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Reduced and Full Rank QR Factorization
Numerical Linear Algebra
Isaac Amornortey Yowetu
NIMS-GHANA
September 26, 2020
Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples
Introduction
Suppose we have a matrix A ∈ Rm×n
for m ≥ n, we can have
A to be decomposed or factorized as:
A = QR, where:
Q is orthogonal with orthonormal basis.
R is right upper triangular matrix.
It can therefore written as:
A = QR
a1 a2 · · · an = q1 q2 · · · qn





r11 r12 · · · r1n
0 r22 · · · r2n
...
...
...
...
0 0 · · · rnn





Isaac Amornortey Yowetu NIMS-GHANA
Reduced and Full Rank QR Factorization
Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples
Reduced Vs Full Rank Factorization
Suppose we have matrix A ∈ R3×2
, we can have the following:
Reduced Rank QR Factorization


a11 a12
a21 a22
a31 a32

 =


q11 q12
q21 q22
q31 q32

 r11 r12
0 r22
Full Rank QR Factorization


a11 a12
a21 a22
a31 a32

 =


q11 q12 q13
q21 q22 q23
q31 q32 q33




r11 r12
0 r22
0 0


We shall consider this in details with an example.
Isaac Amornortey Yowetu NIMS-GHANA
Reduced and Full Rank QR Factorization
Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples
Example 1
Question
Consider a matrix A =


1 2
0 1
1 0

. Using any method you like,
determined reduced and full rank factorization of A.
A = ˆQ ˆR and A = QR
Isaac Amornortey Yowetu NIMS-GHANA
Reduced and Full Rank QR Factorization
Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples
Solution: Using Gram-Schmidt Process
Let a1 =


1
0
1

 and a2 =


2
1
0

 and suppose we have
{v1, v2, v3} ∈ V , then
v1 = a1 =


1
0
1


q1 =
v1
||v1||
=
√
2
2


1
0
1

 =


√
2
2
0√
2
2


Remark: r11 = ||v1|| =
√
2
Isaac Amornortey Yowetu NIMS-GHANA
Reduced and Full Rank QR Factorization
Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples
v2 = a2 − projq1 (a2)
= a2 − r12(q1)
v2 =


2
1
0

 −
√
2


√
2
2
0√
2
2

 =


1
1
−1


q2 =
v2
||v2||
=
√
3
3


1
1
−1

 =



√
3
3√
3
3
−
√
3
3



Remark: r12 = <q1,a2>
<q1,q1>
=
√
2 and r22 = ||v2|| =
√
3
Isaac Amornortey Yowetu NIMS-GHANA
Reduced and Full Rank QR Factorization
Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples
Reduced Rank QR Factorization
A = ˆQ ˆR


1 2
0 1
1 0

 =



√
2
2
√
3
3
0
√
3
3√
2
2
−
√
3
3



√
2
√
2
0
√
3
Isaac Amornortey Yowetu NIMS-GHANA
Reduced and Full Rank QR Factorization
Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples
To find the full rank QR Factorization, we let q3 =


q13
q23
q33


such that
< q1, q3 > = 0 (1)
< q2, q3 > = 0 (2)
< q3, q3 > = 1 (3)
Where < q1, q3 >= 0 can be expressed as:
√
2
2
q13 + 0 · q23 +
√
2
2
q33 = 0 (4)
and < q2, q3 > can also as be written as:
Isaac Amornortey Yowetu NIMS-GHANA
Reduced and Full Rank QR Factorization
Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples
√
3
3
q13 +
√
3
3
q23 −
√
3
3
q33 = 0 (5)
We can further reduce eqn(4) and (5) respectively as:
q13 + q33 = 0 (6)
q13 + q23 − q33 = 0 (7)
From eqn(6), we let q33 = −t and q13 = t.
Replacing q13 and q33 into eqn (7), q23 = −2t. Hence,
v3 = t


1
−2
−1

 and t is a dummy variable which can take any
number such that t ∈ R.
Here, we consider t = 1.
Isaac Amornortey Yowetu NIMS-GHANA
Reduced and Full Rank QR Factorization
Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples
Full Rank QR Factorization
Hence,
q3 = v3
||v3||
=



√
6
6
−
√
6
3
−
√
6
6



A = QR


1 2
0 1
1 0

 =



√
2
2
√
3
3
√
6
6
0
√
3
3
−
√
6
3√
2
2
−
√
3
3
−
√
6
6





√
2
√
2
0
√
3
0 0


Isaac Amornortey Yowetu NIMS-GHANA
Reduced and Full Rank QR Factorization

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Reduced and Full Rank QR factorization

