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Linear Algebra, Theory And Applications, 2012a

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5.7. THE QR FACTORIZATION 131<br />

so it is symmetric. Now to show it is orthogonal,<br />

(<br />

I − 2vv<br />

T )( I − 2vv T ) = I − 2vv T − 2vv T +4vv T vv T<br />

= I − 4vv T +4vv T = I<br />

because v T v = v · v = |v| 2 =1. Therefore, this is an example of an orthogonal matrix.<br />

Consider the following problem.<br />

Problem 5.7.3 Given two vectors x, y such that |x| = |y| ≠0but x ≠ y and you want an<br />

orthogonal matrix Q such that Qx = y and Qy = x. The thing which works is the Householder<br />

matrix<br />

Q ≡ I − 2 x − y<br />

2<br />

(x − y)T<br />

|x − y|<br />

Here is why this works.<br />

Q (x − y) = (x − y) − 2 x − y<br />

|x − y| 2 (x − y)T (x − y)<br />

= (x − y) − 2 x − y<br />

|x − y| 2 |x − y|2 = y − x<br />

Hence<br />

Q (x + y) = (x + y) − 2 x − y<br />

|x − y| 2 (x − y)T (x + y)<br />

= (x + y) − 2 x − y<br />

2<br />

((x − y) · (x + y))<br />

|x − y|<br />

= (x + y) − 2 x − y (<br />

|x − y| 2 |x| 2 −|y| 2) = x + y<br />

Qx + Qy = x + y<br />

Qx − Qy = y − x<br />

Adding these equations, 2Qx =2y and subtracting them yields 2Qy =2x.<br />

A picture of the geometric significance follows.<br />

x<br />

y<br />

The orthogonal matrix Q reflectsacrossthedottedlinetakingx to y and y to x.<br />

Definition 5.7.4 Let A be an m × n matrix. Then a QR factorization of A consists of two<br />

matrices, Q orthogonal and R upper triangular (right triangular) having all the entries on<br />

the main diagonal nonnegative such that A = QR.

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