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

The minimizer is the arithmetic meanIt remains a positive self-adjoint trace-class operator and hence a covariance operator. The Hilbert-Schmidt inner product gives, for every Hilbert-Schmidt operator $C$,because ${\sum }_k(C_k-\overline{C})=0$. Thus $\overline{C}$ is the unique minimizer, including when the minimization is restricted to covariance operators.

For factorizations $C_i=L_i{L}_i^{\ast }$ and $C_j=L_j{L}_j^{\ast }$, the Procrustes distance between covariance operators is the following infimum over unitary operators $R$:Unitary invariance of the Hilbert-Schmidt norm givesThe polar decomposition of a bounded operator and trace duality implywhere the last equality expresses the trace norm as the sum of the singular values. Taking the infimum proves

The two rank-one operators have orthogonal ranges. Their Hilbert-Schmidt inner product is zero, while $\Vert C_i{\Vert }_{\mathrm{HS}}={\lambda }_i$, so the Hilbert-Schmidt distance between covariance operators isFor each $C_i$, the positive square root of an operator is $C_i^{1/2}=\sqrt{{\lambda }_i}{e}_i\otimes e_i$. Orthogonality therefore gives the square-root distance between covariance operatorsFinally $C_2^{1/2}{C}_1^{1/2}=0$, so every singular value in the Procrustes cross-term vanishes. Hence