Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Polar decomposition for bounded operators

Statement

Assume AC. Every bounded operator T on a nonzero complex Hilbert space has a unique partial isometry U with T=U∣T∣ and ker⁡U=ker⁡T; its initial space is ran⁡∣T∣‾ and its final space is ran⁡T‾.

Facts & Assumptions

[A1]

∣T∣ is positive, ∣T∣2=T∗T, ∥∣T∣x∥=∥Tx∥ and ker⁡∣T∣=ker⁡T (Absolute value of a bounded operator).

[A2]

(ran⁡S)⊥=ker⁡S∗ and ran⁡S‾=(ker⁡S∗)⊥, so for S=∣T∣ the closure of the range is (ker⁡∣T∣)⊥=(ker⁡T)⊥ (Kernel–range orthogonality for Hilbert adjoints).

[A3]

For the closed subspace M the space decomposes as H=M⊕M⊥ and the orthogonal projection PM is the linear self-adjoint idempotent with range M and kernel M⊥ (Orthogonal decomposition by a closed subspace, The Hilbert orthogonal projection onto a closed subspace).

[A4]

U is a partial isometry when it vanishes on ker⁡U and is isometric on (ker⁡U)⊥, with initial space (ker⁡U)⊥ and final space ran⁡U; a bounded operator isometric on a closed subspace and zero on its orthogonal complement is a partial isometry (Isometry coisometry and partial isometry, Partial isometry characterizations).

[A5]

A bounded linear map that is isometric on a subspace extends uniquely to an isometry on its closure, since the Hilbert space is complete and the extension is obtained by limits of Cauchy images; the operator norm controls such extensions (Hilbert space, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A6]

AC is the hypothesis of the square-root and Hilbert-space suppliers (The Axiom of Choice).

Proof

technique · direct

Given: A nonzero complex Hilbert space H and a bounded operator T∈B(H), with ∣T∣=(T∗T)1/2 and M:=ran⁡∣T∣‾.

1.1A1A2A3

The subspace M is closed with M=(ker⁡∣T∣)⊥=(ker⁡T)⊥, and H=M⊕M⊥.

1.2A1algebra

The assignment V(∣T∣x):=Tx on ran⁡∣T∣ is well defined, because ∣T∣x=∣T∣y implies x−y∈ker⁡∣T∣=ker⁡T and hence Tx=Ty, and it is isometric, because ∥V(∣T∣x)∥=∥Tx∥=∥∣T∣x∥.

2.1step 1.2A5

V extends uniquely to a bounded linear isometry U on the closure M of its domain: an isometry on a dense subspace is uniformly continuous, its images of Cauchy sequences are Cauchy and converge by completeness, and the limit is independent of the sequence.

3.1step 2.1step 1.1A3A4A5

Extend U to H=M⊕M⊥ by U=0 on M⊥; then U is bounded and linear, ker⁡U=M⊥=ker⁡T, and U is isometric on M=(ker⁡U)⊥, so U is a partial isometry with initial space M and final space ran⁡U=ran⁡T‾.

3.2step 2.1step 1.2

T=U∣T∣: for every x one has ∣T∣x∈ran⁡∣T∣⊆M and U(∣T∣x)=V(∣T∣x)=Tx by construction.

4.1step 3.1step 3.2step 1.1A1

Uniqueness: if W is a partial isometry with T=W∣T∣ and ker⁡W=ker⁡T, then on the dense subspace ran⁡∣T∣ of M one has W(∣T∣x)=Tx=U(∣T∣x); both W and U vanish on M⊥=ker⁡T and both are continuous, so W and U agree on M and on M⊥, hence W=U.

5.1step 3.1step 3.2step 4.1A6∎

Therefore U is the unique partial isometry with T=U∣T∣ and ker⁡U=ker⁡T, with initial space ran⁡∣T∣‾ and final space ran⁡T‾, as asserted.

Depends on

Used by

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Sources