TITLE: if matrix such $AA^T=A^2$ then $A$ is symmetric?
QUESTION [11 upvotes]: 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

REPLY [4 votes]: Just to add a different approach:

We wish to prove the more general fact for complex matrices obtained replacing transpose with Hermitian transpose.
use a Schur decomposition to reduce to the case of $A$ upper triangular.
For $A$ upper triangular, computing the (1,1) terms of $AA^T$ and $A^2$ yields $A_{11}^2 = \sum_{j=1}^n |A_{1j}|^2$. By comparing absolute values, it is easier to see that $A_{11}$ must be real and $A_{1j}=0$ for all $j\neq 1$.
go on like this by induction, each time proving that all terms in the $i$-th row are zero apart from the one on the diagonal
You have proved that your upper triangular $A$ is real diagonal, hence Hermitian.