Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 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, VoW = V T W = WT V. Goal: Learn about orthogonal matrices and the matrix of orthogonal projections.

Previous Assignment Page 199 Problem 14 Using paper and pencil, perform the Gram-Schmidt process on the sequences of vectors given in Exercises 1 through 14. 1 0 1 7 7 8 1 2 1 7 7 6 V1 V2 V3

W1 = V1 = 1 7 1 V2oW1 0 100 1 -1 W2 = V2 - ---------- W1 = 7 - --- 7 = 0 W1oW1 2 100 1 1 7 7 0

W3 = V3oW1 V3oW2 1 100 1 0 -1 0 V3 - ----- W1 - ----- W2 = 8 - ------- 7 - ----- 0 = 1 W1oW1 W2oW2 1 100 1 2 1 0 6 7 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. (i) AT A = A AT = I (ii) A -1 = A T (iii) | A V | = | V | A preserves lengths. (iv) If A and B are orthogonal, then AB is orthogonal. (v) If A is orthogonal, then A -1 is orthogonal. Proof: Suppose [V1 V2 ... Vn ] is an orthogonal matrix.

Then |-------V1 T-------| | . . . | | 1 | |-------V2 T-------| | . . . | | 1 | | . | | V1 V2 ... Vn| = | . | | . | | . . . | | . | | . | | . . . | | . | |-------Vn T-------| | | | 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) | A V | 2 = A V o A V = (A V) T A V = V T A T A V = VT 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.

Part (v) If A is orthogonal, then A A T = I and so the columns of A T are also orthonormal.

The matrix of an orthogonal projection. Consider a subspace V or R n 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 | 1 | | 1 | ½ | 1 | ½ | -1 | | 1 | | -1 |

Solution | ½ ½ | | ½ 0 0 ½ | | ½ - ½ || ½ ½ ½ ½ | = | 0 ½ ½ 0 | | ½ - ½ || ½ - ½ - ½ ½ | | 0 ½ ½ 0 |