if matrix such $AA^T=A^2$ then $A$ is symmetric? let matrix $A_{n\times n}$ is  real matrix,such $AA^T=A^2$, The transpose of matrix $A$ is written $A^T$,
show that :
the matrix $A$ is Symmetric matrices
maybe this problem have more methos,because it is know that if matrix $A$ is symmetric,then we have
$AA^T=A^2$,But for my problem,I can't prove it.Thank you

The solution suggested by user1551 is very nice, but, just for fun, here is a 2nd approach.
Assume that $AA^T = AA$. First we prove the following:

Claim: $AA = \underbrace{AA^T}_P = A^T A^T =  \underbrace{A^TA}_Q$
Proof: Transposing  $AA^T = AA$ gives $AA^T = A^T A^T$. Now we just need to worry about $Q$. With some fiddling around, we notice
$$B^2 = A^TAA^TA = A^TA^TA^TA =  AAA^TA =AAAA= AA^T AA^T = Q^2.$$
Since $P,Q$ are positive matrices, $P^2 = Q^2 \Rightarrow P=Q$ (uniqueness of positive square roots) and the claim is proved.

Since $A$ has real entries, $A^T$ equals its conjugate transpose $A^*$ so we have proved that $A$ is normal. At this point the problem is quite easy since  $A = U D U^*$ for some diagonal $D$ and some unitary $U$. From $AA^* = A^2$ follows $DD^* = D^2$ which implies $D$ has real entries i.e. $D=D^*$. From here we calculate
$$ A^T = A^* = (UDU^*)^* = UD^*U^* = UDU^* = A$$
so $A$ is symmetric.
