Polar decomposition of a bounded operator ($T=U|T|$)

Every bounded operator on a Hilbert space has a polar decomposition $T=U|T|$, where $|T|=(T^{\ast }T{)}^{1/2}$ and $U$ is a partial isometry with $\ker U=\ker T$. On $\mathrm{im}|T|$, it is defined by $U(|T|x)=Tx$.

This is the bounded-operator case of polar decomposition.

For a bounded linear operator $L$ on a Hilbert space, its operator absolute value is the positive square root of an operatorIt appears in the polar decomposition of a bounded operator $L=U|L|$. The identity $\Vert |L|x{\Vert }^2=⟨L^{\ast }Lx,x⟩=\Vert Lx{\Vert }^2$ shows that $|L|$ retains the size of the action of $L$, while $U$ supplies its isometric part.

For a compact operator, the nonzero eigenvalues of $|L|$, counted with multiplicity, are the singular values of $L$. When $L$ is trace-class, $\mathrm{tr}|L|=\Vert L{\Vert }_1$ is its trace norm. An arbitrary bounded linear operator need not have a discrete sequence of singular values in this sense or a finite trace norm; those conclusions require their stated hypotheses.

Let $P_n:H\to H_n$ be coisometries onto finite-dimensional Hilbert spaces, with $Q_n=P_n^{\ast }{P}_n\to I$ strongly, and let $A$ be a unitary operator on $H$. The square compression $T_n=P_{n}A{P}_n^{\ast }$ has a singular value decomposition $T_n=V_n{\Sigma }_n{W}_n^{\ast }$. Thus $U_n=V_n{W}_n^{\ast }$ is a unitary operator on $H_n$ extending the polar factor, including a choice on its null space.

Because the codomains vary, the convergence statement in $H$ isIndeed, $P_n^{\ast }{T}_n^{\ast }{T}_n{P}_n=Q_n{A}^{\ast }{Q}_{n}A{Q}_n\to I$ strongly, with uniformly bounded operator norms. Polynomial approximation to the square-root function gives $P_n^{\ast }|T_n|P_n\to I$ strongly. Since $T_n=U_n|T_n|$,Finally, $P_n^{\ast }{T}_n{P}_n=Q_{n}A{Q}_n\to A$ strongly. Including the lift $P_n^{\ast }$ makes these comparisons between operators on the same Hilbert space.