Given a positive definite matrix $A\in\mathbb{R}^{n\times n}$ with eigenvalues $\mu_{1}\geq \mu_{2}\geq\cdots\mu_n>0$ and a diagonal matrix $\Lambda\in\mathbb{R}^{n\times n}$ with diagonals $\lambda_{1}\geq \lambda_{2}\geq\cdots\lambda_n>1$.
My question is: could we find the lower/upper bound of the eigenvalue of $A\Lambda+\Lambda A$ based on $\mu_i$ and $\lambda_i$? Is $A\Lambda+\Lambda A$ always positive definite? If not, what additional assumption is needed for the positive definiteness?
Any help would be highly appreciated!
For a quick way to upper-bound the eigenvalues of $A\Lambda + \Lambda A$, recall that the eigenvalues of any symmetric $S$ satisfies $|{\lambda_{\text{min}}(S)}|, |\lambda_{\text{max}}(S)| \leq \| S \|$, the operator norm. Hence, we can conclude that the eigenvalues of $A \Lambda + \Lambda A$ have magnitudes bounded by $$ \lvert|A \Lambda + \Lambda A\rvert| \leq \lvert| A \Lambda \rvert| + \lvert| \Lambda A \rvert| \leq 2 \lvert|A\rvert| \cdot \lvert| \Lambda \rvert| \leq 2 \mu_1 \lambda_1. $$
Here's an example where $A\Lambda + \Lambda A$ is not PD. Take $$ A = \begin{bmatrix} 3.26 & 1.79 & 1.42 & -2.16 \\ 1.79 & 1.76 & 1.9 & -0.2 \\ 1.42 & 1.9 & 10.34 & 3.05 \\ -2.16 & -0.2 & 3.05 & 3.58 \end{bmatrix}, \qquad \Lambda = \begin{bmatrix} 5 \\ & 4 \\ & & 3 \\ & & & 2 \end{bmatrix}, $$ then $\lambda_{\text{min}}(A) = 0.02449$, but $\lambda_{\text{min}} (A \Lambda + \Lambda A) = -1.1334$. To make the sum PD, one idea is to add some regularization term $(A\Lambda + \Lambda A) + \gamma I$ for sufficiently large $\gamma \geq 0$.