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SDA preserves PSD blocks exactly and exposes a finite-precision symmetry defect

Example

For one exact SDA step, take $A = I_{2}$ , $G_{0} = diag (1, 0)$ , and $H_{0} = [\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}]$ .

Facts & Assumptions

Given: The matrices above and the recurrence in Finite-budget structure-preserving doubling iteration for the DARE.

[F1]

Finite-budget structure-preserving doubling iteration for the DARE: $W = (I + G H)^{- 1}$ , $G_{1} = G + A W G A^{*}$ , and $H_{1} = H + A^{*} H W A$ .

[F2]

The DARE doubling recurrence composes two Riccati steps and preserves positive semidefiniteness: when $G, H ⪰ 0$ , the exact inverse exists and the updated $G_{1}, H_{1}$ remain positive semidefinite.

Proof

1.1F1F2givenalgebra

The inputs are positive semidefinite, and $I + G_{0} H_{0} = [\begin{array}{cc} 2 & 1 \\ 0 & 1 \end{array}], W_{0} = [\begin{array}{cc} \frac{1}{2} & - \frac{1}{2} \\ 0 & 1 \end{array}] .$ Using [F1] gives $A_{1} = W_{0}, G_{1} = [\begin{array}{cc} \frac{3}{2} & 0 \\ 0 & 0 \end{array}], H_{1} = [\begin{array}{cc} \frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & \frac{3}{2} \end{array}] .$ Both updated symmetric blocks are positive semidefinite; $G_{1}$ is diagonal PSD and $H_{1} = \frac{3}{2} (1, 1)^{T} (1, 1)$ .

2.1F1F2step 1.1algebra∎

To illustrate the diagnostic, perturb only one stored off-diagonal entry by $ϵ > 0$ and report ${\tilde{H}}_{1} = [\begin{array}{cc} \frac{3}{2} & \frac{3}{2} + ϵ \\ \frac{3}{2} & \frac{3}{2} \end{array}]$ . Its symmetry defect is $∥ {\tilde{H}}_{1} - {\tilde{H}}_{1}^{*} ∥_{F} = \sqrt{2} ϵ .$ Even its symmetrization has determinant $\frac{9}{4} - (\frac{3}{2} + \frac{ϵ}{2})^{2} < 0$ , so this perturbed stored block is not positive semidefinite. It must be reported and rechecked rather than treated as covered by the exact-arithmetic PSD invariant.

Depends on

Used by

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Sources

Federico Poloni, Lecture notes on Riccati-type matrix equations (standard reference, not scraped)