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

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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...
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