LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Woodbury inverse for a diagonal-plus-low-rank update

Statement

Let $D \in C^{N \times N}$ be invertible, let $U, V \in C^{N \times r}$ , and set $W = I_{r} + V^{*} D^{- 1} U$ , where $^{*}$ denotes conjugate transpose. Then $D + U V^{*}$ is invertible if and only if $W$ is invertible. When these matrices are invertible,

(D + U V^{*})^{- 1} = D^{- 1} - D^{- 1} U W^{- 1} V^{*} D^{- 1} .

Statement references

Gu, Goel, and Ré, “Efficiently Modeling Long Sequences with Structured State Spaces,” Appendix C.2–C.3

Facts & Assumptions

Given: Finite integers $N, r \geq 1$ , matrices $D \in C^{N \times N}$ and $U, V \in C^{N \times r}$ , with $D$ invertible, and $W : = I_{r} + V^{*} D^{- 1} U$ .

[A1]

The inverse $D^{- 1}$ exists, so all displayed matrix products are defined.

[A2]

$W$ is an $r \times r$ matrix; a square linear map in finite dimension is invertible once its kernel is zero.

Proof

1.1A1algebra

Assume $W$ is invertible and define $X : = D^{- 1} - D^{- 1} U W^{- 1} V^{*} D^{- 1}$ . Expansion gives $(D + U V^{*}) X = I + U (V^{*} D^{- 1} - W^{- 1} V^{*} D^{- 1} - V^{*} D^{- 1} U W^{- 1} V^{*} D^{- 1}) = I$ , because $(I_{r} + V^{*} D^{- 1} U) W^{- 1} V^{*} D^{- 1} = W W^{- 1} V^{*} D^{- 1} = V^{*} D^{- 1}$ .

2.1A1step 1.1algebra

Expanding in the other order gives $X (D + U V^{*}) = I + D^{- 1} U (V^{*} - W^{- 1} V^{*} - W^{- 1} V^{*} D^{- 1} U V^{*}) = I$ , because $W^{- 1} (I_{r} + V^{*} D^{- 1} U) V^{*} = W^{- 1} W V^{*} = V^{*}$ . Thus $X$ is the two-sided inverse of $D + U V^{*}$ , proving this direction and the formula.

3.1A1A2step 1.1algebra∎

Conversely assume $D + U V^{*}$ is invertible and let $w \in C^{r}$ satisfy $W w = 0$ . Set $x = D^{- 1} U w$ . Then $(D + U V^{*}) x = U (I_{r} + V^{*} D^{- 1} U) w = U W w = 0$ , so invertibility gives $x = 0$ . Hence $U w = D x = 0$ , and therefore $W w = w + V^{*} D^{- 1} U w = w$ . Since $W w = 0$ , we have $w = 0$ . Thus $W$ has zero kernel; by [A2] it is invertible.

Used by

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Gu, Goel, and Ré, Efficiently Modeling Long Sequences with Structured State Spaces, Appendix C.2–C.3 (standard reference, not scraped)