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The square-root distance between covariance operators isWriting $A_i=C_i^{1/2}$, minimizing ${\sum }_i\Vert A_i-A{\Vert }_{\mathrm{HS}}^2$ gives $A=n^{-1}{\sum }_i{A}_i$. Since this average is positive,the square-root barycenter of covariance operators.

For positive trace-class operators, the square-root distance between covariance operators isIf $C_i=L_i{L}_i^{\ast }$ are Hilbert-Schmidt factorizations, the Procrustes distance between covariance operators iswhere $\mathcal{O}(H)$ is the group of orthogonal operators on the real separable Hilbert space $H$.

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