Characterization of Partial Isometries

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Theorem

Let $\struct {\HH_1, \innerprod \cdot \cdot_1}$ and $\struct {\HH_2, \innerprod \cdot \cdot_2}$ be Hilbert spaces.

Let $T : \HH_1 \to \HH_2$ be a bounded linear transformation.


The following statements are equivalent:

$(1) \quad$ $T$ is a partial isometry
$(2) \quad$ $T = T T^\ast T$
$(3) \quad$ $T^\ast T$ is a Hilbert space projection
$(4) \quad$ $T T^\ast$ is a Hilbert space projection.


Proof

$(2)$ implies $(3)$

Suppose that $T = T T^\ast T$.

We then have $T^\ast T = T^\ast \paren {T T^\ast T} = \paren {T^\ast T}^2$.

From Product of Element in *-Star Algebra with its Star is Hermitian, $T^\ast T$ is Hermitian.

Hence from Characterization of Projections, $T^\ast T$ is a Hilbert space projection.

$\Box$


$(3)$ implies $(2)$

Suppose that:

$T^\ast T$ is a Hilbert space projection

We then have, for $x \in \HH_1$:

\(\ds \norm {T x}^2_2\) \(=\) \(\ds \innerprod {T x} {T x}_2\) Definition of Inner Product Norm
\(\ds \) \(=\) \(\ds \innerprod {T^\ast T x} x_1\) Definition of Adjoint Linear Transformation
\(\ds \) \(=\) \(\ds \innerprod {\paren {T^\ast T}^2 x} x_1\) $T^\ast T$ is idempotent
\(\ds \) \(=\) \(\ds \innerprod {T^\ast T x} {T^\ast T x}_1\) Definition of Adjoint Linear Transformation, Definition of Hermitian Operator
\(\ds \) \(=\) \(\ds \norm {T^\ast T x}^2\) Definition of Inner Product Norm

We then have, for $x \in \HH_1$:

$\innerprod {T x} {T T^\ast T x} = \innerprod {T^\ast T x} {T^\ast T x} = \norm {T^\ast T x}^2$

from the definition of the adjoint and inner product norm.

From Square of Inner Product Norm of Sum, we have:

$\norm {T \paren {I - T^\ast T} x}^2 = \norm {T x}^2 - 2 \map \Re {\innerprod {T x} {T T^\ast T x} } + \norm {T^\ast T x}^2$

We have:

$\innerprod {T x} {T T^\ast T x} = \norm {T^\ast T x}^2 = \norm {T x}^2$

So we have:

$2 \map \Re {\innerprod {T x} {T T^\ast T x} } + \norm {T^\ast T x}^2 = -2 \norm {T x}^2 + \norm {T x}^2 = -\norm {T x}^2$

and hence:

$\norm {T \paren {I - T^\ast T} x}^2 = 0$

Hence from Norm Axiom $\text N 1$: Positive Definiteness, we have:

$T \paren {I - T^\ast T} x = {\mathbf 0}_\HH$ for each $x \in \HH_1$.

We conclude that:

$T = T T^\ast T$

$\Box$


$(3)$ implies $(4)$

Suppose that:

$T^\ast T$ is a Hilbert space projection

We then have:

\(\ds \paren {T T^\ast}^3\) \(=\) \(\ds \paren {T T^\ast} \paren {T T^\ast} \paren {T T^\ast}\)
\(\ds \) \(=\) \(\ds T \paren {T^\ast T} \paren {T^\ast T} T^\ast\)
\(\ds \) \(=\) \(\ds T \paren {T^\ast T}^2 T^\ast\)
\(\ds \) \(=\) \(\ds T \paren {T^\ast T} T^\ast\) $T^\ast T$ is idempotent
\(\ds \) \(=\) \(\ds \paren {T T^\ast}^2\)

Hence:

$\paren {T T^\ast}^3 - \paren {T T^\ast}^2 = {\mathbf 0}_{\map B {\HH_2} }$

From the Spectral Mapping Theorem for Polynomials and Spectrum of Zero Vector in Algebra, we have that:

$\set {z^3 - z^2 : z \in \map \sigma {T T^\ast} } = \set 0$

So if $z \in \map \sigma {T T^\ast}$ then $z^3 - z^2 = 0$.

That is, $z = 0$ or $z = 1$.

Then $\map \sigma {T T^\ast} \subseteq \set {0, 1}$.

From Product of Element in *-Star Algebra with its Star is Hermitian, $T T^\ast$ is Hermitian.

Hence $T T^\ast$ is in particular normal.

So from Normal Element of C*-Algebra is Projection iff Spectrum contains only Zero and One, $T T^\ast$ is idempotent.

From Characterization of Projections, $T T^\ast$ is then a Hilbert space projection.

$\Box$


$(4)$ implies $(3)$

To reiterate, the fact that $(3) \implies (4)$ shows:

if $T^\ast T$ is a Hilbert space projection then $T T^\ast$ is a Hilbert space projection.

