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
Although not an exact duplicate, the method for solving An equivalent condition for a real matrix to be skew-symmetric applies. Briefly speaking, for real square matrices, $\langle X,Y\rangle = \operatorname{trace}(Y^TX)$ defines an inner product and symmetric matrices are orthogonal to skew-symmetric matrices. Now, write $A=H+K$, where $H=\frac12(A+A^T)$ is the symmetric part and $K=\frac12(A-A^T)$ is the skew-symmetric part. Then $AA^T=A^2$ implies that $(H+K)K=0$ and in turn $HK=-K^2=K^TK$. Taking trace on both sides, we get $\langle K,H\rangle=\langle K,K\rangle$. Since $\langle K,H\rangle=0$, it follows that $\langle K,K\rangle=0$ and $K=0$, i.e. $A$ is symmetric.
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.
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.