I have two square matrices $P$ and $Q$. Given $P = I - PQ$ I managed to show that $P$ is invertible and that $PQ = QP$. I need to show somehow that if $Q$ is symmetric then $P$ is symmetric as well.

I have tried to develop $(I - PQ)^T$ in hopes that it will equal to $P$ but no such luck.



$$P(I+Q)=I$$ Taking transpose: $$(I+Q)P^T=I$$

Are you able to see why $P$ is symmetric?

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  • $\begingroup$ P(I+Q) = (I+Q)P^t . But we don't know that (I+Q) is invertible and therefore we cannot just assume that P = P^t. Still missing something. :( $\endgroup$ – eventhorizon02 Nov 17 '17 at 9:47
  • $\begingroup$ I see. what is the definition of $(I+Q)$ being invertible? $\endgroup$ – Siong Thye Goh Nov 17 '17 at 9:50
  • $\begingroup$ ok, ok, (I+Q) is the inverse of P. But it appears to the right of P and to the left of P^t. So would it be legal to "divide" by (I+Q) on two different sides? $\endgroup$ – eventhorizon02 Nov 17 '17 at 10:02
  • $\begingroup$ a more proper term is to multiply $(I+Q)^{-1}$. We can post multiply something for one equation and pre multiply something for another equation. $\endgroup$ – Siong Thye Goh Nov 17 '17 at 10:14
  • $\begingroup$ @eventhorizon02 Alternatively, we can use the two equalities given in this answer to say $$P = PI = P(I+Q)P^T = IP^T = P^T$$ $\endgroup$ – Arthur Nov 17 '17 at 11:04

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