From Adjoint is Involutive, we have:

$T^{\ast \ast} = T$

Hence swapping $T$ for $T^\ast$ in the above we have that:

if $T T^\ast$ is a Hilbert space projection then $T^\ast T$ is a Hilbert space projection.

Hence we have that $(4)$ implies $(3)$.

$\Box$


$(2)$ implies $(1)$

Suppose that $T = T T^\ast T$.

We show that in this case, $T^\ast T$ is the orthogonal projection onto $\paren {\map \ker T}^\perp$.

From Product of Element in *-Star Algebra with its Star is Hermitian, $T^\ast T$ is Hermitian.

Further from:

$T = T T^\ast T$

we have that:

$T^\ast T = \paren {T^\ast T}^2$

So $T^\ast T$ is Hermitian and idempotent.

So from Characterization of Projections, we have that $T^\ast T$ is a Hilbert space projection.

Hence from Characterization of Projections, $\Img {T^\ast T}$ is closed.

We just need to show that $\Img {T^\ast T} = \paren {\map \ker T}^\perp$.

From Kernel of Linear Transformation is Orthocomplement of Image of Adjoint, we have:

$\map \ker T = \paren {\Img {T^\ast} }^\perp$

Hence:

$\paren {\map \ker T}^\perp = \paren {\Img {T^\ast} }^{\perp \perp}$

From Double Orthocomplement is Closed Linear Span, we have:

$\paren {\Img {T^\ast} }^{\perp \perp} = \paren {\Img {T^\ast} }^-$

We have:

\(\ds T^\ast\) \(=\) \(\ds T^\ast \paren {T T^\ast}^\ast\) Adjoint of Composition of Linear Transformations is Composition of Adjoints
\(\ds \) \(=\) \(\ds T^\ast T T^\ast\) Adjoint of Composition of Linear Transformations is Composition of Adjoints, Adjoint is Involutive

Hence for each $x \in \HH_1$ we have:

$T^\ast x = \map {T^\ast T} {T^\ast x}$

So $\Img {T^\ast} \subseteq \Img {T^\ast T}$.

Clearly we have $\Img {T^\ast T} \subseteq \Img {T^\ast}$, so we obtain $\Img {T^\ast} = \Img {T^\ast T}$.

Hence we have:

$\paren {\map \ker T}^\perp = \paren {\Img {T^\ast T} }^- = \Img {T^\ast T}$

from Set is Closed iff Equals Topological Closure.

Hence $T^\ast T$ is an orthogonal projection onto $\paren {\map \ker T}^\perp$.

Now, for $x \in \paren {\map \ker T}^\perp$ we have:

$\paren {T^\ast T} x = x$

from Fixed Points of Orthogonal Projection.

We therefore have, for $x \in \paren {\map \ker T}^\perp$:

\(\ds \norm {T x}^2\) \(=\) \(\ds \innerprod {T x} {T x}_2\) Definition of Inner Product Norm
\(\ds \) \(=\) \(\ds \innerprod x {T^\ast T x}_1\) Definition of Adjoint Linear Transformation
\(\ds \) \(=\) \(\ds \innerprod x x_1\)
\(\ds \) \(=\) \(\ds \norm x^2_1\) Definition of Inner Product Norm

Hence:

$\norm {T x} = \norm x$ for each $x \in \paren {\map \ker T}^\perp$.

$\Box$


$(1)$ implies $(3)$

Suppose that:

$T$ is a partial isometry.

Let $P$ be the orthogonal projection onto $\paren {\map \ker T}^\perp$.

From Orthogonal Projection is Projection and Characterization of Projections, $P$ is Hermitian.

We show that $T^\ast T = P$.

Let $x \in \paren {\map \ker T}^\perp$.

We then have:

Then:

\(\ds \innerprod {T^\ast T x} x_1\) \(=\) \(\ds \innerprod {T x} {T x}_2\) Definition of Adjoint Linear Transformation
\(\ds \) \(=\) \(\ds \norm {T x}^2_2\) Definition of Inner Product Norm
\(\ds \) \(=\) \(\ds \norm x^2_1\) Definition of Partial Isometry
\(\ds \) \(=\) \(\ds \norm {P x}^2_1\) Fixed Points of Orthogonal Projection
\(\ds \) \(=\) \(\ds \innerprod {P x} {P x}_1\) Definition of Inner Product Norm
\(\ds \) \(=\) \(\ds \innerprod {P^2 x} x_1\) Definition of Adjoint Linear Transformation
\(\ds \) \(=\) \(\ds \innerprod {P x} x_1\) $P$ is idempotent

From Operator with Zero Numerical Range is Zero Operator: Corollary, we obtain $T^\ast T = P$.

In particular, $T^\ast T$ is a projection by Characterization of Projections.

$\blacksquare$


Sources

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