Polar Decomposition for Bounded Linear Operator on Hilbert Space

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Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\GF$.

Let $T : \HH \to \HH$.


Then there exists a unique partial isometry $U : \HH \to \HH$ such that:

$T = U \cmod T$ and $\map \ker T = \map \ker U$

where $\cmod T$ is the modulus of $T$.

Further, $U^\ast T = \cmod T$.


Proof

Existence

From the definition of the modulus, $\cmod T$ is positive and hence Hermitian.

We have, for each $x \in \HH$:

\(\ds \norm {\cmod T x}^2\) \(=\) \(\ds \innerprod {\cmod T x} {\cmod T x}\) Definition of Inner Product Norm
\(\ds \) \(=\) \(\ds \innerprod {\cmod T^2 x} x\) Definition of Adjoint Linear Transformation
\(\ds \) \(=\) \(\ds \innerprod {\paren {T^\ast T} x} x\) Definition of Modulus of Element of C*-Algebra
\(\ds \) \(=\) \(\ds \innerprod {T x} {T x}\) Definition of Adjoint Linear Transformation, Adjoint is Involutive
\(\ds \) \(=\) \(\ds \norm {T x}^2\)

Hence, since $\cmod T$ and $T$ are linear, we have:

for $x, y \in \HH$ such that $\cmod T x = \cmod T y$, we have $T x = T y$.

We note in particular that $\map \ker T = \map \ker {\cmod T}$.

We can therefore define $\widetilde U_0 : \Img {\cmod T} \to \Img T$ by:

$\map {\widetilde U_0} {\cmod T x} = T x$

for each $x \in \Img {\cmod T}$.

We show that this is linear.

Let $u, v \in \Img {\cmod T}$ and $\lambda \in \GF$.

Then there exists $x, y \in \HH$ such that $u = \cmod T x$ and $v = \cmod T y$.

We then have:

\(\ds \map {\widetilde U_0} {u + \lambda v}\) \(=\) \(\ds \map {\widetilde U_0} {\cmod T x + \lambda \cmod T y}\)
\(\ds \) \(=\) \(\ds \map {\widetilde U_0} {\map {\cmod T} {x + \lambda y} }\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \map T {x + \lambda y}\) definition of $\widetilde U_0$
\(\ds \) \(=\) \(\ds T x + \lambda T y\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \map {\widetilde U_0} {\cmod T x} + \lambda \map {\widetilde U_0} {\cmod T y}\)
\(\ds \) \(=\) \(\ds \widetilde U_0 u + \lambda \widetilde U_0 v\)

so $\widetilde U_0$ is linear transformation.

Since $\norm {T x} = \norm {\cmod T x}$ for each $x \in \HH$, this an linear isometry.

From Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain: Corollary 3, there exists a linear isometry $\widetilde U : \map \cl {\Img {\cmod T} } \to \HH$ extending $\widetilde U_0$.

From Extension of Bounded Linear Transformation from Closed Subspace of Hilbert Space to Whole Space, we can define a bounded linear transformation $U : \HH \to \HH$ by:

$U x = \widetilde U u$ for each $x \in \HH$

where $x = u + v$ for $u \in \map \cl {\Img {\cmod T} }$ and $v \in \map \cl {\Img {\cmod T} }^\bot$.


We show that $U$ is the desired map.

We first show that $U$ is a partial isometry.

We note that since $\widetilde U$ is an isometry, we have $\ker {\widetilde U} = \set { {\mathbf 0}_\HH}$ from Linear Isometry is Injective.

For $u \in \map \cl {\Img {\cmod T} }$, we have $\norm {U u} = \norm {\widetilde U u} = \norm u$.

So $U$ is a linear isometry on $\map \cl {\Img {\cmod T} }$.

We want to show that $\map \ker U^\bot = \map \cl {\Img {\cmod T} }$.

We first show that $\map \ker U = \map \cl {\Img {\cmod T} }^\bot$.

Let $x \in \map \ker U$.

From Direct Sum of Subspace and Orthocomplement we can write $x = u + v$ for $u \in \map \cl {\Img {\cmod T} }$ and $v \in \map \cl {\Img {\cmod T} }^\bot$.

Then we have $U x = \widetilde U u$.

If $u \ne {\mathbf 0}_\HH$, we have $U x \ne {\mathbf 0}_\HH$ since $\ker \widetilde U = \set { {\mathbf 0}_\HH}$, hence $x = v \in \map \cl {\Img {\cmod T} }^\bot$.

So we have $\map \ker U \subseteq \map \cl {\Img {\cmod T} }^\bot$.

The fact that $\map \cl {\Img {\cmod T} }^\bot \subseteq \map \ker U$ is immediate by construction.

So $\map \ker U = \map \cl {\Img {\cmod T} }^\bot$.

Hence from Double Orthocomplement is Closed Linear Span, we obtain:

$\map \ker U^\bot = \map \cl {\Img {\cmod T} }^{\bot \bot} = \map \cl {\Img {\cmod T} }$

Recalling that $U$ is an isometry on $\map \cl {\Img {\cmod T} }$, $U$ is an isometry on $\map \ker U^\bot$.

