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Two accepted local truncations can destroy positive definiteness after accumulation

Example

Let $e_{2} = (0, 1)^{T}$ , $M = diag (1, - 1 / 50)$ , and $D_{1} = D_{2} = (3 / 50) e_{2} e_{2}^{*}$ . The exact sum $A = M + D_{1} + D_{2} = diag (1, 1 / 10)$ is positive definite. If each rank-one update is separately recompressed to rank zero with tolerance $δ = 3 / 50$ before assembly, both local truncations are accepted and the assembled approximation is $M$ , which is indefinite. Each local error $3 / 50$ is below the smaller diagonal coefficient $1 / 10$ of $A$ . The example concerns truncation of low-rank updates before their sum is assembled; it does not claim that every SVD truncation of a completed positive-definite H-matrix destroys positivity.

Facts & Assumptions

Given: The matrices $M, D_{1}, D_{2}$ , tolerance $δ = 3 / 50$ , and rank cap $k = 0$ for the two update compressions.

[F1]

Core-SVD recompression with rank and error tolerances: the procedure accepts a rank-zero output when the discarded singular-value tail is at most $δ$ .

[F2]

Core-SVD recompression is best at the declared Frobenius tolerance: the returned rank-zero approximation has Frobenius error exactly its discarded singular-value tail.

[F3]

Finite tensor arrays and the Frobenius inner product: the Frobenius norm is computed from the squared moduli of all matrix entries.

Verification

1.1givenalgebra

The exact sum is $A = diag (1, - 1 / 50 + 3 / 50 + 3 / 50) = diag (1, 1 / 10)$ . For every nonzero $z = (z_{1}, z_{2})^{T}$ , $z^{*} A z = ∣ z_{1} ∣^{2} + (1 / 10) ∣ z_{2} ∣^{2} > 0$ , so $A$ is positive definite.

1.2givenF1F2F3algebra

Each $D_{i}$ has rank one and its only positive singular value is $3 / 50$ , so its rank-zero discarded tail and Frobenius error are both $3 / 50 = δ$ by [F3]. Thus [F1] accepts each update as zero, and [F2] gives its exact local error.

2.1givenstep 1.1step 1.2algebra∎

After both updates are replaced by zero, the sum is $M$ . Its quadratic form at $e_{2}$ is $e_{2}^{*} M e_{2} = - 1 / 50 < 0$ , so it is not positive definite. The accumulated perturbation is $D_{1} + D_{2} = (3 / 25) e_{2} e_{2}^{*}$ , whose action in this direction exceeds either local error; hence the two separate local tolerances do not preserve definiteness of the assembled sum.

Depends on

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Sources

Ralf Hiptmair, AdvNumCSE Lecture Document, Chapter 2, “Local Low-Rank Compression of Non-Local Operators” (ETH Zürich, AT’23), §2.4.5 (standard reference, not scraped)