Let $A$ be square non-singular matrix of order $n \geq 2$.
If $A$ is symmetric, then $A^2$ is symmetric positive definite.
If $A^2$ is symmetric positive definite, then $A$ is symmetric.
I think I have an idea how to prove (1). Consider $x'A^2x$. If A is symmetric, then $x'A^2x = x'AAx = x'A'Ax = (Ax)'Ax = $ sum of squares that cannot be negative. Hence, $A^2$ is positive semidefinite for sure. Now, given $\det A^2 = ( \det A )^2 \neq 0$, we may conclude that $A^2$ is positive definite (I used the result that states “positive semidefiniteness + non-singularity = positive definiteness”). QED
So, my question is: is my proof correct and how to find a counterexample to (2)?