ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Ridge regularisation damps the small-singular-value directions in a diagonal inverse problem

Example

Let

$A = diag (1, 10^{- 2}), b = [\begin{array}{c} 1 \\ 1 \end{array}], λ = 10^{- 2} .$

Then the pseudoinverse gives $(1, 100)^{T}$ , whereas the Tikhonov solution is

$x_{λ} = [\begin{array}{c} \frac{1}{1 + λ} \\ \frac{10^{- 2}}{10^{- 4} + λ} \end{array}] = [\begin{array}{c} \frac{100}{101} \\ \frac{100}{101} \end{array}] .$

Facts & Assumptions

Given: The diagonal matrix $A$ , the vector $b$ , and the parameter $λ = 10^{- 2}$ .

[L1]

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

Verification

technique · computation

1.1

The singular values are $1$ and $10^{- 2}$ , so the unregularised pseudoinverse multiplies the second component of $b$ by $10^{2}$ , producing the large coefficient $100$ .

algebra

1.2

By [L1], the Tikhonov filter factors are $\frac{1}{1 + λ} = \frac{100}{101}, \frac{10^{- 2}}{10^{- 4} + λ} = \frac{100}{101},$ so the regularised solution is the displayed vector.

L1algebra

2.1

Step 1.2 shows that the small singular direction is strongly damped: the unstable coefficient $100$ is replaced by $100 / 101$ .

step 1.1step 1.2∎

Depends on

Tikhonov regularisation scales each singular component by the filter factor $\sigma/(\sigma^2+\lambda)$

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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