PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31

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.

Tikhonov regularisation scales each singular component by the filter factor $σ / (σ^{2} + λ)$

Statement

Let $F \in {R, C}$ and let $A \in M_{m \times n} (F)$ have singular value decomposition $A = U Σ V^{*}$ , and let $b \in F^{m}$ . If $x_{λ}$ is the Tikhonov minimiser for this $A$ and $b$ at a parameter $λ > 0$ , then

$x_{λ} = \sum_{σ_{i} > 0} \frac{σ_{i}}{σ_{i}^{2} + λ} ⟨ b, u_{i} ⟩ v_{i} .$

Thus the $i$ th singular component is multiplied by the filter factor $σ_{i} / (σ_{i}^{2} + λ)$ .

Facts & Assumptions

Given: A scalar field $F \in {R, C}$ , a singular value decomposition $A = U Σ V^{*}$ , a right-hand side $b \in F^{m}$ , and a parameter $λ > 0$ .

[L1]

$A$ admits a singular value decomposition with left singular vectors $u_{i}$ and right singular vectors $v_{i}$ (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

[L2]

The Tikhonov minimiser is $(A^{*} A + λ I)^{- 1} A^{*} b$ (For every $λ > 0$ , the Tikhonov objective is strictly convex and has the unique minimiser $(A^{*} A + λ I)^{- 1} A^{*} b$ ).

Proof

technique · direct

1.1L1L2algebra

Using [L2] and the SVD from [L1], $x_{λ} = V (Σ^{*} Σ + λ I)^{- 1} Σ^{*} U^{*} b .$

2.1step 1.1algebra

The diagonal matrix $(Σ^{*} Σ + λ I)^{- 1} Σ^{*}$ has diagonal entries $σ_{i} / (σ_{i}^{2} + λ)$ in the nonzero singular directions and $0$ in the zero singular directions. Therefore $x_{λ} = \sum_{σ_{i} > 0} \frac{σ_{i}}{σ_{i}^{2} + λ} ⟨ b, u_{i} ⟩ v_{i} .$

3.1step 2.1∎

This is exactly the stated filter-factor formula.

Depends on

Used by

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Stanford CS205L, Unit 12: Regularization (standard reference, not scraped)

Tikhonov regularisation scales each singular component by the filter factor σ/(σ2+λ)σ/(σ2+λ)