Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 225 / 3 / ii

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- Solution ii

Solution

Because

C_{i} = L_{i} L_{i}^{*}

is a covariance operator,

∥ L_{i} ∥_{HS}^{2} = tr (L_{i} L_{i}^{*}) = tr C_{i} < \infty.

Thus both factors are Hilbert-Schmidt. The product of two Hilbert-Schmidt operators is a trace-class operator, with

∥ L_{2}^{*} L_{1} ∥_{1} \leq ∥ L_{2} ∥_{HS} ∥ L_{1} ∥_{HS} < \infty.

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