Linear Algebra
Rank
The Complete Solution to Ax=b
Math 301
1
Matrix rank
The “true” size of a matrix is determine by the rank of
a matrix.
The rank r of a matrix is defined as the number of
pivots.
Example :
A=

4
4
6
6
3
3
1
1
0 1
0 1
The size of A is R2⇥6 , but its rank is only 1.
The row reduced form reveals the pivots and the rank :

One non-zero
4
6
3
1
0
1
pivot so r = 1.
R=
0 0 0 0 0 0
2
Row reduced form
Consider the matrix
2
2
A=4 1
2
Its reduced row echelon
2
1
R=4 0
0
4
2
4
0
3
0
matrix is
2
0
0
0
1
0
3
2 0
8 0 5
2 1
3
1 0
3 0 5
0 1
Note : Pivot columns
are not necessarily the
first r columns of R.
Be sure you know how
to obtain the Reduced
Row Echelon matrix
The pivot columns in a reduced row matrix are columns
of the identity matrix.
3
Pivot columns
The reduced row matrix
2
1
R=4 0
0
R:
2
0
0
0
1
0
3
1 0
3 0 5
0 1
What is the rank of A? 3 (we have 3 pivot columns)
What space do the special solutions live in?
The special solutions are solutions to Ax = 0, and so
live in R5 (since x 2 R5 ).
How many free variables are there?
2
How may special solutions are there?
2
4
Nullspace matrix
We can construct a matrix whose columns are the special
solutions of the matrix A.
Pivot variables
Free variables
2
6
6
N =6
6
4
2
1
0
0
0
1
0
3
1
0
3
7
7
7
7
5
Special solutions to Ax = 0
are the columns of N.
The free variables are
form rows of the identity.
This is the nullspace matrix and satisfies AN = 0 (and
RN = 0).
The columns of N are linearly independent.
5
Nullspace Matrix
More generally, suppose A 2 Rm⇥n has rank r.
Suppose that all the pivot variables are in the first r
columns.
Then the reduced row form R of A can be written as

r pivot rows
This I is r ⇥ r.
I F
R=
m r zero rows
0 0
r pivot columns
n-r free columns
and the nullspace matrix has the form

r pivot variables
F
N=
n r free variables
I
This I is (n r) ⇥ (n r).
6
Summary
Let A be a Rm⇥n . Then
• The number of pivots (revealed by the “row reduced
form” of the matrix) tells us the rank r of A.
• The number of free variables is (n-r)
• The number of special solutions is (n-r)
• The dimension of the nullspace matrix is (n) x (n-r)
• Both A and the reduced row matrix R satisfy
AN=RN = 0
• The column space of A and the column space of R
are not the same in general.
7
The complete solution
Now we want to solve Ax = b. We use the augmented
matrix idea and reduce and augmented matrix to reduced row echelon form.
⇥
A
b
⇤
For what follows, assume A 2 R
has reduced row form R.
m⇥n
⇥
R
d
⇤
, has rank r and
Four cases to consider :
• A has full column rank (m > r, n = r)
We have already
done this case
• A has full row rank (m = r, n > r)
• A has an invertible (r x r) submatrix, m > r, n > r.
8
• A has both full row and column rank
Full column rank (m > r)
Ax = b,
Full column rank :
The number of
columns equals
the rank
R=
⇥
A

n=r<m
r
r
I
r
0
I
0
m
b
⇤

I
0
d1
d2
If d2 = 0, there is exactly one solution : x = d1
If d2 6= 0, the system has no solution.
The only nullspace solution is x = 0.
9
Full row rank (n > r)
Ax = b,
Full row rank :The
number of rows
equals the rank
R=
⇥
m=r<n
r
⇥
A
I
F
b
⇤
⇤
r
n
I
⇥
I
r
F
F
d
⇤
There are infinitely many solutions.
There are n-r free variables.
The solution has the form x = xp + xN where xp is a
particular solution, and xN is in the nullspace of A.
10
General case (r < m, r < n)
Ax = b,
r < m,
r<n
r
R=

I F
0 0
m
⇥
A
b
⇤
n
r
r
I
F
r
0
0

I
0
F
0
d1
d2
If d2 6= 0, the system has no solution.
If d2 = 0, the system has an infinite number of solutions of the form x = xp + xN .
11
Both full row and column rank
m=n=r
Ax = b
r
R=I
r
I
This is the first case
we considered
A is an invertible matrix
Ax = b has exactly one solution x = A
⇥
⇤
⇥
⇤
I d
A b
The only nullspace solution is x = 0.
1
b.
12
Example
2
2
4 1
2
0
3
0
Find the reduced
2
1 0
R=4 0 1
0 0
The rank is
3
.
0
0
1
4
2
4
3
2
2 6
6
8 56
6
2 4
x1
x2
x3
x4
x5
3
2
3
7
2
7
7=4 1 5
7
5
5
row form for A and b :
3
2
3
0 2
2
1
0 0
3 5
d=4 0 5
3
1 0
0
The matrix has full
The linear system has
1
row
rank.
solution(s).
13
Complete solution
The special solutions s1 and s2 are given
2
2
3
2
1
6 0 7
6 3
6
6
7
7
6 0
0
s1 = 6
s
=
2
6
7
6
4 1 5
4 0
0
1
by
3
7
7
7
7
5
Set the free variables x4 and x5 to zero to get a particular solution : xp = d = (2, 0, 3, 0, 0).
The complete solution is then
x = x p + a s1 + b s2
where a and b are arbitrary scalars.
14