Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 1 / 8 / b / Solution

The polar decomposition of a bounded operator states that every bounded $T:H\to H$ is uniquelywhere $V$ is a partial isometry with initial space $\overline{\mathrm{Ran}|T|}=(\ker T{)}^{\perp }$ and final space $\overline{\mathrm{Ran}T}$. Its initial and final orthogonal projections are $V^{\ast }V$ and $V{V}^{\ast }$.

Existence of $|T|$ was proved in (a). For every $\xi$,Thus the map $|T|\xi \mapsto T\xi$ is well defined and isometric on $\mathrm{Ran}|T|$. It extends uniquely to an isometry between its closed range and $\overline{\mathrm{Ran}T}$. Extend it by zero on $\ker |T|=\ker T$ to obtain $V$. The equation and the projection assertions follow immediately. Any other normalized partial isometry with the same product must agree on the dense range of $|T|$ and vanish on its orthogonal complement, giving uniqueness. The kernel normalization is essential; without it an action on the kernel could be added.

A useful consequence for a Von Neumann algebra $M$ is that the polar factors of $T\in M$ also belong to $M$. The polynomial construction in (a) puts $|T|$ in the norm-closed algebra $M$. For $\epsilon >0$, the inverse of $|T|+\epsilon I$ belongs to $M$ by a convergent Neumann series, andThese operators are contractions, since $\Vert |T|\xi \Vert \le \Vert (|T|+\epsilon I)\xi \Vert$. On vectors $|T|\xi$ the difference from $V$ has norm at most $\epsilon \Vert \xi \Vert$, and on the kernel it is zero. Density and the uniform bound prove the strong limit. Strong closedness of $M$ then gives $V\in M$. This justifies the use of algebra-valued polar factors in Question 7(c).