Math 221: LINEAR ALGEBRA §8-4. QR Factorization Le Chen 1 Emory University, 2020 Fall (last updated on 08/27/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.
The QR Factorization Definition Let A be a real m × n matrix. Then a QR factorization of A can be written as A = QR where Q is an orthogonal matrix and R is an upper triangular matrix.
The QR Factorization Definition Let A be a real m × n matrix. Then a QR factorization of A can be written as A = QR where Q is an orthogonal matrix and R is an upper triangular matrix. Theorem Let A be a real m × n matrix with linearly independent columns. Then A can be written A = QR with Q orthogonal and R upper triangular with non-negative entries on the main diagonal.
. . The result is . . . . . . . . . . . be constructed as Let orthogonal. is . Then be constructed as Let . Defjne respectively. resulting columns . Label the Apply the Gram-Schmidt Process to the columns of Finding the QR Factorization Let A be a real m × n matrix with linearly independent columns A 1 , A 2 , · · · , A n . The following procedure results in the QR factorization.
. . The result is . . . . . . . . . . . be constructed as Let orthogonal. is . Then be constructed as Let . Defjne Finding the QR Factorization Let A be a real m × n matrix with linearly independent columns A 1 , A 2 , · · · , A n . The following procedure results in the QR factorization. 1. Apply the Gram-Schmidt Process to the columns of A . Label the resulting columns B 1 , B 2 , · · · , B n respectively.
. . The result is . . . . . . . . . . . be constructed as Let orthogonal. is . Then be constructed as Let Finding the QR Factorization Let A be a real m × n matrix with linearly independent columns A 1 , A 2 , · · · , A n . The following procedure results in the QR factorization. 1. Apply the Gram-Schmidt Process to the columns of A . Label the resulting columns B 1 , B 2 , · · · , B n respectively. 1 2. Defjne C i = � B i � B i .
. Let The result is . . . . . . . . . . . . be constructed as orthogonal. Finding the QR Factorization Let A be a real m × n matrix with linearly independent columns A 1 , A 2 , · · · , A n . The following procedure results in the QR factorization. 1. Apply the Gram-Schmidt Process to the columns of A . Label the resulting columns B 1 , B 2 , · · · , B n respectively. 1 2. Defjne C i = � B i � B i . � � 3. Let Q be constructed as Q = C 1 C 2 · · · C n . Then Q is
. orthogonal. The result is . . . . . . . . . . . . Finding the QR Factorization Let A be a real m × n matrix with linearly independent columns A 1 , A 2 , · · · , A n . The following procedure results in the QR factorization. 1. Apply the Gram-Schmidt Process to the columns of A . Label the resulting columns B 1 , B 2 , · · · , B n respectively. 1 2. Defjne C i = � B i � B i . � � 3. Let Q be constructed as Q = C 1 C 2 · · · C n . Then Q is 4. Let R be constructed as � B 1 � A 2 · C 1 A 3 · C 1 · · · A n · C 1 � B 2 � A 3 · C 2 · · · A n · C 2 0 � B 3 � · · · A n · C 1 0 0 R = · · · � B n � 0 0 0
. orthogonal. . . . . . . . . . . . Finding the QR Factorization Let A be a real m × n matrix with linearly independent columns A 1 , A 2 , · · · , A n . The following procedure results in the QR factorization. 1. Apply the Gram-Schmidt Process to the columns of A . Label the resulting columns B 1 , B 2 , · · · , B n respectively. 1 2. Defjne C i = � B i � B i . � � 3. Let Q be constructed as Q = C 1 C 2 · · · C n . Then Q is 4. Let R be constructed as � B 1 � A 2 · C 1 A 3 · C 1 · · · A n · C 1 � B 2 � A 3 · C 2 · · · A n · C 2 0 � B 3 � · · · A n · C 1 0 0 R = · · · � B n � 0 0 0 5. The result is A = QR .
Example Let 4 1 A = 2 3 0 1 Find the QR factorization of A.
Example Let 4 1 A = 2 3 0 1 Find the QR factorization of A. Solution First we apply the Gram-Schmidt Process to the columns of A. 4 B 1 A 1 = = 2 0 − 1 1 4 A 2 − A 2 · B 1 − 10 = B 2 � B 1 � 2 B 1 = 3 2 2 = 20 1 0 1
Solution (continued) Next, normalize these vectors. 4 1 1 √ C 1 = � B 1 � B 1 = 2 20 0 − 1 1 1 C 2 � B 2 � B 2 = √ = 2 6 1
Solution (continued) Next, normalize these vectors. 4 1 1 √ C 1 = � B 1 � B 1 = 2 20 0 − 1 1 1 C 2 � B 2 � B 2 = √ = 2 6 1 Then the matrix Q is constructed using these vectors as columns. 4 − 1 2 − 1 √ √ √ √ 20 6 5 6 2 2 1 2 Q = √ √ = √ √ 20 6 5 6 1 1 0 0 √ √ 6 6
Solution (continued) Now construct R: � � B 1 � � A 2 · C 1 R = � B 2 � 0 � √ √ � 20 5 √ = 0 6
Solution (continued) Now construct R: � � B 1 � � A 2 · C 1 R = � B 2 � 0 � √ √ � 20 5 √ = 0 6 Therefore A = QR 2 − 1 � √ √ √ √ 4 1 5 6 � 20 5 1 2 √ √ √ 2 3 = 5 6 0 6 1 0 1 0 √ 6
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