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11.6: Polar decomposition


Continuing the analogy between $$\mathbb{C}$$ and $$\mathcal{L}(V)$$, recall the polar form of a complex number $$z=|z|e^{i\theta}$$, where $$|z|$$ is the absolute value or modulus of $$z$$ and $$e^{i\theta}$$ lies on the unit circle in $$\mathbb{R}^{2}$$. In terms of an operator $$T\in \mathcal{L}(V)$$, where $$V$$ is a complex inner product space, a unitary operator $$U$$ takes the role of $$e^{i\theta}$$, and $$|T|$$ takes the role of the modulus. As in Section11.5, $$T^*T\ge 0$$ so that $$|T|:=\sqrt{T^*T}$$ exists and satisfies $$|T|\ge 0$$ as well.

Theorem 11.6.1. For each $$T\in \mathcal{L}(V)$$, there exists a unitary $$U$$ such that

$T = U |T|.$

This is called the polar decomposition of $$T$$.

Proof. We start by noting that

$\norm{Tv}^2 = \norm{\,|T|\,v}^2,$

since $$\inner{Tv}{Tv} = \inner{v}{T^*Tv} = \inner{\sqrt{T^*T}v}{\sqrt{T^*T}v}$$. This implies that $$\kernel(T) = \kernel(|T|)$$. By the Dimension Formula, this also means that $$\dim(\range(T)) = \dim(\range(|T|))$$. Moreover, we can define an isometry $$S: \range(|T|) \to \range(T)$$ by setting

$S( |T|v) = Tv.$

The trick is now to define a unitary operator $$U$$ on all of $$V$$ such that the restriction of $$U$$ onto the range of $$|T|$$ is $$S$$, i.e.,
$U|_{\range(|T|)} = S.$

Note that $$\kernel(|T|) \bot \range(|T|)$$, i.e., for $$v\in \kernel(|T|)$$ and $$w=|T|u \in \range(|T|)$$,

$\inner{w}{v} = \inner{|T|u}{v} = \inner{u}{|T|v} = \inner{u}{0} = 0$

since $$|T|$$ is self-adjoint.

Pick an orthonormal basis $$e=(e_1,\ldots,e_m)$$ of $$\kernel(|T|)$$ and an orthonormal basis $$f=(f_1,\ldots,f_m)$$ of $$(\range(T))^\bot$$. Set $$\tilde{S} e_i = f_i$$, and extend $$\tilde{S}$$ to all of $$\kernel(|T|)$$ by linearity. Since $$\kernel(|T|)\bot \range(|T|)$$, any $$v\in V$$ can be uniquely written as $$v=v_1+v_2$$, where $$v_1\in \kernel(|T|)$$ and $$v_2\in \range(|T|)$$. Now define $$U:V\to V$$ by setting $$Uv = \tilde{S} v_1 + S v_2$$. Then $$U$$ is an isometry. Moreover, $$U$$ is also unitary, as shown by the following calculation application of the Pythagorean theorem:

\begin{equation*}
\begin{split}
\norm{Uv}^2 &= \norm{\tilde{S}v_1 + S v_2}^2 = \norm{\tilde{S} v_1}^2 + \norm{S v_2}^2\\
&= \norm{v_1}^2 + \norm{v_2}^2 = \norm{v}^2.
\end{split}
\end{equation*}

Also, note that $$U|T|=T$$ by construction since $$U|_{\kernel(|T|)}$$ is irrelevant.

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