5
10
Bases for Row Spaces, Column
Spaces, and Nullspaces

Example: [Basis and Linear Combinations]
Find a subset of the vectors
v1=(1,-2,0,3), v2=(2,-5,-3,6), v3=(0,1,3,0),
v4=(2,-1,4,-7), v5=(5,-8,1,2) that forms a basis for the
space spanned by these vectors.
b) Express each vector not in the basis as a linear
combination of the basis vector
Solution:  1 2 0 2 5  reduced 1 0 2 0 1
a)
a)
 2  5 1  1  8
0
row 
 0 3 3 4
0
1
echelon

form



3
6
0

7
2


0
    

v1 v 2 v 3 v 4 v 5
w1
1 1
0
0
0
1
0
0
0

w2

w3

w4
1
1

0

w5
11
Bases for Row Spaces, Column
Spaces, and Nullspaces
b)
Basis for the column space of matrix vectors w is
{w1,w2,w4} and consequently basis for the column
space of matrix vectors v is {v1,v2,v4}.
Expressing w3 and w5 as linear combinations of the
basis vectors w1,w2, and w4 (dependency equations).
w3 = 2w1 - w2
w5 = w1 + w2 + w4
The corresponding relationships are
v3 = 2v1 – v2
v5 = v1 + v2 + v4
12
Bases for Row Spaces, Column
Spaces, and Nullspaces
 Given a set of vectors S={v1,v2,...,vk) in Rn, the following
procedure produces a subset of these vectors that forms a basis
for span(S) and expresses those vectors of S that are not in the
basis as linear combinations of the basis vectors.
Step 1. Form the matrix A having v1,v2,...,vk as its column
vectors.
Step 2. Reduce the matrix A to its reduced row-echelon form R,
and let w1,w2,...,wk be the column vectors of R.
Step 3. Identify the columns that contain the leading 1’s in R.
The corresponding column vectors of A are the basis vectors for
span(S).
Step 4. Express each column vector of R that does not contain
a leading 1 as a linear combination of preceding column vectors
that do contain leading 1’s.
13
5.6 Rank and Nullity
Four Fundamental Matrix Spaces
 Fundamental matrix spaces:
Row space of A,
Nullspace of A,
Column space of A
Nullspace of AT
 Relationships between the dimensions of
these four vector spaces.
15
Row and Column Spaces have Equal
Dimensions
 Theorem 5.6.1: If A is any matrix, then the
row space and column space of A have the
same dimension.
 The common dimension of the row space and
column space of a matrix A is called the rank
of A and is denoted by rank(A); the dimension
of the nullspace of A is called the nullity of A
and is denoted by nullity(A).
16
Row and Column Spaces have Equal
Dimensions
 Example: [Rank and Nullity of a 4x6 Matrix]
Find the rank and nullity  1
3
of the matrix
A
Solution:
The reduced row-echeclon
form of A is
2

4
1
0

0

0
2 0 4
5  3
7 2 0
1
4 
5 2 4
6
1

9 2 4 4 7 
0  4  28  37 13
1  2  12  16 5 
0 0
0
0
0

0 0
0
0
0
rank(A) = 2 and the corresponding system will be
x1  4 x3  28x4  37x5  13x6  0
x2  2 x3  12x4  16x5  5 x6  0
17
Row and Column Spaces Have Equal
Dimensions
x1  4 x3  28x4  37x5  3x6
x2  2 x3  12x4  16x5  5 x6
The general solution of the system is
x1  4 r  2 8s  3 7t  3u
x2  2 r  1 2s  1 6t  5u
x3  r
x4  s
x5  t
x6  u
18
Row and Column Spaces Have Equal
Dimensions
 x1 
4
 28
37
  13
x 
2
12
16
 5 
 2
 
 
 


 x3 
1 
0
0
 0 
   r    s   t    u

 x4 
0 
1
0
 0 
 x5 
0 
0
1
 0 
 
 
 
 







