If A and B are orthogonal projection matrices, how can I show that trace(AB) <= rank(AB)?

Mainuillato2p 2022-09-26 Answered
If A and B are orthogonal projection matrices, how can I show that trace(AB)≤ rank(AB)?
I was using C-S inequality to get tr ( A B ) t r ( A 2 ) t r ( B 2 ) and I know that t r ( A 2 ) =rank(A). But I can't get the rank of AB.
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Answers (1)

Jaelyn Levine
Answered 2022-09-27 Author has 9 answers
If either A or B is zero, this holds trivially.
Suppose that both A and B are non-zero. It suffices to show that all eigenvalues of AB have magnitude at most equal to 1. To that end, note that if denotes the spectral norm, then we have A = B = 1 , so that A B A B = 1. It follows that all eigenvalues λ of AB satisfy | λ | A B 1. Thus, if AB has rank r and λ 1 , , λ k (with k r) are the non-zero eigenvalues of AB, then we have
tr ( A B ) | tr ( A B ) | = | i = 1 k λ i | i = 1 k | λ i | i = 1 k 1 = k r ,
which is what we wanted.
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