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Partial isometry characterizations

Statement

Assume Countable Choice. For a bounded operator $U$ on a nonzero complex Hilbert space, the partial-isometry condition is equivalent to $U^{*} U$ being the orthogonal projection onto $(\ker U)^{⊥}$ , and then $U U^{*}$ is the orthogonal projection onto $ran U$ ; equivalently $U^{*}$ is a partial isometry.

Facts & Assumptions

[A1]

$U$ is a partial isometry when it vanishes on $\ker U$ and is isometric on the initial space $(\ker U)^{⊥}$ ; an isometry is exactly an operator with $U^{*} U = I$ (Isometry coisometry and partial isometry).

[A2]

$⟨ U^{*} x, y ⟩ = ⟨ x, U y ⟩$ , $U^{* *} = U$ , and $U^{*} U$ is self-adjoint for every bounded $U$ (The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[A3]

The kernel of a bounded operator is closed: if $U x \neq 0$ and $C$ bounds $U$ , the ball about $x$ of radius $∥ U x ∥ / (2 (C + 1))$ misses its kernel. Orthogonal complements are closed linear subspaces (Orthogonal complements are closed), so $H = \ker U \oplus (\ker U)^{⊥}$ (Orthogonal decomposition by a closed subspace). The Hilbert orthogonal projection $P_{M}$ onto a closed subspace $M$ is the linear self-adjoint idempotent with range $M$ and kernel $M^{⊥}$ (The Hilbert orthogonal projection onto a closed subspace, Hilbert projections are linear, self-adjoint and contractive). Conversely, a bounded self-adjoint idempotent $Q$ has closed range $\ker (I - Q)$ (the same kernel argument applies), and $x - Q x$ is perpendicular to its range since $⟨ x - Q x, Q y ⟩ = ⟨ Q (x - Q x), y ⟩ = 0$ . Thus the defining decomposition shows $Q = P_{ran Q}$ .

[A4]

$(ran U)^{⊥} = \ker U^{*}$ and $\overline{ran U} = (\ker U^{*})^{⊥}$ (Kernel–range orthogonality for Hilbert adjoints).

[A5]

Countable Choice is the hypothesis of the adjoint, projection and decomposition suppliers (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[A6]

For a bounded operator $S$ and $C \geq 0$ , $∥ S ∥ \leq C$ is equivalent to $∥ S x ∥ \leq C ∥ x ∥$ for all $x$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum). The pairing is linear in its first argument and conjugate-linear in its second (Real and complex inner product spaces, with the inner product linear in the first argument). For any such sesquilinear form $B$ , direct expansion gives $4 B (x, y) = B (x + y, x + y) - B (x - y, x - y) + i B (x + i y, x + i y) - i B (x - i y, x - i y)$ ; hence a form with zero diagonal is zero.

Proof

technique · direct

Given: A nonzero complex Hilbert space $H$ and a bounded operator $U \in B (H)$ , with $M : = (\ker U)^{⊥}$ .

1.1

If $U$ is a partial isometry, then for $x = m + n$ with $m \in M$ , $n \in \ker U$ one has $U x = U m$ and $∥ U x ∥ = ∥ U m ∥ = ∥ m ∥ = ∥ P_{M} x ∥$ .

A1A3A5algebra

1.2

If $U^{*} U = P_{M}$ , then $U$ vanishes on $\ker U$ and is isometric on $M$ : for $x \in \ker U$ one has $∥ U x ∥^{2} = ⟨ P_{M} x, x ⟩ = 0$ , and for $x \in M$ one has $∥ U x ∥^{2} = ⟨ P_{M} x, x ⟩ = ∥ x ∥^{2}$ .

A2A3algebra

2.1

If $U$ is a partial isometry, put $D = U^{*} U - P_{M}$ . The adjoint and projection identities and step 1.1 give $⟨ D x, x ⟩ = ∥ U x ∥^{2} - ∥ P_{M} x ∥^{2} = 0$ for every $x$ . Applying the expansion in [A6] to $B (x, y) = ⟨ D x, y ⟩$ gives $⟨ D x, y ⟩ = 0$ for all $x, y$ ; taking $y = D x$ gives $D x = 0$ . Hence $U^{*} U = P_{M}$ .

step 1.1A2A3A6

2.2

Conversely, if $U^{*} U = P_{M}$ then $U$ is a partial isometry, since it vanishes on $\ker U$ and is isometric on the initial space $M$ .

step 1.2

3.1

If $U$ is a partial isometry, then $U = U P_{M} = U U^{*} U$ : the first identity follows since $x - P_{M} x \in \ker U$ , and the second uses step 2.1. Let $Q = U U^{*}$ . It is bounded and self-adjoint by [A2], and $Q^{2} = (U U^{*} U) U^{*} = U U^{*} = Q$ . Its range is contained in $ran U$ , while $U = Q U$ gives the reverse inclusion. By [A3], $ran U = ran Q$ is closed and $U U^{*} = P_{ran U}$ .

step 1.1step 2.1A2A3

4.1

If $U$ is a partial isometry, then $U^{*}$ is a partial isometry: [A4] and step 3.1 give $(\ker U^{*})^{⊥} = ran U$ . On this space, write $y = U x$ ; then $∥ U^{*} y ∥ = ∥ U^{*} U x ∥ = ∥ P_{M} x ∥ = ∥ U x ∥ = ∥ y ∥$ . On its kernel $U^{*}$ vanishes by definition.

step 1.1step 2.1step 3.1A1A4

5.1

Conversely, if $U^{*}$ is a partial isometry, apply step 4.1 to the bounded operator $U^{*}$ ; it shows $U^{* *} = U$ is a partial isometry.

step 4.1A2

6.1

Therefore $U$ is a partial isometry exactly when $U^{*} U = P_{(\ker U)^{⊥}}$ , and exactly when $U^{*}$ is a partial isometry; whenever these conditions hold, $ran U$ is closed and $U U^{*} = P_{ran U}$ .

step 2.1step 2.2step 3.1step 4.1step 5.1∎

Depends on

Used by

Polar decomposition for bounded operators Theorem

Dependency tree · two levels

39 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., Chapter IX §3, 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)