Preconditioning and globalizing conjugate gradients in dual space for quadratically penalized nonlinear-least squares problems

Abstract

When solving nonlinear least-squares problems, it is often useful to regularize the problem using a quadratic term, a practice which is especially common in applications arising in inverse calculations. A solution method derived from a trust-region Gauss-Newton algorithm is analyzed for such applications, where, contrary to the standard algorithm, the least-squares subproblem solved at each iteration of the method is rewritten as a quadratic minimization subject to linear equality constraints. This allows the exploitation of duality properties of the associated linearized problems. This paper considers a recent conjugate-gradient-like method which performs the quadratic minimization in the dual space and produces, in exact arithmetic, the same iterates as those produced by a standard conjugate-gradients method in the primal space. This dual algorithm is computationally interesting whenever the dimension of the dual space is significantly smaller than that of the primal space, yielding gains in terms of both memory usage and computational cost. The relation between this dual space solver and PSAS (Physical-space Statistical Analysis System), another well-known dual space technique used in data assimilation problems, is explained. The use of an effective preconditioning technique is proposed and refined convergence bounds derived, which results in a practical solution method. Finally, stopping rules adequate for a trust-region solver are proposed in the dual space, providing iterates that are equivalent to those obtained with a Steihaug-Toint truncated conjugate-gradient method in the primal space.

Appendix: Proof for the Lemma 2.4

Proof

If the singular value decomposition for H is given by

$$ H = [U_1 \mbox{\ \ } U_2 ] \left [\begin{array}{c@{\quad}c}\Sigma_r & 0 \\0 & 0 \\\end{array} \right ] \left [ \begin{array}{c}V_1^T \\ [2pt]V_2^T \\\end{array}\right ],$$

(A.1)

a possible theoretical choice for $\check{H}$ could be

$$\check{H} = \Sigma_rV_1^T.$$

Denoting $\check{R}= ( U_{1}^{T}R^{-1}U_{1} )^{-1}$, direct computations show that

$$B^{-1}+H^TR^{-1}H=B^{-1}+\check{H}^T\check{R}^{-1}\check{H}.$$

Using the assumption (2.23) and denoting $\check{G}=U_{1}^{T} G U_{1}$, we also obtain that $F\check{H}^{T} = B \check{H}^{T} \check{G}$.

The matrix $\check{H}$ is now a r×n matrix of rank r and Lemma 2.3 can then be applied using r, $\check{R}$, $\check{H}$ and $\check{G}$ instead of m, R, H and G, yielding the desired result where ν ₁,…,ν _r the eigenvalues of $\check{G}(I_{r}+\check{R}^{-1}\check{H}B\check{H}^{T})$, replace those of G(I _m +R ⁻¹ HBH ^T) in (2.48). We next investigate the relations between these two sets of eigenvalues.

Using the relation on $\check{H}$, $\check{R}$ and $\check{G}$, it can be shown that

$$ \bigl[U_1U_1^TG\widehat{A}\bigr] U_1 = U_1[\check{G}\check{A}]$$

(A.2)

where $\widehat{A}$ is defined in (2.21). This says that U ₁ is an invariant subspace of $U_{1}U_{1}^{T}G\widehat{A}$ and every eigenvalue of $\check{G}\check{A}$ is an eigenvalue of $U_{1}U_{1}^{T}G\widehat{A}$. Therefore, the nonzero eigenvalues of $U_{1}U_{1}^{T}G\widehat{A}$ are equal to the eigenvalues of $\check{G}\check{A}$ using the fact that $U_{1}U_{1}^{T}G\widehat{A}$ has (m−r) null eigenvalues. We now consider the relations between the eigenvalues of $U_{1}U_{1}^{T}G\widehat{A}$ and $G\widehat{A}$, and start by rewriting these matrices blockwise.

Using the relation (A.1), it can be shown that

$$HBH^T = U_1\Sigma_rV_1^TBV_1\Sigma_rU_1^T.$$

(A.3)

Defining

$$ U_1 = \left [ \begin{array}{c}U_r \\0 \\\end{array} \right ],$$

(A.4)

HBH ^T can thus be rewritten in a block matrix form as

(A.5)

where $M_{r} = U_{r} \Sigma_{r}V_{1}^{T}BV_{1}\Sigma_{r}U_{r}^{T}$ has full rank r. Using the equality (2.23), we can write that

$$HPH^T = HBH^TG.$$

(A.6)

Hence, HBH ^T G is symmetric due to the symmetry of HPH ^T. Using the relation (A.5) and defining

$$G = \left [ \begin{array}{c@{\quad}c}G_r & G_2 \\G_3 & G_{m-r} \\\end{array} \right ] ,$$

(A.7)

where G _r is r×r matrix and G _m−r is a m−r×m−r matrix, we can write HBH ^T G as

$$ HBH^TG = \left [ \begin{array}{c@{\quad}c}M_rG_r & M_rG_2 \\0 & 0 \\\end{array}\right ].$$

(A.8)

From the symmetry of HBH ^T G given by (A.8), M _r G ₂=0 which implies that G ₂=0 since M _r is a full rank matrix. Thus, G has the form

$$ G = \left [ \begin{array}{c@{\quad}c}G_r & 0 \\G_3 & G_{m-r} \\\end{array} \right ].$$

(A.9)

We next derive a block matrix form of $\widehat{A}$. Defining R ⁻¹ as

$$R^{-1} = \left [ \begin{array}{c@{\quad}c}R^{-1}_r & R^{-1}_2 \\R^{-1}_3 & R^{-1}_{m-r} \\\end{array} \right ],$$

(A.10)

and using (A.5) and the definition of $\widehat{A}$, we can write $\widehat{A}$ as

(A.11)

where $\widehat{A}_{r} = I + R^{-1}_{r}M_{r}$. From (A.4), (A.9) and (A.11), we deduce that

$$ G\widehat{A}= \left [ \begin{array}{c@{\quad}c}G_r \widehat{A}_r & 0 \\G_3 \widehat{A}_r & G_{m-r} \\\end{array} \right ] \quad \mbox{and} \quad U_1U_1^TG\widehat{A}= \left [ \begin{array}{c@{\quad}c}G_r \widehat{A}_r & 0 \\0 & 0 \\\end{array} \right ].$$

(A.12)

From (A.12), the eigenvalues of $G\widehat{A}$ are the eigenvalues of G _r A _r and the eigenvalues of G _m−r . Also, the eigenvalues of $U_{1}U_{1}^{T}G\widehat{A}$ are the eigenvalues of G _r A _r and (m−r) null eigenvalues. Therefore, the nonzero eigenvalues of $U_{1}U_{1}^{T}G\widehat{A}$ which are equal to the eigenvalues of $\check{G}\check{A}$ form a subset of the eigenvalues of $G\widehat{A}$. As a result, the eigenvalues of $\check{G}(I_{r}+\check{R}^{-1}\check {H}B\check{H}^{T})$ can be used in (2.47) instead of those G(I _m +R ⁻¹ HBH ^T), which completes the proof. □