  • 1. Reduced and Full Rank QR Factorization Numerical Linear Algebra Isaac Amornortey Yowetu NIMS-GHANA September 26, 2020
  • 2. Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples Introduction Suppose we have a matrix A ∈ Rm×n for m ≥ n, we can have A to be decomposed or factorized as: A = QR, where: Q is orthogonal with orthonormal basis. R is right upper triangular matrix. It can therefore written as: A = QR a1 a2 · · · an = q1 q2 · · · qn      r11 r12 · · · r1n 0 r22 · · · r2n ... ... ... ... 0 0 · · · rnn      Isaac Amornortey Yowetu NIMS-GHANA Reduced and Full Rank QR Factorization
  • 3. Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples Reduced Vs Full Rank Factorization Suppose we have matrix A ∈ R3×2 , we can have the following: Reduced Rank QR Factorization   a11 a12 a21 a22 a31 a32   =   q11 q12 q21 q22 q31 q32   r11 r12 0 r22 Full Rank QR Factorization   a11 a12 a21 a22 a31 a32   =   q11 q12 q13 q21 q22 q23 q31 q32 q33     r11 r12 0 r22 0 0   We shall consider this in details with an example. Isaac Amornortey Yowetu NIMS-GHANA Reduced and Full Rank QR Factorization
  • 4. Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples Example 1 Question Consider a matrix A =   1 2 0 1 1 0  . Using any method you like, determined reduced and full rank factorization of A. A = ˆQ ˆR and A = QR Isaac Amornortey Yowetu NIMS-GHANA Reduced and Full Rank QR Factorization
  • 5. Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples Solution: Using Gram-Schmidt Process Let a1 =   1 0 1   and a2 =   2 1 0   and suppose we have {v1, v2, v3} ∈ V , then v1 = a1 =   1 0 1   q1 = v1 ||v1|| = √ 2 2   1 0 1   =   √ 2 2 0√ 2 2   Remark: r11 = ||v1|| = √ 2 Isaac Amornortey Yowetu NIMS-GHANA Reduced and Full Rank QR Factorization
  • 6. Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples v2 = a2 − projq1 (a2) = a2 − r12(q1) v2 =   2 1 0   − √ 2   √ 2 2 0√ 2 2   =   1 1 −1   q2 = v2 ||v2|| = √ 3 3   1 1 −1   =    √ 3 3√ 3 3 − √ 3 3    Remark: r12 = <q1,a2> <q1,q1> = √ 2 and r22 = ||v2|| = √ 3 Isaac Amornortey Yowetu NIMS-GHANA Reduced and Full Rank QR Factorization
  • 7. Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples Reduced Rank QR Factorization A = ˆQ ˆR   1 2 0 1 1 0   =    √ 2 2 √ 3 3 0 √ 3 3√ 2 2 − √ 3 3    √ 2 √ 2 0 √ 3 Isaac Amornortey Yowetu NIMS-GHANA Reduced and Full Rank QR Factorization
  • 8. Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples To find the full rank QR Factorization, we let q3 =   q13 q23 q33   such that < q1, q3 > = 0 (1) < q2, q3 > = 0 (2) < q3, q3 > = 1 (3) Where < q1, q3 >= 0 can be expressed as: √ 2 2 q13 + 0 · q23 + √ 2 2 q33 = 0 (4) and < q2, q3 > can also as be written as: Isaac Amornortey Yowetu NIMS-GHANA Reduced and Full Rank QR Factorization
  • 9. Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples √ 3 3 q13 + √ 3 3 q23 − √ 3 3 q33 = 0 (5) We can further reduce eqn(4) and (5) respectively as: q13 + q33 = 0 (6) q13 + q23 − q33 = 0 (7) From eqn(6), we let q33 = −t and q13 = t. Replacing q13 and q33 into eqn (7), q23 = −2t. Hence, v3 = t   1 −2 −1   and t is a dummy variable which can take any number such that t ∈ R. Here, we consider t = 1. Isaac Amornortey Yowetu NIMS-GHANA Reduced and Full Rank QR Factorization
  • 10. Introduction Reduced Vs Full Rank Factorization Reduced and Full Rank QR Factorization with Examples Full Rank QR Factorization Hence, q3 = v3 ||v3|| =    √ 6 6 − √ 6 3 − √ 6 6    A = QR   1 2 0 1 1 0   =    √ 2 2 √ 3 3 √ 6 6 0 √ 3 3 − √ 6 3√ 2 2 − √ 3 3 − √ 6 6      √ 2 √ 2 0 √ 3 0 0   Isaac Amornortey Yowetu NIMS-GHANA Reduced and Full Rank QR Factorization