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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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&#039;t get the rank of AB.

Jaelyn Levine · Accepted Answer

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.

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

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