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Example text

1 , . , σr } ≡ {σ1 , . , σr } viewed as sets) and initial tangent directions bˆ 1 , . , bˆ r and cˆ 1 , . , cˆ r , also closed under conjugation. 2. Vr = (σ1 E − A)−1 Bbˆ 1 · · · (σr E − A)−1 Bbˆ r Wr = (σ1 E − AT )−1 CT cˆ 1 · · · (σr E − AT )−1 CT cˆ 1 . 3. while (not converged) a. Ar = WTr AVr , Er = WTr EVr , Br = WTr B, and Cr = CVr b. Compute Y∗ Ar X = diag(λ˜ i ) and Y∗ Er X = Ir where Y∗ and X are the left and right eigenvectors of λ Er − Ar . c. σi ←− −λi (Ar , Er ) for i = 1, . , r, bˆ ∗i ←− eTi Y∗ Br and cˆ i ←− Cr Xei .

Let σ , μ ∈ C be such that s E − A and s Er − Ar are invertible for s = σ , μ . , N ( ) then H( ) (σ )b = Hr (σ )b f or (b) if (μ E − A)−T ET j−1 = 0, 1, . . , M, ( ) then cT H( ) (μ ) = cT Hr (μ )b f or = 0, 1, . . , M − 1; ( ) (c) if both (a) and (b) hold, and if σ = μ , then cT H( ) (σ )b = cT Hr (σ )b, for = 1, . . 2 solves the rational tangential interpolation problem via projection. All one has to do is to construct the matrices Vr and Wr according to the theorem, and the reduced order model is guaranteed to satisfy the interpolation conditions.

Since H and E are real, the eigenvalues of⎡Hx = λ ⎤ Ex occur in complex con0 −I 0 jugate pairs. Furthermore, if we define J = ⎣ I 0 0 ⎦ Then (JH)∗ = JH and 0 0 I (JE)∗ = −JE and it is easy to see that if xT = [vT wT zT ] is a (right) eigenvector associated with λ : Hx = λ Ex, then y∗ = [wT − v∗ z∗ ] is a left eigenvector of the pencil associated with −λ : y∗ H = −λ y∗ E. Thus, if λ is an eigenvalue of Hx = λ Ex then so is each of λ , −λ , and −λ . Thus, the spectral zeros, associated spectral zero directions, and bases for the left and right modeling subspaces can be obtained by means of the above Hamiltonian eigenvalue problem.

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