0 

0

0

 1 

 x6 

Nullity(A)=4
19
Row and Column Spaces Have Equal
Dimensions



Theorem 5.6.2: If A is any matrix, then rank(A) =
rank(AT).
Theorem 5.6.3: [Dimension Theorem for Matrices]
If A is a matrix with n columns, then
rank(A) + nullity(A) = n
Theorem 5.6.4: If A is an mxn matrix, then:
a)
b)
Rank(A) = the number of leading variables in the
solution of Ax = 0.
Nullity(A) = the number of parameters in the general
solution of Ax = 0.
20
Row and Column Spaces Have Equal
Dimensions
 A is an mxn matrix of rank r
Fundamental Space Dimension
Row space of A
r
Column space of A
r
Nullspace of A
n-r
Nullspace of AT
m-r
21
Maximum Value for Rank
 A is an mxn matrix:
rank(A) ≤ min(m,n)
where min(m,n) denotes the smaller of the
numbers m and n if m≠n or their common
value if m=n.
22
Linear Systems of m Equations in n
Unknowns
 Theorem 5.6.5: [The Consistency Theorem]
If Ax = b is a linear system of m equations in
n unknowns, then the following are
equivalent.
a)
b)
c)
Ax = b is consistent
b is in the column space of A.
The coefficient matrix A and the augmented
matrix [A|b] have the same rank.
23
Linear Systems of m Equations in n
Unknowns

Theorem:
If Ax = b is a linear system of m equations in n
unknowns, then the following are equivalent.
a)
b)
c)

Ax = b is consistent for every mx1 matrix b.
The column vectors of A span Rm.
Rank(A) = m
A linear system with more equations than unknowns
is called an overdetermined linear system. The
system cannot be consistent for every possible b.
24
Linear Systems of m Equations in n
Unknowns
 Example: [Overdetermined System]
x1  2 x 2  b1
x1  x 2  b2
x1  x 2  b3
x1  2 x 2  b4
x1  3 x 2  b5
The system is consistent
iff b1, b2, b3, b4, and b5
satisfy the conditions

1
0

0

0
0
0
1
0
0
0
2b2  b1

b2  b1 
b3  3b2  2b1 

b4  4b2  3b1 
b5  5b2  4b1 
2b1  3b2  b3  0
3b1  4b2  b4  0
4b1  5b2  b5  0
b1  5r  4 s, b2  4r  3s,
b3  2r  s, b4  r , b5  s
where r and s are arbit rary
25
Linear Systems of m Equations in n
Unknowns


Theorem 5.6.7: If Ax=b is a consistent linear system of
m equations in n unknowns, and if A has rank r, then the
general solution of the system contains n-r parameters.
Theorem 5.6.8: If A is an mxn matrix, then the following
are equivalent.
a)
b)
c)


Ax=0 has only the trivial solution.
The column vectors of A are linearly independent.
Ax=b has at most one solution (none or one) for every mx1
matrix b.
A linear system with more unknowns than equations is
called an underdetermined linear system.
Underdetermined linear system is consistent if its
solution has at least one parameter → has infinitely
many solution.
26
Summary

Theorem 5.6.9: [Equivalent Statements]
If A is an nxn matrix, and if TA:Rn→Rn is
multiplication by A, then the following are
equivalent.
a)
b)
c)
d)
e)
f)
g)
A is invertible
Ax=0 has only the trivial solution
The reduced row-echelon form of A is In.
A is expressible as a product of elementary matrices.
Ax=b is consistent for every nx1 matrix b
Ax=b has exactly one solution for every nx1 matrix b
Det(A)≠0
27
Summary
h)
i)
j)
k)
l)
m)
n)
o)
p)
q)
The range of TA is Rn
TA is one-to-one
The column vectors of A are linearly independent
The row vectors of A are linearly independent
The column vectors of A span Rn
The row vectors of A span Rn
The column vectors of A form a basis for Rn
The row vectors of A form a basis for Rn
A has rank n
A has nullity 0
28