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Hilbert-space Moore–Penrose inverse

Definition

Assume the Axiom of Countable Choice. Let $X$ and $Y$ be real or complex Hilbert spaces over the same scalar field, and let $K : X \to Y$ be a bounded linear operator. Write $R (K) : = {K x : x \in X}$ for its range, $N (K) : = {x \in X : K x = 0}$ for its kernel, and $M^{⊥} : = {z : ⟨ z, m ⟩ = 0 for every m \in M}$ for the orthogonal complement of a subspace $M$ . The kernel $N (K)$ is closed, so the restriction $\tilde{K} : = K ∣_{N (K)^{⊥}} : N (K)^{⊥} \to R (K)$ is a linear bijection. Define the Hilbert-space Moore–Penrose inverse $K^{†}$ to have domain $D (K^{†}) : = R (K) \oplus R (K)^{⊥}$ and, for the unique decomposition $y = r + z$ with $r \in R (K)$ and $z \in R (K)^{⊥}$ , set $K^{†} y : = {\tilde{K}}^{- 1} r .$ Thus $K^{†} y$ is the unique minimum-norm solution of $K x = r$ , and the orthogonal-range component is mapped to zero. Boundedness of $K^{†}$ on this domain is not part of the definition.

Definition references

Bounded linear operators and their norm bound

Real and complex Hilbert spaces

Positive-definite inner products and induced norms

Orthogonal complements

Orthogonal decomposition by a closed subspace

Pythagoras for orthogonal vectors

Metric closure and approximation by points of a set

The Axiom of Countable Choice used for orthogonal decomposition

Finite-matrix Moore–Penrose definition (context only)

Facts & Assumptions

Given: Countable Choice, real or complex Hilbert spaces $X, Y$ , and a bounded linear operator $K : X \to Y$ .

[F1]

A bounded linear operator has a real bound $C \geq 0$ such that $∥ K x ∥ \leq C ∥ x ∥$ for every $x$ (A bounded linear operator between normed spaces).

[F2]

For a subspace $M$ , $M^{⊥}$ consists of vectors orthogonal to every element of $M$ , and $M \cap M^{⊥} = {0}$ (Orthogonality and the orthogonal complement).

[F3]

Every closed subspace $M$ of a Hilbert space splits the space uniquely as $M \oplus M^{⊥}$ (Orthogonal decomposition by a closed subspace).

[F4]

Finite orthogonal sums satisfy Pythagoras (Pythagoras and finite orthogonal sums).

[F5]

For nonempty $A$ in a metric space, $x \in \overline{A}$ exactly when every ball about $x$ meets $A$ , and $\overline{A}$ is the smallest closed superset (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

[F6]

The inner-product norm is definite: $∥ v ∥ = 0$ implies $v = 0$ (Real and complex inner-product spaces and their induced length).

[A1]

Countable Choice is the hypothesis used by [F3] (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF1F5F6choosealgebra

If $K = 0$ , then $N (K) = X$ is closed. Otherwise choose a bound $C > 0$ from [F1]. For any $x \in \overline{N (K)}$ , [F5] lets us choose, for every $ε > 0$ , an $n_{ε} \in N (K)$ with $∥ x - n_{ε} ∥ < ε$ . Hence $∥ K x ∥ = ∥ K (x - n_{ε}) ∥ \leq C ε$ for every $ε > 0$ , so $∥ K x ∥ = 0$ and $K x = 0$ by [F6]. Thus $\overline{N (K)} \subseteq N (K)$ , and [F5] implies that $N (K)$ is closed.

2.1givenstep 1.1F2F3A1choosealgebra

Fix $r_{0} \in R (K)$ and choose $x_{0} \in X$ with $K x_{0} = r_{0}$ . By [F3] and step 1.1, write $x_{0} = p + n$ with $p \in N (K)^{⊥}$ and $n \in N (K)$ ; then $K p = K x_{0} - K n = r_{0}$ . The restriction $\tilde{K}$ is therefore surjective. If $p, q \in N (K)^{⊥}$ and $K p = K q$ , then $p - q \in N (K) \cap N (K)^{⊥}$ , so $p = q$ by [F2]. Thus $\tilde{K}$ is bijective.

3.1step 2.1F2F4F6algebra

If $x$ is any solution of $K x = r_{0}$ , then $x - p \in N (K)$ . Since $p ⊥ (x - p)$ , [F4] gives $∥ x ∥^{2} = ∥ p ∥^{2} + ∥ x - p ∥^{2} \geq ∥ p ∥^{2}$ , with equality only when $x = p$ . So $p = {\tilde{K}}^{- 1} r_{0}$ is the unique minimum-norm solution.

4.1step 2.1step 3.1F2algebra∎

For $y = r + z \in R (K) \oplus R (K)^{⊥}$ , the decomposition is unique because $R (K) \cap R (K)^{⊥} = {0}$ by [F2], so the definition is single-valued. If $z \in R (K)^{⊥}$ , then $r = 0$ and the unique minimum-norm preimage of $0$ is $0$ by step 3.1; hence $K^{†} z = 0$ . For $y_{i} = r_{i} + z_{i}$ and scalars $a, b$ , the vector $a K^{†} y_{1} + b K^{†} y_{2}$ lies in $N (K)^{⊥}$ and maps under $K$ to $a r_{1} + b r_{2}$ , so uniqueness in step 2.1 gives $K^{†} (a y_{1} + b y_{2}) = a K^{†} y_{1} + b K^{†} y_{2}$ . The definition is therefore a well-defined linear operator on the stated domain.

Depends on

Used by

Noise underestimation defeats a truncated singular-value rule Counterexample
Conditional stability on a power-source class Example
Bakushinskii veto for discontinuous pseudoinverses Theorem
Closed range and boundedness of the Hilbert pseudoinverse Theorem
Conditional convergence of double-discrepancy composite inversion Theorem
Picard range and pseudoinverse-domain criterion Theorem

Dependency tree · two levels

38 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Christian Clason, Regularization of Inverse Problems: Lecture Notes, arXiv:2001.00617v2, Theorem 3.3 (standard reference, not scraped)
Otmar Scherzer and Leonidas Mindrinos, Inverse Problems: Lecture Notes, University of Vienna, 2015 (standard reference, not scraped)