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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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For every λ>0, the Tikhonov objective is strictly convex and has the unique minimiser (A∗A+λI)−1A∗b

Statement

Let F∈{R,C}, let A∈Mm×n(F), let b∈Fm, and let λ∈R satisfy λ>0. Then the Tikhonov objective

Jλ(x)=∥Ax−b∥22+λ∥x∥22

is strictly convex and has the unique minimiser

xλ=(A∗A+λI)−1A∗b.

Facts & Assumptions

Given: A matrix A∈Mm×n(F), a vector b∈Fm, and a real parameter λ>0, where F∈{R,C}.

[L1]

The regularised objective is Jλ(x)=∥Ax−b∥22+λ∥x∥22 (The Tikhonov regularised least-squares objective ∥Ax−b∥22+λ∥x∥22 for λ>0).

Proof

technique · direct
1.1L1L2algebra

By [L2], write A=UΣV∗ and set x=Vy and b^:=U∗b. Since U and V are unitary, Jλ(Vy)=∥Σy−b^∥22+λ∥y∥22.

2.1step 1.1algebra

If σ1,…,σr are the nonzero singular values, then step 1.1 becomes Jλ(Vy)=∑i=1r∣σiyi−b^i∣2+∑i=r+1m∣b^i∣2+λ∑i=1n∣yi∣2. Each variable yi appears in a one-variable quadratic with positive coefficient σi2+λ or λ, so the objective is strictly convex.

3.1step 2.1algebra

Minimising coordinatewise gives yi=σiσi2+λb^i(1≤i≤r),yi=0(r<i≤n). Equivalently, y=(Σ∗Σ+λI)−1Σ∗b^.

4.1L2step 3.1algebra

Returning to x=Vy and using A∗A=VΣ∗ΣV∗, one gets xλ=V(Σ∗Σ+λI)−1Σ∗U∗b=(A∗A+λI)−1A∗b.

5.1step 2.1step 4.1∎

Step 2.1 proves strict convexity and step 4.1 gives the unique minimiser.

Depends on

Used by

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Sources