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
 A: 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.
A: Mike's answer is very nice, but for fun, here is a third approach. It is developed along the same line as Mike's, but using more basic techniques.
Observe that if $W$ is a symmetric or skew-symmetric matrix and $W^2=0$, then $W=0$. This is because $\|Wx\|_2^2=x^TW^TWx=\pm x^TW^2x=0$ for all $x$ and hence $Wx\equiv 0$.
Now, from the condition $AA^T=A^2$, we obtain $AA^T=AA=A^TA^T$ and in turn
\begin{cases}
AA^TAA^T = AAAA = AA^TA^TA,\\
A^TAAA^T = A^TA^TA^TA^T = AAAA = AAA^TA = A^TA^TA^TA = A^TAA^TA.
\end{cases}
Hence $(AA^T-A^TA)^2 = 0$. Since $W_1=AA^T-A^TA$ is symmetric and $W_1^2=0$, we infer that $W_1=0$, i.e. $AA^T=A^TA$.
But then we get $(A-A^T)^2 = AA-AA^T-A^TA+A^TA^T=0$. Since $W_2=A-A^T$ is skew-symmetric and $W_2^2=0$, we conclude that $W_2=A-A^T=0$, i.e. $A$ is symmetric.

Edit: As some users have pointed out that the statement is trivial when $A$ is invertible, we will also develop a proof along this line. By the given conditions, $\ker A=\ker A^T$:


*

*$\ker A \subseteq \ker A^T$ because
$Ax=0 \Rightarrow  \|A^Tx\|^2=x^TAA^Tx=x^TA(Ax)=0.$

*$\ker A^T \subseteq \ker A$ because
$$
A^Tx=0 \Rightarrow  AAx=AA^Tx=0 \Rightarrow Ax\in\ker A\subseteq\ker A^T \Rightarrow  \|Ax\|^2=x^T(A^T(Ax))=0.
$$


Therefore, by an orthonormal change of basis, we may assume that $A=\pmatrix{B&0\\ 0&0_{k\times k}}$, where $B$ is invertible, $BB^T=B^2$ and $k$ is the nullity of $A$. Hence $B^T=B^{-1}(BB^T)=B^{-1}(B^2)=B$ and $A$ is symmetric.
A: 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.
A: 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.
A: Given $AA^t=A^2$. Premultiply by $A^{-1}$. You obtain $A^t=A$ which means that the matrix is symmetric. $AA^{-1}=A^{-1}A=I_d\,\,$identity matrix)
{The above result is true only if $A$ is invertible because only then  premultiplying by $A^{-1}$ is defined..}
