THE QR METHOD A = Q 1 R 1

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1 THE QR METHOD Given a square matrix A, form its QR factorization, as Then define A = Q 1 R 1 A 2 = R 1 Q 1 Continue this process: for k 1(withA 1 = A), A k = Q k R k A k+1 = R k Q k Then the sequence {A k } will usually converge to something from which the eigenvalues can be computed easily. Note first that A 2 is similar to A (and A k+1 is similar to A k ): R 1 = Q T 1 A A 2 = R 1 Q 1 = Q T 1 AQ 1 Thus A 2 is similar to A with an orthogonal similarity transformation.

2 EXAMPLE A = The true eigenvalues are λ 1 = λ 2 = λ 3 = For k =20,wehaveA 20 is approximately the matrix which gives the correct answers to within an error of A complete table of the first 20 iterates is given in the data file on the web page. For later use, λ 2 λ 1 =.4506, λ 3 λ 1 =.1253, λ 3 λ 2 =.2782

3 IMPLEMENTATION ISSUES The repeated QR factorizations can be quite expensive. For that reason it is important to first convert A to a simpler form using orthogonal similarity transformations. For A symmetric, we convert to tridiagonal matrix. For A nonsymmetric, we convert A to an upper Hessenberg matrix: U T AU = H = The QR factorization of H is usually carried out with rotation matrices R k,l rather than with Householder matrices, as using the rotation matrices is generally slightly more efficient. This leads to R = R n 1,n R 1,2 H, Q = ( R n 1,n R 1,2 ) T

4 R 1,2 is used to convert to zero the element in the (2,1) position of H; R 2,3 is used to subsequently convert to zero the element in the (2,3) position of R 1,2 H;and this is continued for each column. In doing H = QR, H 1 = RQ it is important to know that H 1 is again a Hessenberg matrix. To see this, we must examine carefully the process by which H = QR is computed. Let me consider the case of n =4. Then Then H = R = R 3,4 R 2,3 R 1,2 H 0 Q = ( R 3,4 R 2,3 R 1,2 ) T = R T 1,2 R T 2,3 RT 3,4

5 R 1,2, R 2,3, R 3,4 have the form , , Then for RQ = RR T 1,2 RT 2,3 RT 3,4, RR T 1,2 = =

6 RR T 1,2 RT 2,3 = = RR T 1,2 RT 2,3 RT 3,4 = = Thus H 1 = RQ = RR1,2 T RT 2,3 RT 3,4 is in upper Hessenberg form. By induction, all of the matrices produced in the iteration are in upper Hessenberg form.

7 CONVERGENCE If the eigenvalues of A are distinct and satisfy λ 1 > λ 2 > > λ n > 0 Then we can prove for A m = Q m R m that lim n A m = D as upper triangular matrix. If A is symmetric, then {A m } converges to a diagonal matrix D. In both cases, A m D c max 1 i n 1 A proof is sketched in the text. λ i+1 λ i For the case of eigenvalues not distinct, the convergence is more complicated.

8 EXAMPLE Recall A = λ 2 λ 1 =.4506, λ 3 λ 1 =.1253, λ 3 λ 2 =.2782 In that case, look at the ratios of the elements of A m to the corresponding components of A m 1,forthe off-diagonal elements. For the case of A 6 to A 5,we have A 6./ A 5 =

9 SYMMETRIC MATRICES As shown earlier, if A is symmetric, then we can find an orthogonal U for which U T AU = T a symmetric tridiagonal matrix. Now consider the QR factorization of T : T = Q 1 R 1, T 2 = R 1 Q 1 = Q T 1 TQ 1 This can be used to show T 2 is also symmetric. Since T is trivially Hessenberg, it follows from our earlier work that T 2 is also Hessenberg. Since T 2 is also symmetric, this implies T 2 is tridiagonal, the same as T. If the eigenvalues of T are distinct, then T m D, a diagonal matrix.

10 But consider the example [ 0 1 T = 1 0 Then T = QR implies Q = T and R = I, giving the QR factorization of T. ThenintheQRmethod, ] T m = T, m 1 and {T m } does not converge to a diagonal matrix. In general, T m D D q in which each D i is either a scalar (and an eigenvalue) or a 2 2matrix (from which the two eigenvalues can be determined easily).

11 ACCELERATION The QR method is too slow as I have defined it, especially if some of the eigenvalues are close together. In practice, the method is accelerated. I describe one such acceleration method. Write T m = Define α (m) 1 β (m) β (m) 1 α (m) 2 β (m) β (m). 2 α (m) 3 β (m) β (m) n 2 α(m) n 1 0 β (m) n 1 T m α n (m) I = Q m R m T m+1 = R m Q m + α n (m) I β(m) n 1 α n (m)

12 Then R m = Q T m T m+1 = Q T m ( ( T m α (m) n T m α (m) n ) I ) I Q m + α n (m) I = Q T mt m Q m Again T m+1 is similar to T m. For convergence, we can show that one of the new coefficients β (m) n 1 or β(m) n 2 converges to zero extremely rapidly. If it is β (m) n 1,then we will have α n (m) is converging to an eigenvalue of T. When β (m) n 1 is sufficiently small, set λ n = α n (m) ; and then we delete the last row and column of T m and continue the same process with the new smaller matrix. When β (m) n 2 is converging to zero, we end up with a 2 2matrix and we proceed in much the same manner as before.

13 EXAMPLE With acceleration, T 1 = T = λ 1 = λ 2 = λ 3 = T 2 = T 3 =

14 T 4 = In contrast, without acceleration, T 4 =

15 ERROR CONSIDERATIONS For the error resulting from dropping out an element of the matrix, let T be a tridiagonal matrix and let T the matrix obtained by deleting the element β n 1 from the (n 1,n) and (n, n 1) positions of T. Let { λ j } and { λ j } denote the associated eigenvalues. Then from the Wielandt-Hoffman theorem, n j=1 ( λj λ j ) F ( T T ) =sqrt(2) β n 1 Thus there is not much difference in the eigenvalues if β n 1 is a small number.

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