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.



Contributors

• Isaiah Lankham, Mathematics Department at UC Davis
• Bruno Nachtergaele, Mathematics Department at UC Davis
• Anne Schilling, Mathematics Department at UC Davis

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