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 $\overline{ran ∣ T ∣}$ and its final space is $\overline{ran T}$ .

Facts & Assumptions

[A1]

$∣ T ∣$ is positive, $∣ T ∣^{2} = T^{*} T$ , $∥ ∣ T ∣ x ∥ = ∥ T x ∥$ and $\ker ∣ T ∣ = \ker T$ (Absolute value of a bounded operator).

[A2]

$(ran S)^{⊥} = \ker S^{*}$ and $\overline{ran S} = (\ker S^{*})^{⊥}$ , 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 $H = M \oplus M^{⊥}$ and the orthogonal projection $P_{M}$ 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 $T \in B (H)$ , with $∣ T ∣ = (T^{*} T)^{1 / 2}$ and $M : = \overline{ran ∣ T ∣}$ .

1.1

The subspace $M$ is closed with $M = (\ker ∣ T ∣)^{⊥} = (\ker T)^{⊥}$ , and $H = M \oplus M^{⊥}$ .

A1A2A3

1.2

The assignment $V (∣ T ∣ x) : = T x$ on $ran ∣ T ∣$ is well defined, because $∣ T ∣ x = ∣ T ∣ y$ implies $x - y \in \ker ∣ T ∣ = \ker T$ and hence $T x = T y$ , and it is isometric, because $∥ V (∣ T ∣ x) ∥ = ∥ T x ∥ = ∥ ∣ T ∣ x ∥$ .

A1algebra

2.1

$V$ 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.

step 1.2A5

3.1

Extend $U$ to $H = M \oplus M^{⊥}$ 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 $ran U = \overline{ran T}$ .

step 2.1step 1.1A3A4A5

3.2

$T = U ∣ T ∣$ : for every $x$ one has $∣ T ∣ x \in ran ∣ T ∣ \subseteq M$ and $U (∣ T ∣ x) = V (∣ T ∣ x) = T x$ by construction.

step 2.1step 1.2

4.1

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 $W (∣ T ∣ x) = T x = U (∣ T ∣ x)$ ; both $W$ and $U$ vanish on $M^{⊥} = \ker T$ and both are continuous, so $W$ and $U$ agree on $M$ and on $M^{⊥}$ , hence $W = U$ .

step 3.1step 3.2step 1.1A1

5.1

Therefore $U$ is the unique partial isometry with $T = U ∣ T ∣$ and $\ker U = \ker T$ , with initial space $\overline{ran ∣ T ∣}$ and final space $\overline{ran T}$ , as asserted.

step 3.1step 3.2step 4.1A6∎

Depends on

Used by

Polar decomposition of the unilateral shift Example

Dependency tree · two levels

31 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

John B. Conway, A Course in Functional Analysis, 2nd ed., Polar Decomposition 3.11, printed pp.239–243 (standard reference, not scraped)
Theo Bühler and Dietmar Salamon, Functional Analysis, §5.3, printed pp.235–245 (standard reference, not scraped)