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.

The absolute value of a linear operator $L$ is the positive square root of an operator $L^{†}L$. Its eigenvalues are the singular values of $L$, its trace is the trace norm of $L$, and it appears in the polar decomposition of a bounded operator $L=U|L|$.

If $T_n=P_{n}A{P}_n^{\ast }$ is a square finite compression with singular value decomposition $T_n=V_n{\Sigma }_n{W}_n^{\ast }$, then $U_n=V_n{W}_n^{\ast }$ is a unitary extension of its polar factor. When $A$ is unitary and $P_n^{\ast }{P}_n\to I$ strongly, $|T_n|\to I$ strongly and hence $U_n{P}_n\to A$ strongly.