Low Rank ADI Solution of Sylvester Equation via Exact Shifts

Neglecting the second order terms and subtracting the unperturbed Sylvesterequation from the perturbed one yieldA δX − δX B ≈ G δF ∗ + δG F ∗ − δA X + X δB. (3.13)Note that we can approximate the solution δX of (3.13) by δX ≈ δX 1 +δX 2 +δX 3 , whereA δX 1 − δX 1 B = −δA X, (3.14)A δX 2 − δX 2 B = −Xδ B, (3.15)A δX 3 − δX 3 B = (G δF ∗ + δG F ∗ ). (3.16)For (3.14), we again choose parameterseither α i = λ i for i = 1, . . . , m, or β j = µ j for j = 1, . . . , n,to get⎛∑n 0δX 1 = S ⎝ (µ j − λ j )ˆΦj=1(j) ˆΨ(j)⎞⎠ T −1 , n 0 = min{m, n}, (3.17)whereˆΦ (j) = diagandˆΨ (j) = XT diagDefineand( )σ(1, j − 1)j−1σ(m, j − 1), . . . , S −1 ∏ λ i − λ sδA, σ(i, j − 1) =λ 1 − µ j λ m − µ j s=1λ i − µ s( )τ(1, j − 1)j−1τ(n, j − 1)∏ µ i − µ s, . . . , , τ(i, j − 1) = .µ 1 − λ j µ n − λ j s=1µ i − λ s( )σ(1, j − 1) σ(m, j − 1)∆ Φ (j) = diag, . . . , , (3.18)λ 1 − µ j λ m − µ j( )τ(1, j − 1) τ(n, j − 1)∆ Ψ (j) = diag, . . . , , (3.19)µ 1 − λ j µ n − λ jσ (j)max = ‖∆ Φ (j)‖ = max|σ(i, j − 1)|i |λ i − µ j |, τ (j)max = ‖∆ Ψ (j)‖ = max|τ(i, j − 1)|.i |µ i − λ j |(3.20)11