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As $λ ↓ 0$ , the Tikhonov minimisers converge to the Moore--Penrose solution $A^{+} b$

Statement

Let $x_{λ}$ be the Tikhonov minimiser for $A x = b$ . Then

$\lim_{λ ↓ 0} x_{λ} = A^{+} b .$

Facts & Assumptions

Given: A matrix $A$ , a vector $b$ , and the Tikhonov minimisers $x_{λ}$ .

[L1]

In singular-value coordinates,

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

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

[L2]

$A^{+} b$ is the minimum-norm least-squares solution (For every right-hand side $b$ , $A^{+} b$ is the unique least-squares solution of minimum Euclidean norm).

[L3]

The Moore--Penrose pseudoinverse exists (Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse).

Proof

technique · direct

1.1L1L3algebra

In the same SVD coordinates as [L1], define $B : = V Σ^{+} U^{*}$ by reciprocating each positive singular value and leaving the zero block fixed. Direct diagonal multiplication verifies all four Penrose equations, so uniqueness in [L3] gives $B = A^{+}$ . Consequently $A^{+} b = \sum_{σ_{i} > 0} \frac{1}{σ_{i}} ⟨ b, u_{i} ⟩ v_{i} .$

2.1L1step 1.1algebra

Step 1.1 and [L1] show that the coefficient of $v_{i}$ in $x_{λ} - A^{+} b$ is $(\frac{σ_{i}}{σ_{i}^{2} + λ} - \frac{1}{σ_{i}}) ⟨ b, u_{i} ⟩ = - \frac{λ}{σ_{i} (σ_{i}^{2} + λ)} ⟨ b, u_{i} ⟩$ for each nonzero singular value, while the zero-singular-value coefficients are $0$ in both vectors.

3.1step 2.1algebra

For each fixed nonzero $σ_{i}$ , $σ_{i} / (σ_{i}^{2} + λ) \to 1 / σ_{i}$ as $λ ↓ 0$ , so every coefficient from step 2.1 tends to $0$ . Because there are only finitely many singular directions, $∥ x_{λ} - A^{+} b ∥_{2} \to 0$ .

4.1L2step 3.1∎

Therefore $x_{λ} \to A^{+} b$ , the Moore--Penrose minimum-norm least-squares solution from [L2].

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Sources

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

As λ↓0λ↓0, the Tikhonov minimisers converge to the Moore--Penrose solution A+bA+b