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The Index of Invariance and its Implications for a Parameterized Least Squares Problem (2008.11154v3)

Published 25 Aug 2020 in math.NA, cs.NA, and math.OC

Abstract: We study the problem $x_{b,\omega} := \text{arg min}{x \in \mathcal{S}} |(A + \omega I){-1/2} (b - Ax)|_2$, with $A = A*$, for a subspace $\mathcal{S}$ of $\mathbb{F}n$ ($\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$), and $\omega > -\lambda{min}(A)$. We show that there exists a subspace $\mathcal{Y}$ of $\mathbb{F}n$, independent of $b$, such that ${x_{b,\omega} - x_{b,\mu} \mid \omega,\mu > -\lambda_{min}(A)} \subseteq \mathcal{Y}$, where $\dim(\mathcal{Y}) \leq \dim(\mathcal{S} + A\mathcal{S}) - \dim(\mathcal{S}) = \mathbf{Ind}A(\mathcal{S})$, a quantity which we call the index of invariance of $\mathcal{S}$ with respect to $A$. In particular if $\mathcal{S}$ is a Krylov subspace, this implies the low dimensionality result of HaLLMan & Gu (2018). The problem is also such that when $A$ is positive and $\mathcal{S}$ is a Krylov subspace, it reduces to CG for $\omega = 0$ and to MINRES for $\omega \to \infty$. We study several properties of $\mathbf{Ind}_A(\mathcal{S})$ in relation to $A$ and $\mathcal{S}$. We show that the dimension of the affine subspace $\mathcal{X}_b$ containing the solutions $x{b,\omega}$ can be smaller than $\mathbf{Ind}A(\mathcal{S})$ for all $b$. However, we also exhibit some sufficient conditions on $A$ and $\mathcal{S}$, under which $\mathcal{X} := \text{Span}{{x{b,\omega} - x_{b,\mu} \mid b \in \mathbb{F}n, \omega,\mu > -\lambda_{min}(A)}}$ has dimension equal to $\mathbf{Ind}A(\mathcal{S})$. We then study the injectivity of the map $\omega \mapsto x{b,\omega}$, leading us to a proof of the convexity result from HaLLMan & Gu (2018). We finish by showing that sets such as $M(\mathcal{S},\mathcal{S}') = {A \in \mathbb{F}{n \times n} \mid \mathcal{S} + A\mathcal{S} = \mathcal{S}'}$, for nested subspaces $\mathcal{S} \subseteq \mathcal{S}' \subseteq \mathbb{F}n$, form smooth real manifolds, and explore some topological relationships between them.

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