Suppose that $A$ is a real square matrix with transpose $A^T$. I am trying to show that, in general, $$A^TA\neq AA^T.$$ As an example, consider the matrix $A=\begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}$. Then $A^T=\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$.
Hence $AA^T=\begin{pmatrix} 1 & 2 \\ 2 & 13 \end{pmatrix}$ and $A^TA=\begin{pmatrix} 5 & 6 \\ 6 & 9 \end{pmatrix}$, which are clearly not equal. I am wondering how this generalises to the entries of my matrix $A$. For instance, is this always true if $A$ is not symmetric?
The square matrix that satisfy $AA^T=A^TA$ is known as normal matrix.
Clearly every symmetric matrix is normal but not the converse.
For an example choose,
$A=\begin{pmatrix} 1&1&0 \\ 0&1&1 \\ 1&0&1\end{pmatrix}$
Then $A$ is normal matrix (verify!) but clearly $A$ is not symmetric.