The two rank-one operators have orthogonal ranges. Their Hilbert-Schmidt inner product is zero, while $\lVert C_i\rVert_{\mathrm{HS}}=\lambda_i$, so the <Hilbert-Schmidt distance between covariance operators> is
$$
d_L(C_1,C_2)=\sqrt{\lambda_1^2+\lambda_2^2}.
$$
For 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 operators>
$$
d_R(C_1,C_2)=\sqrt{\lambda_1+\lambda_2}.
$$
Finally $C_2^{1/2}C_1^{1/2}=0$, so every singular value in the Procrustes cross-term vanishes. Hence
$$
d_P(C_1,C_2)=\sqrt{\lambda_1+\lambda_2}.
$$

Solution

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

{title2=$T^{1/2}$}

Every bounded <positive operator> $T$ has a unique positive operator $T^{1/2}$ satisfying $(T^{1/2})^2=T$. The <spectral theorem for normal operators on a separable Hilbert space> constructs it by applying the scalar square-root function to the spectrum.

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 Incoming links: Positive square root of an operator