4.6 Rank
We introduce a new space: the row space.
Definition: The row space of an m x n matrix
A is the set (Row A) of all linear combinations
of the rows of A.
Example: Let
3
6 
 1 2
A   2  5  6  12
 1  3  3  6 
Here r1 =(–1, 2, 3, 6)
r2 =(2, –5, –6, –12)
r3 =(1, –3, –3, –6)
It is natural to express the vectors in Row A
as row vectors.
Row A = Span{r1, r2, r3}.
Claim: The row space of an m x n matrix A is
is a subspace of Rn.
Proof: Note that the rows of A are the
columns of AT, and AT is an n x m matrix, so
Col AT is a subspace of Rn by Th. 4.3.
Col AT = Row A
 Row A is a subspace of Rn. ■
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Note: when we use row operations to reduce
a matrix A to a matrix B, we are taking linear
combinations of rows of A to get B. We could
reverse this process to get back to A from B.
So, Row A = Row B.
Example: Let
  1 2 3 6
3
6 
 1 2
 0  1 0 0


B

A   2  5  6  12


 0 0 0 0
 1  3  3  6 
These matrices are row equivalent. We can
use this fact to find bases and dimensions of
Row A, Col A, and Nul A.
Row A
Since Row A = Row B,
and we can see that the basis for Row B is
{(–1, 2, 3, 6), (0, –1, 0, 0)}, the basis for
Row A is {(–1, 2, 3, 6), (0, –1, 0, 0)}, and
dim Row A = 2.
2
Col A
Since B shows that columns 1 and 2 are
linearly independent, a basis for Col A is
 1  2  
    
 2 ,  5 
 1   3 
    
Note: these rows come from the matrix A, and
dim Col A = 2.
Nul A
Row reduce the augmented matrix [A 0]
corresponding to the equation Ax = 0.
1 0  3  6 0


[ A 0] ~ [ B 0] ~ 0 1 0
0 0
0 0 0
0 0
We get the solution
x1  3x3  6 x4
x2  0
x3 and x4 are free
or
3
 3
6 
0 
0 
x  x3    x4  
1
0 
 
 
0
 
1
3 6 
    
0 0 
 , 
1 0 
So,    
0 1 
is a basis for Nul A, and dim Nul A = 2
Note:
dim Col A = the number of pivot columns in A
= the number of non-zero rows in B
= dim Row A.
dim Nul A = the number of free variables in
the system associated to Ax = 0.
= the number of “non-pivot” columns
in A.
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Definition: The rank of an m x n matrix A is
the dimension of the column space of A.
rank A = dim Col A
= the number of pivot columns in A
= the number of non-zero rows in B
= dim Row A.
►rank A + dim Nul A = n,
Where n is the number of columns in A.
Theorem 4.14: (The Rank Theorem)
The dimensions of the column space and the
row space of an m x n matrix A are equal.
This common dimension, the rank of A, also
equals the number of pivot positions in A and
satisfies the equation:
rank A + dim Nul A = n.
Since Row A = Col AT,
rank A = rank AT.
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Example: Suppose a 5x8 matrix A has rank 5.
Find
a) dim Nul A
b) dim Row A
c) rank AT
d) is Col A = R5?
a) By Th. 4.14,
rank A + dim Nul A = n
Here,
n = 8 and rank A = 5, so
5 + dim Nul A = 8
dim Nul A = 3.
b) rank A = dim Col A
= the number of pivot columns in A
= the number of non-zero rows in B
= dim Row A, so
dim Row A = 5
c) rank A = rank AT, so rank AT = 5.
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d) Since rank A = dim Col A
= the number of pivots in A
= 5,
there is a pivot in every row of A. Thus, the
columns of A span R5 by Th. 1.4.
So, Col A contains 5 vectors that span R5.
By the Basis Theorem (Th. 4.12), the vectors
in Col A are a basis for R5 and Col A = R5.
Example: For a 9 x 12 matrix A, find the
smallest possible value of dim Nul A.
Again, By Th. 14,
rank A + dim Nul A = n
dim Nul A = n – rank A
So, the largest value of rank A is the number
of linearly independent columns of A.
A can have at most 9 linearly independent
columns since there can be at most 9 pivots
positions in A.
So, the smallest possible value of
dim Nul A is 12 – 9 = 3.
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Visualizing Row A and Nul A
Ex: Let
1 0  1 
A

2
0

2


►A basis for Nul A is
0 1 
    
1, 0 
0 1  , so Nul A is a plane in R3.
    
►A basis for Row A is
 1  
  
 0  
1  , so Row A is a line in R3.
  
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►A basis for Col A is
 1  
   , so Col A is a line in R2.
  2 
►A basis for Nul AT is
   2 
  
T
2
,
so
Nul
A
is
a
line
in
R
.
1
  
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The Rank Theorem provides us with a
powerful tool for finding information about
systems of equations.
Example: A scientist solves a homogeneous
system of 50 equations in 54 variables and
finds that exactly 4 of the unknowns are free
variables. Can the scientist be certain that
any associated non-homogenous system
(with the same coefficients) has a solution?
Recall:
►rank A = dim Col A = number of pivots in A
►dim Nul A = the number of free variables in
Ax = 0.
Since A is 50 x 54
By the Rank Theorem,
rank A + dim Nul A = 54
so rank A = 50.
Thus,  a pivot in every row of A, so every
non-homogenous system Ax = b has a
solution.
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This lets us add six statements to the IMT:
m. The columns of A form a basis for Rn.
n. Col A = Rn.
o. dim Col A = n.
p. rank A = n.
q. Nul A = {0}.
r. dim Nul A = 0.
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