We can define a (not necessarily symmetric) real $n \times n$ matrix $A$ to be positive definite if ${\bf x}^T A {\bf x} \geq 0$ for all ${\bf x} \in \mathbb{R}^n$ and equality holds if and only if $x={\bf 0}$.
Then
$$A = \begin{pmatrix}
1& 2a\\
0& 1
\end{pmatrix}$$
is positive definite if and only if $|a|<1$. And so is the transpose. But, if you try to compute the product of some of these matrices, you'll find a product to two which is no longer definite.
Here's a non-symmetric positive definite matrix which squares to a indefinite matrix:
$$L = \begin{pmatrix}
1& 1\\
-1& 1
\end{pmatrix}.$$
For symmetric positive definite matrices, as Jules says in the comments above, the product of two symmetric matrices need not be symmetric. Indeed, if $A$, $B$, and $AB$ are symmetric, then we must have that $AB = (AB)^T = B^T A^T = BA$. However, since every matrix commutes with itself, the square of a positive definite matrix is positive definite.
This note may be helpful regarding whether or not to insist that the matrix be defined as a symmetric matrix. The example below is inspired by this note.
A search of "non symmetric positive definite matrix" will yield this and many other discussions including some here on math.stackexchange.
One final example: consider $q(x,y) = {\bf x}^T A {\bf x} = x^2 + 2axy + y^2$. We find that the symmetric matrix associated to this quadratic form is
$$M = \begin{pmatrix}
1& a\\
a& 1
\end{pmatrix},$$
so that $q(x,y) = {\bf x}^T M {\bf x}$. This is positive definite if and only if its determinant is positive (which proves the claim about the matrix $A$ above).