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Hilbert projections are linear, self-adjoint and contractive

Statement

Assume the Axiom of Countable Choice. Let $M$ be a closed linear subspace of a real or complex Hilbert space $H$ and let $P_{M}$ be the Hilbert orthogonal projection onto $M$ . Then:

$P_{M}$ is linear and idempotent, $ran P_{M} = M$ and $\ker P_{M} = M^{⊥}$ ;
$⟨ P_{M} x, y ⟩ = ⟨ x, P_{M} y ⟩$ for all $x, y \in H$ , that is $P_{M}$ is self-adjoint;
$∥ P_{M} x ∥ \leq ∥ x ∥$ for every $x$ , so $P_{M}$ is a bounded linear operator of norm at most $1$ ; and if $M \neq {0}$ , then $∥ P_{M} ∥ = 1$ .

Facts & Assumptions

[A1]

$P_{M} x \in M$ and $x - P_{M} x \in M^{⊥}$ , and a vector in $M^{⊥}$ is orthogonal to every vector of $M$ (The Hilbert orthogonal projection onto a closed subspace, Orthogonality and the orthogonal complement).

[A2]

$M$ and $M^{⊥}$ are linear subspaces, so they are closed under sums and scalar multiples (Linear subspace of a vector space, Orthogonality and the orthogonal complement).

[A3]

For pairwise orthogonal vectors, $∥ u + v ∥^{2} = ∥ u ∥^{2} + ∥ v ∥^{2}$ (Pythagoras and finite orthogonal sums).

[A4]

A bounded linear operator has finite operator norm, and $∥ T ∥ = \sup {∥ T x ∥ : ∥ x ∥ \leq 1}$ with $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A5]

Countable Choice is the assumption under which the projection is defined (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: Countable Choice, a closed linear subspace $M$ of a Hilbert space $H$ and the projection $P_{M}$ .

1.1A1A2A5

Linearity: for scalars $a, b$ and $x, x^{'}$ , the vector $a P_{M} x + b P_{M} x^{'}$ lies in $M$ and $(a x + b x^{'}) - (a P_{M} x + b P_{M} x^{'}) = a (x - P_{M} x) + b (x^{'} - P_{M} x^{'})$ lies in $M^{⊥}$ , so by the defining property of $P_{M}$ it equals $P_{M} (a x + b x^{'})$ ; idempotence follows because $P_{M} x \in M$ has zero orthogonal component, so $P_{M} P_{M} x = P_{M} x$ .

2.1step 1.1A1A2

Range and kernel: $P_{M} x \in M$ always, $P_{M} m = m$ for $m \in M$ because $m - m = 0 \in M^{⊥}$ , and $P_{M} x = 0$ exactly when $x = x - 0 \in M^{⊥}$ ; hence $ran P_{M} = M$ and $\ker P_{M} = M^{⊥}$ .

3.1step 2.1A1A2

Self-adjointness: writing $y = P_{M} y + (y - P_{M} y)$ and using additivity in the second argument together with $y - P_{M} y \in M^{⊥}$ and $P_{M} x \in M$ gives $⟨ P_{M} x, y ⟩ = ⟨ P_{M} x, P_{M} y ⟩$ , and symmetrically $⟨ x, P_{M} y ⟩ = ⟨ P_{M} x, P_{M} y ⟩$ ; hence the two pairings agree.

4.1step 3.1A3A4algebra∎

Contractivity: $x = P_{M} x + (x - P_{M} x)$ is a sum of orthogonal vectors, so Pythagoras gives $∥ x ∥^{2} = ∥ P_{M} x ∥^{2} + ∥ x - P_{M} x ∥^{2} \geq ∥ P_{M} x ∥^{2}$ , hence $∥ P_{M} x ∥ \leq ∥ x ∥$ and $∥ P_{M} ∥ \leq 1$ by the definition of the operator norm; if $M \neq {0}$ choose $0 \neq m \in M$ , then $P_{M} m = m$ gives $∥ P_{M} ∥ \geq ∥ P_{M} m ∥ / ∥ m ∥ = 1$ , so $∥ P_{M} ∥ = 1$ .

Hilbert projections are linear, self-adjoint and contractive

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources