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Mathematics LibreTexts

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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