Positive definiteness of matrix A. P is a symmetric positive definite matrix. I want to know under what condition of matrix A will the matrix A'P+PA be positive definite, assuming A,P are square matrices of same order. 
A'P+PA is a symmetric matrix 
proof:
 let w=A'P then 
W'=(A'P)'=PA
W=(W+W')/2 + (W-W')/2  where (W+W')/2 is symmetric and (W-W')/2 is skew symmetric
therefore A'P+PA is symmetric (equals symmetric part w+w')
Basically I want to clarify if both A and P must be symmetric positive definite matrix for A'P+PA to be positive definite (or positive semi def). I am interested  in knowing the properties of A which would ensure positive definiteness (or positive semi def) of A'P+PA. 
 A: I think for positive definiteness of $A^T P + PA$, the matrix $A$ needs to be positive definite. Let us assume that $A$ has one negative eigenvalue such that $\exists q: \; A q = \lambda q$ for $\lambda<0$. Then using the same $q$, to compute the quadratic form of $A^T P + PA$, we have 
$$ q^T (A^T P + PA) q = q^T A^T P q + q^T P A q = \lambda q^T P q + q^T P \lambda q  = 2 \lambda q^T P q < 0,$$
where the $<0$ follows since $\lambda<0$ and $q^T P q>0$ (due to positive definiteness of $P$).
Note that this only shows that positive definiteness is necessary, not that it is sufficient (though I think it should be). 
As for semidefiniteness, it is clear that if $A$ is semidefinite, it has at least one zero eigenvalue. Then, with the same argument as above, we can also show that $A^T P + P A $ can become zero for the corresponding eigenvector of $A$.
edit: As pointed out in the comments, from the argument above it is only necessary that the eigenvalues of $A$ are (non-negative) positive. Positive (semi)-definite matrices satisfy this but the converse is not true, matrices with strictly positive eigenvalues can be indefinite.
