Linear Algebra 201-105-RE
Vincent Carrier
Row, Column, Null Spaces
Let A be an m × n matrix and let
Ri = ith row of A
for i = 1, 2, . . . m,
Cj = j th column of A
for j = 1, 2, . . . n.
Example: If
"
A =
1 2 3
4 5 6
#
,
then
"
R1 = [ 1 2 3 ],
C1 =
R2 = [ 4 5 6 ],
1
4
"
#
,
C2 =
2
5
#
"
,
C3 =
3
6
#
.
Consider the following definitions.
1)
Row Space of A :
RS(A) = span(R1 , R2 , . . . , Rm ).
2)
Column Space of A :
CS(A) = span(C1 , C2 , . . . , Cn ).
3)
Null Space of A :
NS(A) = {X ∈ Rn : AX = 0}.
It is readily seen that RS(A) is a subspace of Rn and that CS(A) is a subspace of Rm .
Theorem: If A is an m × n matrix, then NS(A) is a subspace of Rn .
Proof:
1) Assume that X1 , X2 ∈ NS(A). Then X1 + X2 ∈ NS(A) since
A(X1 + X2 ) = AX1 + AX2
= 0+0
= 0.
2) Assume that k ∈ R, X ∈ NS(A). Then kX ∈ NS(A) since
A(kX) = k(AX)
= k(0)
= 0.
Consider the 4 × 5 matrix A and its reduced row-echelon form A∗ :



A=


1 −1 −2 −1 −6
−1
0
3 −2 −2 


−3
1
8 −1
8 
2 −1 −5
0 −7



A∗ = 

1
0
0
0

0 −3 0 −4
1 −1 0 −1 

.
0
0 1
3 
0
0 0
0
Let us now address the problem of finding the dimension and a basis for the three vector
spaces associated to A:
RS(A), CS(A), NS(A).
1) Row Space of A
We have
ERO
A
−→
A∗
and thus
RS(A) = RS(A∗ ).
This means that RS(A) and RS(A∗ ) have the same dimension and basis.
Thus,
dim(RS(A)) = 3 = number of leading ones
Basis = {(1, 0, −3, 0, −4), (0, 1, −1, 0, −1), (0, 0, 0, 1, 3)}.
The dimension of RS(A) is called the rank of the matrix, written Rank(A). It is
always equal to the number of leading ones in the reduced row-echelon form of A.
2) Column Space of A
Since elementary row operations do not affect relationships between columns, we
have that dim(CS(A)) = dim(CS(A∗ )). Furthermore, the columns forming a basis
for CS(A∗ ) are the same columns forming a basis for CS(A).
Thus,
dim(CS(A)) = 3 = number of leading ones
Basis = {(1, −1, −3, 2), (−1, 0, 1, −1), (−1, −2, −1, 0)}.
We notice that dim(CS(A)) = dim(RS(A)) = Rank(A).
3) Null Space of A
The homogeneous linear system AX = 0 must be solved.



[A|0] = 








x1
x2
x3
x4
x5

1 −1 −2 −1 −6 0
−1
0
3 −2 −2 0
−3
1
8 −1
8 0
2 −1 −5
0 −7 0







 = 




3s + 4t
s+t
s
−3t
t




 =






s












3
1
1
0
0
−→








 + t




1
0
0
0
4
1
0
−3
1

0 −3 0 −4 0
1 −1 0 −1 0 

.
0
0 1
3 0 
0
0 0
0 0







s, t ∈ R.
Thus,
dim(NS(A)) = 2 = number of parameters
Basis = {(3, 1, 1, 0, 0), (4, 1, 0, −3, 1)}.
The dimension of NS(A) is called the nullity of the matrix, written Nullity(A). It
is always equal to the number of parameters in the solution set of the homogeneous
linear system AX = 0.
Theorem: Let A be an m × n matrix. Then
Rank(A) + Nullity(A) = n.
Proof:
Rank(A) = number of leading ones in the RREF of A
Nullity(A) = number of parameters in the solution set of the HLS AX = 0.
Theorem: Let A be an m × n matrix. Then
Rank(AT ) = Rank(A).
Proof:
Rank(AT ) = dim(RS(AT )) = dim(CS(A)) = Rank(A).
Example: Let A be a 5 × 6 matrix. Find Nullity(A) if Nullity(AT ) = 3.
If A is a 5 × 6 matrix, then AT is a 6 × 5 matrix.
Rank(A) = Rank(AT ) = 5 − Nullity(AT ) = 5 − 3 = 2.
Nullity(A) = 6 − Rank(A) = 6 − 2 = 4.
Linear Systems
Let A be an m × n matrix. The linear system AX = B is said to be consistent if its
solution set is not empty.
The product AX can be seen as






···
···
...
a1n
a2n
..
.
am1 am1 · · ·
amn
a11
a21
..
.
a12
a22
..
.

x1

  x2
 .
 .
 .
xn






 = x1 




a11
a21
..
.
am1






 + x2 




a12
a22
..
.
am2






 + · · · + xn 




Thus,
AX = linear combination of the columns of A.
This means that
the linear system AX = B is consistent
←→
B ∈ CS(A).
The following table summarizes the possible cases for an n × n matrix A.
Linear System AX = B
Rank(A) < Rank([A|B])
Inconsistent
Rank(A) = Rank([A|B]) < n
Infinitely Many Solutions
Rank(A) = Rank([A|B]) = n
Unique Solution
a1n
a2n
..
.
amn



.

