E for nding an m m orthogonal matrix q such that qa t

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ormations satisfy: 1. For all real values of 6= 0, H( !) = H(!): 2. H is symmetric. 3. H is orthogonal as we easily check T !!T HT H = H2 = (I ; 2 !T ! )(I ; 2 !!! ) !T or T T !!T HT H = I ; 4 !T ! + 4 !! T !!2 = I: (! ! ) 4. H(!)v = v, if and only if vT ! = 0. 5. H(!)! = ;!. 6. Let u = v + ! and vT ! = 0, then H(!)(v + !) = v ; !: Thus, the Householder transformation re ects ! + v in a plane through the origin perpendicular to !. 16 In order to perform the QR reduction (7.2.16), let a1 be the rst column of A and choose ! = a1 1e1 where e1 is the rst column of the identity matrix and q aT a1 : 1= 1 Letting H(1) = H(!), we have 2 6 H(1) A = 6 6 4 0 ... 0 3 1 ... 7 7 7= 5 0 1 (v(1) )T = A(1) : A(1) The same process is repeated on the (m ; 1) (n ; 1) portion of A. Thus, if a(1) is 1 (1) the rst column of A , choose !(1) = a(1) 1 e (1) 21 where e(1) is the rst column of the (m ; 1) (m ; 1) identity matrix. The transformation 1 (2) H is 1 0 H(2) = 0 H(!(1) ) and the result after an application of H(2) is 2 3 6 6 =6 6 6 4 0 (2) (1) 00 HA...
View Full Document

Ask a homework question - tutors are online