Eigenvalues of the sum of the products of a diagonal matrix and PD matrix $A\Lambda+\Lambda A$ 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!
 A: 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 $\gamma \geq 0$.
