Here's my proof for the following: Given A,P $\in Mn(R)$, show that A symmetric $=> B=P^TAP $ symmetric
A symmetric $=> A=A^T$ symmetric
$B=P^TAP$
$(P^TAP)=P^TAP$
$(P^TAP)=(P^TAP)^T$
$(P^TAP)=P^{TT}A^TP^T$
$(P^TAP)=PA^TP^T$
Recall $A=A^T$
$(P^TAP)=PAP^T$
is there something I missed or is there a more thorough proof?