The minimizer is the arithmetic mean
$$
\overline C=\frac1n\sum_{k=1}^nC_k.
$$
It remains a positive self-adjoint trace-class operator and hence a <covariance operator>. The <Hilbert-Schmidt inner product> gives, for every Hilbert-Schmidt operator $C$,
$$
\sum_{k=1}^n\lVert C_k-C\rVert_{\mathrm{HS}}^2
=\sum_{k=1}^n\lVert C_k-\overline C\rVert_{\mathrm{HS}}^2
+n\lVert C-\overline C\rVert_{\mathrm{HS}}^2,
$$
because $\sum_k(C_k-\overline C)=0$. Thus $\overline C$ is the unique minimizer, including when the minimization is restricted to covariance operators.

Solution

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

{title2=$\langle A,B\rangle_{\mathrm{HS}}$}

For <Hilbert-Schmidt operators> $A$ and $B$, the Hilbert-Schmidt inner product is $\langle A,B\rangle_{\mathrm{HS}}=\operatorname{tr}(A^*B)$. Its induced norm is the <Hilbert-Schmidt operator>[Hilbert-Schmidt norm].

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 Incoming links: Hilbert-Schmidt inner product