5.2 Rank of a Matrix
Set-up
Recall block
multiplication:
A  C 1
C2
...
 R1 
R 
2
C n    
 ... 
R 
 m 
x 1 
 
x2

  C
A
1
... 
x 
 n 
C2
y 1
R1 
R 
2
y m    y 1 R1  y 2 R 2  ...  y m R m
 ... 
R 
 m 
y2
...
...
x 1 
 
x2

  x C  ...  x C
Cn 
1 1
n
... 
x 
 n 
Theorem 1
The following conditions are equivalent for an n x n matrix A:
1. The rows of A are linearly independent in n.
2. The rows of A span n.
3. The columns of A are linearly independent in n.
4. The columns of A span n.
5. A is invertible.
Proof of Theorem 1
To prove that 5 statements are equivalent, we can prove that
(1)(2)(3)(4)(5)(1) or could do:
(1)(2)(5)(3)(4)(1) which we will do.
Proof: (1)  (2) given the rows are linearly independent in n
and we know dim n = n, we know then that the rows must
form a basis, and thus span n.
Proof of Theorem 1 (cont)
(2)  (5) If we can find a matrix B such that BA = I, then we
know A is invertible.
Let row i of B be K = [k1 k2 … kn]
R 
Then I = BA results in:
R 
Row i of I = row i of BA = KA = k k ... k    k R  ...  k R
1
2
1
2
n
... 
R 
 n 
1
1
n
Since the rows of A span n, we know that such ki exist (we know we
can create any vector in n using linear combinations of the rows in A).
Therefore, each row of B can be found making A invertible.
n
Proof of Theorem 1 (cont)
We will now show (5)(1) and leave (3) and (4) for hw.
(5) (1): Let k1R1+…+knRn = 0 where Ri is the ith row of A.
Let K = [k1 k2 … kn] so KA = k1R1+…+knRn = 0
Right multiply by A-1 to get K=0, which means k1=…=kn = 0
which means that the rows of A are linearly independent.
A use of Theorem 1
So now to show that an (n x n) matrix is invertible, we just
need to show that the rows are linearly independent or that they
span n.
Or to show that the rows are linearly independent or that the
rows span n, we could show that the matrix is invertible.
Example 1
Show that {(1,2,-1), (3,1,-4), (1,1,7)} is a basis for 3.
We just need to show that the matrix with these vectors as
rows is invertible which can be done by showing that it has
rank 3, or by showing that the determinant is not zero, etc.
Definition
If A is an m x n matrix, the rows of A are vectors in n, and
the subspace of n spanned by these rows is called the row
space of A (row A).
The space spanned by the columns of A is called the column
space of A (col A).
Theorem 2
A (m x n), U (p x m),V (n x q)
1. row(UA)  row A with equality if U is invertible (& square)
2. col(AV)  col A with equality if V is invertible (& square)
Proof of Theorem 2
Proof: R1,R2,…,Rm are rows of A
row i of mtx U is: U i  u i 1 u i 2 ... u im 
row i of UA = UiA=  R1 
R 
2

u i1 u i 2 ... u im    u i1 R 1  u i2 R 2  ...  u im R m
 ... 
 
R m 
So the rows of UA are in the row space of A, which means that
row (UA)  row A.
If U is invertible: row A = row[U-1(UA)]  row UA (same
logic as above) so row A = row (UA).
(same for part 2)
Theorem 3-Rank Theorem
A (m x n)
Then, dim(row A) = dim(col A)
Also, A can be carried to row echelon form matrix R, and if r
is the number of nonzero rows in R,
1. The r nonzero rows are a basis of (row A)
2. If the leading 1’s are in columns j1,j2,…,jr, then
{j1,j2,…,jr} is a basis of (col A)
Theorem 3-Proof
R = UA (U is the product of elementary matrices and is
invertible).
row A = row R (by theorem 2) since U is invertible
For a matrix in row echelon form, the rows form a basis for
the rows since they span the rows and are linearly
independent. Since row A = row R, the nonzero rows also
form a basis for A.
Theorem 3-Proof (cont)
(2)
A  C 1
C2
R  UA  U C1
...
Cn 
C2
...
C n   U C 1
UC 2
...
UC n 
B={UCj1, UCj2,…,UCjr} is the set of columns of R w/ a
leading 1.
In homework, you will prove that for a matrix is ref, the
columns with leading 1’s form a basis of col R (since
they’re li). So B is a basis for R.
Theorem 3-Proof (cont)
Since {UCj1, UCj2,…,UCjr} is LI, a

U C j1
C j2
...

