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PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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Tikhonov regularisation scales each singular component by the filter factor σ/(σ2+λ)

Statement

Let F{R,C} and let AMm×n(F) have singular value decomposition A=UΣV, and let bFm. If xλ is the Tikhonov minimiser for this A and b at a parameter λ>0, then

xλ=σi>0σiσi2+λb,uivi.

Thus the ith singular component is multiplied by the filter factor σi/(σi2+λ).

Facts & Assumptions

Given: A scalar field F{R,C}, a singular value decomposition A=UΣV, a right-hand side bFm, and a parameter λ>0.

[L1]

A admits a singular value decomposition with left singular vectors ui and right singular vectors vi (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).

Proof

technique · direct
1.1

Using [L2] and the SVD from [L1], xλ=V(ΣΣ+λI)1ΣUb.

L1L2algebra
2.1

The diagonal matrix (ΣΣ+λI)1Σ has diagonal entries σi/(σi2+λ) in the nonzero singular directions and 0 in the zero singular directions. Therefore xλ=σi>0σiσi2+λb,uivi.

step 1.1algebra
3.1

This is exactly the stated filter-factor formula.

step 2.1

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