An element $x$ of a unital <C-star algebra> is positive when $x=y^*y$ for some $y$, equivalently when $x=x^*$ and $\sigma(x)\subseteq[0,\infty)$. The continuous functional calculus for the nonnegative function $t\mapsto\sqrt t$ defines a positive element $x^{1/2}$ satisfying $(x^{1/2})^2=x$. If a positive $z$ also satisfies $z^2=x$, functional calculus for $z$ gives $z=\sqrt{z^2}=\sqrt x$, proving uniqueness. For a positive operator $T=S^*S$ on $\ell^2$,
$$
\langle T\xi,\xi\rangle=\|S\xi\|^2\geq0.
$$

For arbitrary $T$, put $|T|=(T^*T)^{1/2}$. Then
$$
\||T|\xi\|^2
=\langle T^*T\xi,\xi\rangle
=\|T\xi\|^2,
$$
so $\ker|T|=\ker T$. Define
$$
U_0(|T|\xi)=T\xi
$$
on $\operatorname{im}|T|$. The kernel identity makes this well-defined, and the norm identity makes it an isometry. Extend it continuously to
$$
\overline{\operatorname{im}|T|}=(\ker|T|)^\perp=(\ker T)^\perp
$$
and set it equal to zero on $\ker T$. The resulting $U$ is a <partial isometry>, has $\ker U=\ker T$, and satisfies the <polar decomposition of a bounded operator> $T=U|T|$.

Finally $U^*U$ is the <orthogonal projection> onto $(\ker T)^\perp$, which contains the range of $|T|$. Therefore
$$
U^*T=U^*U|T|=|T|.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 106 / 3 / a / Solution

{wiki}

An operator $U$ on a Hilbert space is a partial isometry when it is isometric on $(\ker U)^\perp$. Equivalently, $U^*U$ is the orthogonal projection onto this initial space.


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 Incoming links: Partial isometry