---
title: Ch6-3
tags: Linear algebra
GA: G-77TT93X4N1
---

# Chapter 6 extra note 3

> polar decomposition

## The modulus of an operator

Given $T\in\mathcal{L}(V)$, where $V$ is an inner product space, there exists four linear transformations as
* $T:V\to V$;
* $T^*:V\to V$;
> The adjoint linear transformation of $T$.
* $T^*T:V\to V$;
> $T^*T\ge 0$.
* $|T|:V\to V$.
> $|T|\ge 0$ and $\|T{\bf v}\|=\||T|{\bf v}\|$, for all ${\bf v}\in V$.

**Proposition:**
* $\text{Ker}(T) = \text{Ker}(T^*T) = \text{Ker}(|T|)$.
* $\text{Ran}(T^*) = \text{Ran}(T^*T) = \text{Ran}(|T|)$.

---

## Polar decomposition of an operator

**Theorem:**
Let $T\in\mathcal{L}(V)$, where $V$ is an inner product space. Then $T$ can be represented as
$$
\tag{1}
T = U|T|=U\sqrt{T^*T},
$$
where $U$ is an unitary operator.

* proof:
> Let us firstly define a map $U_1:\text{Ran}(|T|)\to\text{Ran}(T)$ such that
> $$
> U_1({\bf x}) = T({\bf v}), \quad \text{where }|T|{\bf v}={\bf x}.
> $$
>
> claim: $U_1$ is well-defined:
> > Given ${\bf x}\in \text{Ran}(|T|)$, suppose there exist ${\bf v}_1, {\bf v}_2\in V$ such that $|T|{\bf v}_1=|T|{\bf v}_2={\bf x}$, then ${\bf v}_1-{\bf v}_2\in \text{Ker}(|T|)=\text{Ker}(T)$.
> >
> > So there is a ${\bf u}\in\text{Ker}(T)$ such that ${\bf v}_1={\bf v}_2+{\bf u}$.
> > We then have
> > $$
> > \tag{2}
> > T({\bf v}_1) = T({\bf v}_2+{\bf u})=T({\bf v}_2)+T({\bf u})=T({\bf v}_2).
> > $$
> > So $U_1({\bf x})=T({\bf v}_1)=T({\bf v}_2)$, and $U_1$ is well-defined.
>
> claim: $U_1$ is linear:
> > Given ${\bf x}_1$, ${\bf x}_2\in \text{Ran}(|T|)$, $\alpha_1,\alpha_2\in\mathbb{F}$, there exists ${\bf v}_1$, ${\bf v}_2\in V$ such that
> > $$
> > \tag{3}
> > |T|{\bf v}_1={\bf x}_1, \quad |T|{\bf v}_2={\bf x}_2,
> > $$
> > so that
> > $$
> > \tag{4}
> > U_1({\bf x}_1)=T{\bf v}_1, \quad U_1({\bf x}_2)=T{\bf v}_2.
> > $$
> > Since $|T|$ is linear,
> > $$
> > \tag{5}
> > |T|(\alpha_1{\bf v}_1+\alpha_2{\bf v}_2)=\alpha_1{\bf x}_1+\alpha_2{\bf x}_2.
> > $$
> > Therefore,
> > $$
> > \tag{6}
> > \begin{align}
> > U_1(\alpha_1{\bf x}_1+\alpha_2{\bf x}_2)&=T(\alpha_1{\bf v}_1+\alpha_2{\bf v}_2)\\
> > &=\alpha_1 T({\bf v}_1)+\alpha_2T({\bf v}_2)\\
> > &=\alpha_1 U_1({\bf v}_1)+\alpha_2 U_1({\bf v}_2).
> > \end{align}
> > $$
>
> claim: $U_1$ is an isometry.
> > Given ${\bf x}\in\text{Ran}(|T|)$, there exists ${\bf v}\in V$ such that $|T|{\bf v} = {\bf x}$, and $U_1({\bf x})=T({\bf v})$. So,
> > $$
> > \tag{7}
> > \|U_1 {\bf x}\|=\|T {\bf v}\|=\||T|{\bf v}\|=\|{\bf x}\|.
> > $$
>
> claim: $U_1$ is invertible:
> > Since
> > $$
> > \text{dim}(\text{Ran}(|T|))=\text{dim}(\text{Ran}(T^*))=\text{dim}(\text{Ran}(T)).
> > $$
> > So, by [proposition](https://hackmd.io/@teshenglin/2024LA2_ch5_6), $U_1$ is invertible.
>
> claim: $U_1 |T| = T$
> > Given ${\bf v}\in V$, let $|T|{\bf v}={\bf x}$. Then
> > $$
> > \tag{8}
> > U_1|T|{\bf v} = U_1{\bf x}=T{\bf v}.
> > $$
>
> As $U_1$ is only defined on $\text{Ran}(|T|)=\text{Ran}(T^*)$, it is an unitary operator from $\text{Ran}(|T|)$ to $\text{Ran}(T^*)$, but it is not an unitary operator from $V$ to $V$.
>
> Since $T\in\mathcal{L}(V)$, $\text{dim}(\text{Ker}(T^*))=\text{dim}(\text{Ker}(T))$, we can find an unitary operator $U_2\in\mathcal{L}(\text{Ker}(T), \text{Ker}(T^*))$.
> > The simplest construction is to find orthonormal basis for $\text{Ker}(T)$ and $\text{Ker}(T^*)$, and then map basis to basis one-by-one.
>
> Finally, since $V=\text{Ker}(T)\oplus\text{Ran}(T^*)$, we define a map $U\in\mathcal{L}(V)$ by, given ${\bf v}\in V$, we have ${\bf v}={\bf v}_K+{\bf v}_R$, where ${\bf v}_K\in \text{Ker}(T)$ and ${\bf v}_R\in\text{Ran}(T^*)$. Then
> $$
> \tag{9}
> U({\bf v}) = U_1{\bf v}_R + U_2{\bf v}_K.
> $$
> Once can check easily that $U$ is an unitary operator and $U|T| = T$.

---

**Proposition:**
Let $T\in\mathcal{L}(V)$, where $V$ is an inner product space. $T$ is invertible if and only if there is an unique isometry $U$ such that $T=U|T|$.
* proof:
> ($\Rightarrow$)
> By polar decomposition, there exists $U\in\mathcal{L}(V)$ such that $T=U|T|$.
> Since $\text{Ker}(T)=\text{Ker}(|T|)$, $T$ is invertible implies $|T|$ is invertible. So $|T|^{-1}$ exists and is unique.
> Therefore, $T=U|T|$ implies $U=T|T|^{-1}$ which is unique.
>
> ($\Leftarrow$)
> We want to show that if $U$ is unique, then $T$ is invertible. (P$\to$Q)
> Instead, we show that if $T$ is not invertible, then $U$ is not unique. (~Q$\to$ ~P)
>
> Suppose $T$ is not invertible, $\text{dim}(\text{Ker}(T))>0$.
> Then, follow the construction of $U_2$ in the proof of polar decomposition, we should be able to find different $U_2$ such that they are all unitary transformation in $\mathcal{L}(\text{Ker}(T), \text{Ker}(T^*))$, and by combining with $U_1$ they form an unitary operator $U$.