Exercise 10.A.12

Answers

Proof. Every orthogonal projection is positive (see 7.32) and equal to its square (see 6.55). We have

{(PQP)}^{∗} = P^{∗} Q^{∗} P^{∗} = PQP

and

⟨ PQPv, v ⟩ = ⟨ QPv, Pv ⟩ \geq 0,

where the inequality above follows because $Q$ is positive. Hence $PQP$ is positive. Thus

trace (PQ) = trace (P^{2} Q) = trace (PQP) \geq 0 .

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celiopassos

2017-10-06 00:00