4.16 The Gram Schmidt Process with Complex Vectors
When we use the QR algorithm with matrices of complex numbers we will be finding the QR factorization
of matrices with complex numbers. We can either use Gram Schmidt orthonormalization or Householder
triangularization to do the QR factorization. These are the same as in the real case with two modifications.
The definition of the length and inner product for vectors of complex numbers need the following
z1
w1
z2
w2
modifications. Let z = . and w = . be vectors of complex numbers. Then
.
.
zn
wn



 
 
 
_ _
_
_T
z* = (z1, z2,…, zn) = (z ) = conjugate transpose of z
|| z || =
| z1 |2 + | z2 |2 + … + | zn |2 =
z*z
_
_
_
z . w = z1w1 + z2w2 + … + znwn = z*w = complex inner product of z and w
Example 1. Let z =
 2 - 3i  and w =  1 - 6i  . Then
 5 + 4i 
 7 + 8i 
z* = ( 2 + 3i, 5 – 4i)
|| z || =
| 2 + 3i |2 + | 5 – 4i |2 =
4 + 9 + 25 + 16 =
54
z . w = (2 + 3i)(1 - 6i) + (5 - 4i)(7 + 8i) = 2 – 12i + 3i + 18 + 35 + 40i – 28i + 32
= 87 + 3i
With these modifications, the Gram Schmidt is the same as in the real case. Let's illustrate this with an
example.
Example 2. Let A =
 0 - 1 . Do one step of the QR algorithm with shift  = 3i.
4 0
We need to find A1 = RQ + I where QR = A - I is the QR factorization of A - I. One has
A - I = A - 3iI =
 - 3i - 1 
 4 - 3i 
We need to find the QR factorization of this matrix. Let's use the Gram – Schmidt process. We begin with
Q =
 - 3i - 1 
 4 - 3i 
Step 1. Set R11 = || Q●,1 || and Q●,1 =
R = I =
Q●,1
. In our example,
R11
- 3i
R11 = || Q●,1 || = ||  4  || = 5
Q●,1 =
1 0
0 1
Q●,1
1 - 3i
= 5  4 
R11
4.16 - 1
Now
 - 53i - 1 

Q =  4


3i
 5

R =
5 0
0 1
Step 2. Set R12 = Q●,1 . Q●,2 and Q●,2 = Q●,2 – R12Q●,1. In our example,
R12 = Q●,1 . Q●,2
-1
-1
1 - 3i
1
1
= 5  4  .  - 3i  = 5 (3i, 4)  - 3i  = 5 (- 3i –12i) = - 3i
Q●,2 = Q●,2 – R12Q●,1 =
 - 1  - (-3i)
 - 3i 
(15) 
- 3i 
4  =
 -1  +
 - 3i 
Now
Q =
(15) 
- 3i
4
4 - 3i 
Step 3. Set R22 = || Q●,2 || and Q●,2 =
R =
5 - 3i
0 1 
Q●,2
. In our example,
R22
4
R22 = || Q●,2 || = || - 3i  || = 1
Q●,2 =
Q●,2
4
1
= Q●,2 = 5  - 3i 
R22
Now
Q =
(15) 
- 3i
4
4 - 3i 
R =
5 - 3i
0 1 
- 3i - 1
So the QR factorization of A - I = A - 3iI =  4 - 3i  is
 - 53i
- 3i - 1 

A - I = A - 3iI =  4 - 3i  = QR =  4

 5
4
5 

- 3i 
5 
5 - 3i
0 1 
So
5 - 3i  5
= RQ + I = 0 1   4

- 3i
A1

5
4
5 

- 3i 
5 
 - 27i
5
=  4

 5
4.16 - 2
11
5
- 3i
5




(15)  12i 
9