C jr is .LI
a 1UC j 1  a 2 UC j 2  ...  a r UC jr  0  a 1  ...  a r  0
U (a 1 C j1  a 2 C j 2  ...  a r C jr )  0
(a 1C j1  a 2 C j 2  ...  a r C jr )  0  a 1  ...  a r  0
So {Cj1, Cj2,…,Cjr} is LI and it spans A so it is a basis of
col A.
dim(rowA)=r=dim(colA) and is called the rank•
Corollary
If A can be carried to R in ref by elementary row ops, then the
rank of A is equal to the number of nonzero rows of R.
Example
Find basis for row and col space of A and find rank
2

2

A 
4

0
4
6
1
3
5
9
1
1
8 

2

10 

2 
Note that the basis of row space comes from ref, and
the basis of col space comes from initial state.
Corollaries
Cor 2: If A is mxn, then rank A ≤m and rank A≤ n
Cor 3: If A is any matrix, rank A = rank AT
Cor 4: A (m x n), U (m x m) and V (n x n) invertible, then
rank A = rank (UA) = rank (AV) ( by theorem 2)
Cor 5: A (n x n) is invertible iff rank A = n.
Example
Find basis of U=span {(1,-1,0,3),(2,1,5,1),(4,-2,5,7)}
Can just write in matrix form and row reduce.
Theorem 4
A (m x n) has rank r iff there exist invertible matrices U (mxm)
and V(nxn) such that
0 
I
r
UAV  
0

0 
Where Ir is the r x r identity matrix
Proof the row operations that take A to R (rref) also carry Im to
Invertible U such that UA = R so  A I   R U 
m
R has r nonzero rows, which contain columns of Ir.
Theorem 4
Then take RT and put it in rref using U1 so then
I r
U 1 R  
 0
T
0 

0 
Let V=U1T and we have:
UAV  RU 1
T
I r
T T
 (U 1 R )  
0
T
0 
I r
  
0 
0
0 

0 
And also rank A = r if these U and V exist (by Cor 4)
Use of this
Now we can find these U and V which will reduce A to
this form.
1) Find U from  A I   R U 
m
2) Find V from
R
T
I r
I n   
0
0 

0 

V 

T
Example
Find invertible matrices U and V such that
I r
UAV  
0
 1
A  
 2
0 
, r  rankA
0 
1
2
 1

4 
Definition
A (mxn). The set of solutions to AX=0 is a subspace of n
called the null space of A (null A)
Theorem 5
A (m x n) and r = rank A. Then dim (null A) = n-r
Proof:
Recall that when you solve a homogeneous system, you either
get a unique solution if the A is invertible (in which case the set
of solutions contains no parameters and thus dim (nullA) = 0.
Or, A could reduce to have fewer equations than variables in
which case the solution will be of a form like:
x 1 
 
x 
 2 

x 3 

2 
 
s 0 
 

1 

 1 
 
t 1
 

 2 

Which has dim = 2 = # parameters
So we just need to find the number of
parameters.
Theorem 5-Proof-formal
A (m x n) and r = rank A. Then dim (null A) = n-r
Proof:
0 
I
null (A) = {X | AX = 0}. Let UAV  
,U ,V ..invertible
r
0
0 
null(A) = null(UA) since U is invertible
dim(null(UA)) = dim(null(UAV)) since V invertible
So dim (null(A))=dim(null(UAV))
UAV is in rref so dim (null(UAV)) = n-r (# var-rank)=#
parameters.
Example
Find the basis of the nullspace of
3

2

A 
4

1
1
0
2
1
1 

1

1 

1 
Just find the sol’n to the homogeneous system and then
the dimension is the number of parameters, and the
vectors will form the basis.`
Theorem 6
The following conditions are equivalent for an (m x n) matrix.
1. Rank A = m
2. The rows of A are LI
3. If YA = 0 with Y in m, then Y=0
4. AAT is invertible
Proof: (1)(2) rank A = m implies the m rows form a basis for
row A which implies that the rows are LI
(2)(3) YA = y1R1+y2R2+…+ymRm = 0 implies y1=y2=..=ym= 0
Since the rows of A are LI, so Y = 0.
Theorem 6 (proof cont)
(3)(4) We will show that Y(AAT) implies Y=0
(YA)(YA)T = YAATYT =0YT=0
So YA = 0 (since (YA)(YA)T =0 and a AAT=0 implies A=0)
so Y = 0 (given in 3)
(4)(1) Just need to show that {R1,R2,…,Rm} is LI
Let x1R1+…+xmRm = 0.
X   x 1 ... x m 
So XA = 0.
So XAAT = 0
So X=0 since AAT is invertible
So all x’s are 0 and rows are li•