If B is a symmetric matrix such that ABA is well-defined, then show that ABA=0 if and only if B is of the form C-PACPA for some symmetric matrix C.

PA here is the projection matrix of A.

  • $\begingroup$ is this homework? $\endgroup$
    – Dayne
    Oct 10, 2020 at 12:23
  • $\begingroup$ Yes, it is part of my homework. The teacher did not discuss this clearly, so I couldn't quite get a grasp of it. I understood the concept of projection matrices $\endgroup$
    – user530299
    Oct 10, 2020 at 12:29
  • $\begingroup$ but the part "iff B is of the form C-PCP" just confuses me. $\endgroup$
    – user530299
    Oct 10, 2020 at 12:30
  • 1
    $\begingroup$ This assertion is not generally true. Consider the case $$B=\pmatrix{1&0\\0&1};\quad A=\pmatrix{0&1\\0&0}.$$ I suspect you intend to refer to $ABA^\prime$ and, for $P_A$ to be meaningful, you either need to assume $A$ also is symmetric or else replace $A$ by $(A+A^\prime)/2$ everywhere. $\endgroup$
    – whuber
    Oct 10, 2020 at 14:38


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