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                                                                     10.2. The QR Method
                              entry of index (l, m) is thus multiplied at each step by a unit number z lm ,
                              which is not necessarily equal to one if l< m. Hence, the part above the
                              diagonal of A k does not converge.
                                Summing up, a convergence theorem may concern only the diagonal of
                              A k and what is below it.
                              Lemma 10.2.1 Let A ∈ GL n (K) be given, with K = IR or CC.Let
                              A k = Q k R k be the sequence of matrices given by the QR algorithm. Let
                              us define P k = Q 0 ··· Q k−1 and U k = R k−1 ··· R 0 .Then P k U k is the QR
                              factorization of the kth power of A:
                                                          k
                                                         A = P k U k .
                                Proof
                                From (10.2), we have A k = P −1 AP k ;that is, P k A k = AP k .Then
                                                         k
                                          P k+1 U k+1 = P k Q k R k U k = P k A k U k = AP k U k .
                                                   k
                              By induction, P k U k = A . However, P k ∈ U n and U k is triangular, with a
                              positive real diagonal, as a product of such matrices.
                              Theorem 10.2.1 Let A ∈ GL n (CC) be given. Assume that the moduli of
                              the eigenvalues of A are distinct:
                                                |λ 1 | > |λ 2 | > ··· > |λ n |  (> 0).
                              In particular, the eigenvalues are simple, and thus A is diagonalizable:
                                                  A = Y  −1  diag(λ 1 ,... ,λ n )Y.
                              Assume also that Y admits an LU factorization. Then the strictly lower
                              triangular part of A k converges to zero, and the diagonal of A k converges
                              to D := diag(λ 1 ,... ,λ n ).

                                Proof
                                Let Y = LU be the factorization of Y . We also make use of the QR
                                                                            k
                                                                   k
                              factorization of Y  −1 : Y  −1  = QR.Since A = Y  −1 D Y ,wehave P k U k =
                                             k
                                   k
                              Y  −1 D Y = QRD LU.
                                            k
                                The matrix D LD −k  is lower triangular with unit numbers on its di-
                              agonal. By assumption, its strictly lower part tends to zero (because
                                                              k
                              each term is multiplied by (λ i /λ j ) ,where |λ i /λ j | < 1). Therefore,
                               k
                              D LD  −k  = I n + E k with E k → 0 n as k → +∞. Hence, P k U k =
                                                                   k
                                                                                      k
                                           k
                              QR(I n + E k )D U = Q(I n + RE k R −1 )RD U = Q(I n + F k )RD U, where
                              F k → 0 n .Let O k T k = I n + F k be the QR factorization of I n + F k .By
                              continuity, O k and T k both tend to I n .Then
                                                                      k
                                                   P k U k =(QO k )(T k RD U).
                              The first product is a unitary matrix, while the second is a triangular
                              one. Let |D| be the “modulus” matrix of D (whose entries are the moduli
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