Math 307
Spring, 2003
Hentzel
Time: 1:10-2:00 MWF
Room: 1324 Howe Hall
Instructor: Irvin Roy Hentzel
Office 432 Carver
Phone 515-294-8141
E-mail: hentzel@iastate.edu
http://www.math.iastate.edu/hentzel/class.307.ICN
Text: Linear Algebra With Applications,
Second Edition Otto Bretscher
Monday, March 24 Chapter 5.3
Page 209 Problems 2,6,8,12,34
Main Idea: Orthonormal vectors make
Orthogonal matrices.
Key Words: QR-Factorization, (A B)T = BT AT,
Transpose,
T
T
VoW = V W = W V.
Goal: Learn about orthogonal matrices and
the matrix of orthogonal projections.
Previous Assignment
Page 199 Problem 14
Using paper and pencil, perform the GramSchmidt process on the sequences of
vectors given in Exercises 1 through 14.
1
7
1
7
V1
0
7
2
7
V2
1
8
1
6
V3
W1 = V1 = 1
7
1
7
V2oW1
0 100 1
-1
W2 = V2 - ---------- W1 = 7 - --- 7 = 0
W1oW1
2 100 1
1
7
7
0
W3 =
V3oW1
V3
V3oW2
1
100 1
0 -1
0
- ----- W1 - ----- W2 = 8 - ------- 7 - ----- 0 = 1
W1oW1
W2oW2
1
6
100 1
7
2
1 0
0 -1
Basis
1
-1
0
1/10 7 1/Sqrt[2] 0 1/Sqrt[2] 1
1
1
0
7
0
-1
Page 199 Problem 28
Using paper and pencil, find the QR
factorizations of the matrices in Exercises
15 through 28.
| 1/10 -1/Sqrt[2]
0 || 10 10
10 | | 1 0 1 |
| 7/10
0 1/Sqrt[2] || 0 Sqrt[2] 0 | = | 7 7 8 |
| 1/10 1/Sqrt[2] 0 || 0 0 Sqrt[2]| | 1 2 1 |
| 7/10
0 -1/Sqrt[2] || 0 0
0 | | 7 7 6|
Page 199 Problem 40
Consider an invertible nxn matrix A whose
columns are orthogonal, but not necessarily
orthonormal. What does the QR Factorization
of A look like?
Since A is invertible, the columns of A are non
zero.
If A = [ C1 C2 .... Cn ] of lengths d1, d2, ... dn
then
A is invertible, the columns of A are nonzero.
If A = [ C1 C2 .... Cn ] of lengths d1, d2, ... dn
then
| d1
|
| d2
|
|
.
|
A = [1/d1 C1 1/d2 C2 ... 1/dn Cn ] |
.
|
|
.
|
|
dn|
Vectors V1 V2 ... Vn are orthonormal if
Vi o Vj = 0 for i =/= j
and
Vi o Vi = 1 for all i.
Then nxn matrix is called orthogonal if its
columns are orthonormal vectors.
Properties of orthogonal matrices.
T
T
(i) A A = A A = I
(ii) A -1 = A T
(iii) | A V | = | V |
A
preserves lengths.
(iv) If A and B are orthogonal, then AB is
orthogonal.
-1
(v) If A is orthogonal, then A is orthogonal.
Proof: Suppose [V1 V2 ... Vn ] is an orthogonal
matrix.
Then
|-------V1 T-------|
T
|-------V2 -------|
|
.
|
|
.
|
|
.
|
|-------Vn T-------|
|.
.
|.
.
| V1 V2
|.
.
|.
.
|
. |
. |
... Vn| =
. |
. |
|
|1
|
| 1
|
|
.
|
|
. |
|
. |
|
1|
We show Part (i).
This is just the statement that
Vi o Vj = 0 if i =/= j and
Vi o Vi = 1 for all i.
This shows that A T A = I.
But then A T is A -1 and so it works on the
other side as well giving A A T = I.
Part (ii) is a consequence of Part (i).
Part (iii)
2
|AV| = AVoAV
= (A V) T A V
T
T
=V A AV
T
=V V
=|V| 2
Part (iv) If A and B are orthogonal, then
(AB) T (AB) = B T A T A B = I. Thus the
columns of AB are also orthonormal.
T
Part (v) If A is orthogonal, then A A = I and so
T
the columns of A are also orthonormal.
The matrix of an orthogonal projection.
n
Consider a subspace V or R with orthonormal basis
V1, V2 ... Vm. The matrix of the orthogonal projection
onto V is
|
|
A A T where A = | V1
|
|
.
.
.
.
. |
. |
V2 ... Vm |
. .
. |
. .
. |
Page 209 Example 7.
Find the matrix of the orthogonal projection
onto the subspace of R 4 spanned by
½
|1|
|1|
|1|
|1|
| 1|
½ | -1 |
| -1 |
| 1|
Solution
|½ ½|
|½
| ½ - ½ || ½ ½ ½ ½ | = | 0
| ½ - ½ || ½ - ½ - ½ ½ | | 0
|½ ½|
|½
0 0 ½ |
½ ½ 0 |
½ ½ 0 |
0 0 ½ |