I know it is hard to talk about the positive definiteness of matrices summation, but I keep running into this kind of question during my research.
First, to clarify, when I say nonsymmetric PD matrix, I mean $\forall x\in R^n, x^\top Ax=x^\top(\frac{1}{2}A+\frac{1}{2}A^\top)x>0$. I define $A=CX+I$, X is a symmetric M-matrix, and C is a positive definite diagonal matrix. So we know that there exists a C such that $CX+XC$ is positive definite, and the existence of identity gives some flexibility to play with C and maintains the positive definiteness of $CX+XC+2I$.
Now I have some C fixed, which made $CX+XC+2I$, can I determine how much I can change the K (a diagonal matrix with nonnegative entries), such that $A^TA+KA+AK$ is positive definite? I have played with some numerical test with my C and X, I found that if the element of K is not too different, (too different is saying, suppose $K_{11}=1, K_{22}=2,..., K_{66}=100,...$, all diagonal entries except $K_{66}$ is less than 10, $K_{66}$ is too different with others), then this matrix's smallest eigenvalue is positive. Even if some entries of $K$ or all entries of $K$ are zero. Sorry, this part of the description is not so rigorous. Hopefully, it delivered my message.
I want to know if there is any method can help me find a pathway to characterize the flexibility I have to play around with K such that $A^TA+KA+AK$ remains PD. I expect the flexibility to be something like K's largest eigenvalue and the smallest eigenvalue can not be too different, the difference is bounded by something of A.
Any intuition, relaxation, references are welcomed. Thank you so much for your attention and time.