Let $X$ be a positive symmetric matrix. $X$ can be understood as an inverse of a reduced incidence matrix of a tree-structured network. Hence, for the first row, $$\sum_{k:k\sim 1} x_{ik}$$ includes $x_{10}$, $0$ means the root. However, for the reduced matrices, its column is deleted, so there is no index $0$ anywhere else in the matrix, and which makes the matrix nonsingular and positive definite (PD).
Some nodes are connected. We use $I \sim j$ to say that $i\neq j$ and $i$ and $j$ are neighbors. We know that $X$ is a PD matrix. For PD, we only consider real numbers, not complex numbers.
$$\begin{equation} [X]_{ij} = \begin{cases} \sum\limits_{k:k\sim i} x_{ik} & \text{if $i=j$}\\ -x_{ij} & \text{if $i\sim j$}\\ 0 & \text{otherwise} \end{cases} \end{equation}$$ We also have diagonal matrices $C$ and $K$, we want to somehow bound $C$ and $K$ to make
$$[K(CX+I)+(XC+I)K]$$
a PD matrix.
Edit: Or relax it to $(CX+I)^\top(CX+I)+K(CX+I)+(XC+I)K$ and makes it PD.
Edit 2: A trivial answer is $K=C^{-1}$, I want something beyond that, like K can be how much different with $C^{-1}$.
Actually, in my case, $C$ has relatively some entries, so in practice, it shows like a PD. I want some hints so that I can rigorously prove it with some bounds on $C$ or $K$ or both.
Thank you in advance for any possible help.
As you have noticed and also pointed out by the others, the reduced matrix $X$, apart from being symmetric positive semidefinite, has some special structures: it is diagonally dominant and it is an M-matrix. In what follows, however, I will find a sufficient condition for $K(CX+I)$ to be “non-symmetric positive definite” using only the positive semidefiniteness of $X$, without regard to these special structures. Since this sufficient condition is applicable to a generic positive semidefinite matrix $X$, you may probably find it too limiting.
Let $D=KC$. The symmetric part of $K(CX+I)$ is then $\frac12(DX+XD+2K)$. It is positive definite if and only if $$ \lambda_\min(DX+XD+2K)>0.\tag{1} $$ Let $k$ and $c$ be the diagonals of $K$ and $C$ respectively. Their Hadamard product $d=k\circ c$ is the diagonal of $D$. Let $e$ be the vector of ones. Then $DX+XD=(de^T+ed^T)\circ X$ is a principal submatrix of $(de^T+ed^T)\otimes X$. Therefore $\lambda_\min(DX+XD)\ge\lambda_\min\left((de^T+ed^T)\otimes X\right)$ and $$ \begin{align} &\lambda_\min(DX+XD+2K)\\ &\ge\lambda_\min(DX+XD)+2\lambda_\min(K)\\ &\ge\lambda_\min\left((de^T+ed^T)\otimes X\right)+2\lambda_\min(K)\\ &=\lambda_\min(de^T+ed^T)\lambda_\max(X)+2\lambda_\min(K)\quad\text{(because $\lambda_\min(de^T+ed^T)\le0$)}\\ &=-(\|e\|_2\|d\|_2-e^Td)\lambda_\max(X)+2\lambda_\min(K).\tag{2}\\ \end{align} $$ Thus $(1)$ holds if the expression on line $(2)$ is positive, i.e., it holds if the following sufficient condition is satisfied: $$ (\|e\|_2\|d\|_2-e^Td)\lambda_\max(X)<2\min_{1\le i\le n}k_i.\tag{3} $$ If the computation of $\lambda_\max(X)$ is too expensive, you may replace $\lambda_\max(X)$ by some larger but easily computable quantity, such as $\|X\|_\infty=\max_{1\le i\le n}\sum_{j=1}^n|x_{ij}|$, to obtain another sufficient condition with a stronger requirement.