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Rank-deficient data do not identify both parameters, while LM has a unique damped step

Example

Let $r (a, b) = (a + b - 1, 2 (a + b - 1))^{T}$ . The data depend only on $a + b$ , so they do not identify $a$ and $b$ separately. At $(0, 0)$ the Gauss-Newton Jacobian is rank deficient and its linearized least-squares problem has many minimizers. With $D = I$ , damping $λ = 1$ gives a unique trial step. For trust radius $Δ = 1$ , ratio thresholds $η_{1} = 1 / 4, η_{2} = 3 / 4$ , lower damping $λ_{\min} = 1 / 2$ , update factor $γ = 2$ , and one allowed trial, that step is accepted with ratio $1$ .

Facts & Assumptions

Given: The residual model, starting point $(0, 0)$ , and LM parameters in the Example.

[F1]

Gauss-Newton linearized least-squares step with QR/SVD rank safeguards: with a supplied exact reduced QR factorization, a zero triangular diagonal returns rank-deficient rather than an asserted unique Gauss-Newton step.

[F2]

Finite-budget Levenberg-Marquardt with a trust-ratio safeguard: with a supplied exact reduced QR factorization of the augmented matrix, a feasible LM trial with positive predicted reduction and $ρ \geq η_{1}$ is accepted; a high-ratio boundary step is the only accepted branch that decreases damping.

[F3]

A damped least-squares step solves a scaled trust-region model: with nonsingular $D$ and $λ > 0$ , the damped normal system has a unique step and its predicted reduction is positive when $J^{T} r \neq 0$ .

[F4]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : $∥ (u, v) ∥_{2} = \sqrt{u^{2} + v^{2}}$ in two dimensions.

Verification

1.1givenF1algebra

At $(0, 0)$ , $r = (- 1, - 2)^{T}$ , $J = [\begin{array}{cc} 1 & 1 \\ 2 & 2 \end{array}]$ , and $J^{T} J = [\begin{array}{cc} 5 & 5 \\ 5 & 5 \end{array}]$ . The two columns of $J$ coincide, so its rank is $1 < 2$ . Supply the exact reduced QR $Q = \frac{1}{\sqrt{5}} [\begin{array}{cc} 1 & - 2 \\ 2 & 1 \end{array}]$ and $R = [\begin{array}{cc} \sqrt{5} & \sqrt{5} \\ 0 & 0 \end{array}]$ , which satisfies $Q^{T} Q = I$ and $Q R = J$ . The exact QR branch therefore returns rank-deficient. The objective is $\frac{5}{2} (a + b - 1)^{2}$ , whose minimizers are the full line $a + b = 1$ ; hence the parameters are not separately identifiable. The linearized problem at $(0, 0)$ is minimized whenever $p_{1} + p_{2} = 1$ , so it has no unique Gauss-Newton step.

2.1givenF2F3F4algebra∎

For $D = I$ and $λ = 1$ , the augmented matrix and residual are $A_{1} = [J; I]$ and $s_{1} = [r; 0]$ . Supply the exact reduced QR $Q = [\begin{array}{cc} 1 / \sqrt{6} & 1 / \sqrt{66} \\ 2 / \sqrt{6} & 2 / \sqrt{66} \\ 1 / \sqrt{6} & - 5 / \sqrt{66} \\ 0 & 6 / \sqrt{66} \end{array}]$ and $R = [\begin{array}{cc} \sqrt{6} & 5 / \sqrt{6} \\ 0 & \sqrt{66} / 6 \end{array}]$ , which satisfies $Q^{T} Q = I$ and $A_{1} = Q R$ . The damped system is $[\begin{array}{cc} 6 & 5 \\ 5 & 6 \end{array}] p = (5, 5)^{T}$ , whose unique solution is $p = (5 / 11, 5 / 11)^{T}$ . The linear model residual is $r + J p = (- 1 / 11, - 2 / 11)^{T}$ , so $m_{0} (0) = 5 / 2$ , $m_{0} (p) = 5 / 242$ , and $pred = 300 / 121 > 0$ . Since the residual is affine, the true residual at $(0, 0) + p$ equals that linear model residual; therefore $ared = pred$ and $ρ = 1$ . Also $∥ p ∥_{2} = 5 \sqrt{2} / 11 < 1 = Δ$ , so the trial is feasible and is not a boundary step. By [F2]–[F3] it is accepted and $λ$ stays $1$ .

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Sources

Jorge Nocedal and Stephen J. Wright, Numerical Optimization, second edition (standard reference, not scraped)
David Bindel, CS 4220: Numerical Analysis, Gauss-Newton and nonlinear least squares (standard reference, not scraped)