1542 the upper hessenberg form actually when using

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15.4.2 The Upper Hessenberg Form Actually, when using the QR algorithm, contrary to what I have done above, you should always deal with a matrix which is similar to the given matrix which is in upper Hessenberg form. This means all the entries below the sub diagonal equal 0. Here is an easy lemma. Lemma 15.4.5 Let A be an n × n matrix. Then it is unitarily similar to a matrix in upper Hes- senberg form and this similarity can be computed. Proof: Let A be an n × n matrix. Suppose n > 2 . There is nothing to show otherwise. A = ( a b d A 1 ) where A 1 is n 1 × n 1. Consider the n 1 × 1 matrix d . Then let Q be a Householder reflection such that Q b = ( c 0 ) c Saylor URL: The Saylor Foundation
318 NUMERICAL METHODS FOR SOLVING THE EIGENVALUE PROBLEM Then ( 1 0 0 Q ) ( a b d A 1 ) ( 1 0 0 Q ) = ( a b Q c QA 1 Q ) By similar reasoning, there exists an n 1 × n 1 matrix U = ( 1 0 0 Q 1 ) such that ( 1 0 0 Q 1 ) QA 1 Q ( 1 0 0 Q 1 ) = · · · . . . . . . 0 · · · Thus ( 1 0 0 U ) ( a b Q c QA 1 Q ) ( 1 0 0 U ) will have all zeros below the first two entries on the sub diagonal. Continuing this way shows the result. The reason you should use a matrix which is upper Hessenberg and similar to A in the QR algorithm is that the algorithm keeps returning a matrix in upper Hessenberg form and if you are looking for block upper triangular matrices, this will force the size of the blocks to be no larger than 2 × 2 which are easy to handle using the quadratic formula. This is in the following lemma. Lemma 15.4.6 Let { A k } be the sequence of iterates from the QR algorithm, A 1 exists. Then if A k is upper Hessenberg, so is A k +1 . Proof: The matrix is upper Hessenberg means that A ij = 0 whenever i j 2. A k +1 = R k Q k where A k = Q k R k . Therefore A k R 1 k = Q k and so A k +1 = R k Q k = R k A k R 1 k Let the ij th entry of A k be a k ij . Then if i j 2 a k +1 ij = n p = i j q =1 r ip a k pq r 1 qj It is given that a k pq = 0 whenever p q 2 . However, from the above sum, p q i j 2 , and so the sum equals 0. Example 15.4.7 Find the solutions to the equation x 4 4 x 3 + 8 x 2 8 x + 4 = 0 using the QR algorithm. This is the characteristic equation of the matrix 4 8 8 4 1 0 0 0 0 1 0 0 0 0 1 0 Saylor URL: The Saylor Foundation
15.4. THE QR ALGORITHM 319 Since the constant term in the equation is not 0, it follows that the matrix has an inverse. It is already in upper Hessenberg form. Lets apply the algorithm. 4 8 8 4 1 0 0 0 0 1 . 0 0 0 0 0 1 0 55 = 7 . 516 2 × 10 9 3 . 033 3 × 10 10 4 . 509 7 × 10 10 3 . 006 5 × 10 10 7 . 516 2 × 10 9 2 . 254 9 × 10 10 2 . 979 6 × 10 10 1 . 503 2 × 10 10 3 . 758 1 × 10 9 7 . 516 2 × 10 9 7 . 516 2 × 10 9 2 . 684 4 × 10 8 6 . 710 9 × 10 7 3 . 489 7 × 10 9 6 . 979 3 × 10 9 6 . 979 3 × 10 9 Then when you take the QR factorization of this, you find Q = Q = 0 . 666 65 0 . 605 55 0 . 407 45 0 . 151 21 0 . 666 65 0 . 305 4 0 . 446 60 0 . 512 69 0 . 333 33 0 . 592 53 6 . 411 2 × 10 2 0 . 730 54 5 .

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