= Solution
The <polar decomposition of a bounded operator> states that every bounded $T:H\to H$ is uniquely
$$
\boxed{T=V|T|,\qquad |T|=(T^*T)^{1/2},\qquad \ker V=\ker T,}
$$
where $V$ is a <partial isometry> with initial space $\overline{\operatorname{Ran}|T|}=(\ker T)^\perp$ and final space $\overline{\operatorname{Ran}T}$. Its initial and final <orthogonal projections> are $V^*V$ and $VV^*$.
Existence of $|T|$ was proved in (a). For every $\xi$,
$$
\||T|\xi\|^2=\langle T^*T\xi,\xi\rangle=\|T\xi\|^2.
$$
Thus the map $|T|\xi\mapsto T\xi$ is well defined and isometric on $\operatorname{Ran}|T|$. It extends uniquely to an isometry between its closed range and $\overline{\operatorname{Ran}T}$. 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 $\varepsilon>0$, the inverse of $|T|+\varepsilon I$ belongs to $M$ by a convergent Neumann series, and
$$
T(|T|+\varepsilon I)^{-1}\longrightarrow V\quad\text{strongly}.
$$
These operators are contractions, since $\||T|\xi\|\leq\|(|T|+\varepsilon I)\xi\|$. On <vectors> $|T|\xi$ the difference from $V$ has <norm> at most $\varepsilon\|\xi\|$, 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).
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