An operator is trace class exactly when it factors as $T=AB$ with $A$ and $B$ Hilbert--Schmidt. The factors can be chosen so that $\|T\|_1=\|A\|_{\mathrm{HS}}\|B\|_{\mathrm{HS}}$ by using the <polar decomposition of a bounded operator>.

Hilbert-Schmidt factorization of a trace-class operator

For factorizations $C_i=L_iL_i^*$ and $C_j=L_jL_j^*$, the <Procrustes distance between covariance operators> is the following infimum over unitary operators $R$:
$$
d_P(C_i,C_j)=\inf_R\lVert L_i-L_jR\rVert_{\mathrm{HS}}.
$$
Unitary invariance of the Hilbert-Schmidt norm gives
$$
\lVert L_i-L_jR\rVert_{\mathrm{HS}}^2
=\lVert L_i\rVert_{\mathrm{HS}}^2+\lVert L_j\rVert_{\mathrm{HS}}^2
-2\operatorname{Re}\operatorname{tr}(L_i^*L_jR).
$$
The <polar decomposition of a bounded operator> and trace duality imply
$$
\sup_R\operatorname{Re}\operatorname{tr}(L_i^*L_jR)
=\lVert L_j^*L_i\rVert_1
=\sum_{k=1}^{\infty}\sigma_k,
$$
where the last equality expresses the <trace norm> as the sum of the <singular values>. Taking the infimum proves
$$
d_P(C_i,C_j)^2
=\lVert L_i\rVert_{\mathrm{HS}}^2+\lVert L_j\rVert_{\mathrm{HS}}^2-2\sum_{k=1}^{\infty}\sigma_k.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 225 / 3 / b / Solution

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
$$
U^*T=U^*U|T|=|T|.
$$

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

{title2=$T=U|T|$}
{wiki=Polar_decomposition}

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


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 Incoming links: Polar decomposition of a bounded operator