So $U$ is a partial isometry.


We now verify that $T = U \cmod T$.

For each $x \in \HH$, we have:

\(\ds U \cmod T x\) \(=\) \(\ds \widetilde U \cmod T x\) since $\cmod T x \in \Img {\cmod T}$
\(\ds \) \(=\) \(\ds \widetilde U_0 \cmod T x\) since $\cmod T x \in \Img {\cmod T}$
\(\ds \) \(=\) \(\ds T x\) by definition of $\widetilde U_0$

So $T = U \cmod T$.


We show that $\map \ker U = \map \ker T$.

We have already shown that $\map \ker U = \map \cl {\Img {\cmod T} }^\bot$.

From Orthocomplement of Closure, we have $\map \cl {\Img {\cmod T} }^\bot = \Img {\cmod T}^\bot$.

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

$\Img {\cmod T}^\bot = \ker {\cmod T}$

since $\cmod T$ is Hermitian.

Hence we have $\map \ker U = \map \ker {\cmod T}$.

We have already shown that $\map \ker {\cmod T} = \map \ker T$, so we have $\map \ker U = \map \ker T$.


Lastly we show that $U^\ast T = \cmod T$.

Let $z \in \Img {\cmod T}$.

Then there exists $y \in \HH$ such that $z = \cmod T y$.

We then have:

\(\ds \innerprod {U^\ast T x} z\) \(=\) \(\ds \innerprod {U^\ast T x} {\cmod T y}\)
\(\ds \) \(=\) \(\ds \innerprod {T x} {U \cmod T y}\) Definition of Adjoint Linear Transformation, Adjoint is Involutive
\(\ds \) \(=\) \(\ds \innerprod {T x} {T y}\) since $T = U \cmod T$ as already shown
\(\ds \) \(=\) \(\ds \innerprod {T^\ast T x} y\) Definition of Adjoint Linear Transformation
\(\ds \) \(=\) \(\ds \innerprod {\cmod T^2 x} y\) Definition of Modulus of Element of C*-Algebra
\(\ds \) \(=\) \(\ds \innerprod {\cmod T x} {\cmod T y}\) since $\cmod T$ is Hermitian
\(\ds \) \(=\) \(\ds \innerprod {\cmod T x} z\)

for all $z \in \Img {\cmod T}$.

Now take $z \in \map \cl {\Img {\cmod T} }$.

Then there exists a sequence $\sequence {z_n}_{n \mathop \in \N}$ in $\Img {\cmod T}$ such that $z_n \to z$.

Then we have:

$\innerprod {U^\ast T x} {z_n} = \innerprod {\cmod T x} {z_n}$ for each $n \in \N$.

From Inner Product is Continuous, we obtain:

$\innerprod {U^\ast T x} z = \innerprod {\cmod T x} z$ for each $z \in \map \cl {\Img T}$

by taking $n \to \infty$.

For $z \in \HH$, write $z = u + v$ for $u \in \map \cl {\Img T}$ and $v \in \map \cl {\Img T}^\bot$ as before.

We've already established that $\map \cl {\Img T}^\bot = \map \ker U$, and so:

$\innerprod {U^\ast T x} v = \innerprod {T x} {U v} = 0$

We also have:

$\innerprod {\cmod T x} v = 0$

by the definition of orthocomplement.

We then have:

\(\ds \innerprod {U^\ast T x} z\) \(=\) \(\ds \innerprod {U^\ast T x} u + \innerprod {U^\ast T x} v\)
\(\ds \) \(=\) \(\ds \innerprod {U^\ast T x} u\)
\(\ds \) \(=\) \(\ds \innerprod {\cmod T x} u\) we have equality for $z \in \map \cl {\Img T}$
\(\ds \) \(=\) \(\ds \innerprod {\cmod T x} u + \innerprod {\cmod T x} v\)
\(\ds \) \(=\) \(\ds \innerprod {\cmod T x} z\)

for each $z \in \HH$.

In particular we have:

$\innerprod {U^\ast T x - \cmod T x} {U^\ast T x - \cmod T x} = 0$

So we have $U^\ast T x = \cmod T x$.

So $U^\ast T = \cmod T$.

$\Box$

Uniqueness

Take the $U$ constructed above.

Suppose that $W$ is another partial isometry with:

$T = U \cmod T = W \cmod T$

with $\map \ker T = \map \ker U = \map \ker W$.

Hence $U$ and $W$ coincide on $\Img {\cmod T}$.

From Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain: Corollary, $U$ and $W$ coincide on $\map \cl {\Img {\cmod T} }$.

We established earlier in the proof that $\map \ker U = \map \cl {\Img {\cmod T} }^\bot$, and so $\map \ker W = \map \cl {\Img {\cmod T} }^\bot$ as well.

Hence $U$ and $W$ also coincide on $\map \cl {\Img {\cmod T} }^\bot$.

Then for $x \in \HH$ we can write $x = u + v$ for $u \in \map \cl {\Img {\cmod T} }$ and $v \in \map \cl {\Img {\cmod T} }^\bot$, we have:

$W x = W u + W v = W u = U u + U v = U x$

So $W = U$.

$\blacksquare$


Sources

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