Theorem: Polar Decomposition for Bounded Linear Operator on Hilbert Space

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 $x \in \HH$:
{{begin-eqn}}
{{eqn	| l = \norm {\cmod T x}^2
		| r = \innerprod {\cmod T x} {\cmod T x}
		| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| r = \innerprod {\cmod T^2 x} x
		| c = {{Defof|Adjoint Linear Transformation}}
}}
{{eqn	| r = \innerprod {\paren {T^\ast T} x} x
		| c = {{Defof|Modulus of Element of C*-Algebra}} 
}}
{{eqn	| r = \innerprod {T x} {T x}
		| c = {{Defof|Adjoint Linear Transformation}}, Adjoint is Involutive 
}}
{{eqn	| r = \norm {T x}^2
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn	| l = \map {\widetilde U_0} {u + \lambda v}
		| r = \map {\widetilde U_0} {\cmod T x + \lambda \cmod T y}
}}
{{eqn	| r = \map {\widetilde U_0} {\map {\cmod T} {x + \lambda y} }
		| c = {{Defof|Linear Transformation}}
}}
{{eqn	| r = \map T {x + \lambda y}
		| c = definition of $\widetilde U_0$ 
}}
{{eqn	| r = T x + \lambda T y
		| c = {{Defof|Linear Transformation}}
}}
{{eqn	| r = \map {\widetilde U_0} {\cmod T x} + \lambda \map {\widetilde U_0} {\cmod T y}
}}
{{eqn	| r = \widetilde U_0 u + \lambda \widetilde U_0 v
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn	| l = U \cmod T x
		| r = \widetilde U \cmod T x
		| c = since $\cmod T x \in \Img {\cmod T}$
}}
{{eqn	| r = \widetilde U_0 \cmod T x
		| c = since $\cmod T x \in \Img {\cmod T}$
}}
{{eqn	| r = T x
		| c = by definition of $\widetilde U_0$
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn	| l = \innerprod {U^\ast T x} z
		| r = \innerprod {U^\ast T x} {\cmod T y}
}}
{{eqn	| r = \innerprod {T x} {U \cmod T y}
		| c = {{Defof|Adjoint Linear Transformation}}, Adjoint is Involutive
}}
{{eqn	| r = \innerprod {T x} {T y}
		| c = since $T = U \cmod T$ as already shown 
}}
{{eqn	| r = \innerprod {T^\ast T x} y
		| c = {{Defof|Adjoint Linear Transformation}}
}}
{{eqn	| r = \innerprod {\cmod T^2 x} y
		| c = {{Defof|Modulus of Element of C*-Algebra}}
}}
{{eqn	| r = \innerprod {\cmod T x} {\cmod T y}
		| c = since $\cmod T$ is Hermitian
}}
{{eqn	| r = \innerprod {\cmod T x} z
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn	| l = \innerprod {U^\ast T x} z
		| r = \innerprod {U^\ast T x} u + \innerprod {U^\ast T x} v
}}
{{eqn	| r = \innerprod {U^\ast T x} u
}}
{{eqn	| r = \innerprod {\cmod T x} u
		| c = we have equality for $z \in \map \cl {\Img T}$
}}
{{eqn	| r = \innerprod {\cmod T x} u + \innerprod {\cmod T x} v
}}
{{eqn	| r = \innerprod {\cmod T x} z
}}
{{end-eqn}}
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 $U^\ast T = \cmod T$.
{{qed|lemma}}