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Low-rank ADI has an exact residual factor identity

Statement

For every completed iterate $j$ of Low-rank ADI for a continuous Lyapunov equation, with $X_{0} = 0$ and $R_{0} = B$ ,

A X_{j} + X_{j} A^{*} + B B^{*} = R_{j} R_{j}^{*} .

Facts & Assumptions

Given: A completed ADI run with admissible shifts, factors $V_{j}$ , coefficients $c_{j} = - 2 Re p_{j}$ , and iterates from the algorithm.

[F1]

Low-rank ADI for a continuous Lyapunov equation: $V_{j} = (A + p_{j} I)^{- 1} R_{j - 1}$ , $X_{j} = X_{j - 1} + c_{j} V_{j} V_{j}^{*}$ , and $R_{j} = R_{j - 1} + c_{j} V_{j}$ .

[A1]

For induction at a fixed $j \geq 1$ , assume $A X_{j - 1} + X_{j - 1} A^{*} + B B^{*} = R_{j - 1} R_{j - 1}^{*}$ .

Proof

1.1F1basealgebra

At $j = 0$ , $A X_{0} + X_{0} A^{*} + B B^{*} = B B^{*} = R_{0} R_{0}^{*}$ by the initialization.

1.2A1ihF1algebra

Fix $j \geq 1$ , write $p = p_{j}$ , $c = c_{j}$ , $V = V_{j}$ , and $R = R_{j - 1}$ . The shifted solve gives $A V = R - p V$ , and taking adjoints gives $V^{*} A^{*} = R^{*} - \overline{p} V^{*}$ .

2.1A1F1step 1.2algebra

Using $c = - (p + \overline{p})$ and [F1], expand the new Lyapunov residual as $A X_{j} + X_{j} A^{*} + B B^{*} = A X_{j - 1} + X_{j - 1} A^{*} + B B^{*} + c ((A V) V^{*} + V (V^{*} A^{*}))$ . Substituting step 1.2 and [A1] yields $R R^{*} + c (R V^{*} + V R^{*}) + c^{2} V V^{*}$ .

3.1F1step 1.1step 1.2step 2.1discharge-inductionalgebra∎

The expression in step 2.1 factors as $(R + c V) (R + c V)^{*} = R_{j} R_{j}^{*}$ by [F1]. Together with base step 1.1, the induction step in 1.2–2.1 proves the identity for every completed iterate.

Depends on

Used by

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Sources

Valeria Simoncini, Computational Methods for Linear Matrix Equations, SIAM Review 58(3), 2016 (standard reference, not scraped)
Elias Jarlebring, Numerical Methods for Lyapunov Equations, lecture notes in numerical linear algebra, 2017 (standard reference, not scraped)