First, notice that the product is not necessarily symmetric, except if the matrices commute,
Suppose M and N two symmetric positive-definite matrices and λ ian eigenvalue of the product MN. Then it's possible to show that λ>0 and thus MN has positive eigenvalues. Now, take M symmetric positive-definite and N symmetric negative-definite. As a result, apply the previous result to -(MN) then MN have negative eigenvalues. I hope this could be fairly clear,
Remember: positive or negative-definite is not a matrix property but it only applies to quadratic forms, which are naturally described only by symmetric matrices.
For instance, a way to establish positive definiteness of a quadratic form is to find this symmetric matrix representing it and test whether its eigenvalues are all positive.
But there exists infinitely many matrices representing a particular quadratic form, all with and exactly one of them is symmetric. Thus it's possible to have non-symmetric definite matrices.
Furthermore, it could be showed that for a not necessarily symmetric matrix to be
positive definite it's necessary but not sufficient that its real eigenvalues are all positive.
There are good answers, yet, to complete Fabrizio’s answer, the symmetry in positive definite matrices is a property with which we got used only because it appears in many examples.
However, symmetry is NOT needed for a matrix to be positive definite.
As people mentioned, the property comes from the quadratic form, which is defined to be positive definite, namely, the scalar product r=x'Mx>0 for any vector x≠0.
Because the result r is scalar, we clearly have r=r'.
Therefore, even if M is not symmetric, we may still have r=x'Mx=x'M'x >0. When M is symmetric, this is clear, yet iin general, it may also happen if M≠M'.
Of course, if the nonsymmetric matrix M is positive definite, so is its symmetric component Ms=( M+M')/2.