11.6: Polar decomposition
( \newcommand{\kernel}{\mathrm{null}\,}\)
Continuing the analogy between C and L(V), recall the polar form of a complex number z=|z|eiθ, where |z| is the absolute value or modulus of z and eiθ lies on the unit circle in R2. In terms of an operator T∈L(V), where V is a complex inner product space, a unitary operator U takes the role of eiθ, and |T| takes the role of the modulus. As in Section11.5, T∗T≥0 so that |T|:=√T∗T exists and satisfies |T|≥0 as well.
Theorem 11.6.1. For each T∈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
‖Tv‖2=‖|T|v‖2,
since ⟨Tv,Tv⟩=⟨v,T∗Tv⟩=⟨√T∗Tv,√T∗Tv⟩. This implies that null(T)=null(|T|). By the Dimension Formula, this also means that dim(range(T))=dim(range(|T|)). Moreover, we can define an isometry S:range(|T|)→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 null(|T|)⊥range(|T|), i.e., for v∈null(|T|) and w=|T|u∈range(|T|),
⟨w,v⟩=⟨|T|u,v⟩=⟨u,|T|v⟩=⟨u,0⟩=0
since |T| is self-adjoint.
Pick an orthonormal basis e=(e1,…,em) of null(|T|) and an orthonormal basis f=(f1,…,fm) of (range(T))⊥. Set ˜Sei=fi, and extend ˜S to all of null(|T|) by linearity. Since null(|T|)⊥range(|T|), any v∈V can be uniquely written as v=v1+v2, where v1∈null(|T|) and v2∈range(|T|). Now define U:V→V by setting Uv=˜Sv1+Sv2. Then U is an isometry. Moreover, U is also unitary, as shown by the following calculation application of the Pythagorean theorem:
‖Uv‖2=‖˜Sv1+Sv2‖2=‖˜Sv1‖2+‖Sv2‖2=‖v1‖2+‖v2‖2=‖v‖2.
Also, note that U|T|=T by construction since U|null(|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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