Consider the following $M \times M$ matrix $$\mathbf A = \sum_{k=1}^K a_k \mathbf h_k \mathbf h_k^H, \quad (M \geq K)$$ where $a_k$'s are real values and $\mathbf h_k$'s are $M \times 1$ randomly generated vectors, e.g., complex Gaussian random vector. Intuitively, since each $\mathbf H_k$ is a random matrix, it is expected that $\mathbf A$ has rank $K$ and its $K$ non-zero eigenvalues are distinct, i.e., $\lambda_1(\mathbf A) > \cdots > \lambda_K(\mathbf A)$ with probability one. I verified it experimentally by generating $\mathbf A$ randomly several times. However, I cannot provide mathematical proof or reasonable explanation.
In fact, I can prove it for $a_k \geq 0$ for $\forall k$. We first assume that $\mathbf F = [\sqrt a_1 \mathbf h_1, \cdots, \sqrt a_K \mathbf h_K]$ has rank $K$ and $K$ distinct non-zero singular values $\sigma_1(\mathbf F) > \cdots > \sigma_K(\mathbf F)$, which is reasonable because $\mathbf h_k$'s are random vectors. Then, we have $\mathbf A = \mathbf F \mathbf F^H$, which has $K$ distinct eigenvalues $\sigma_1^2(\mathbf F) > \cdots > \sigma_K^2(\mathbf F)$.
However, I cannot use this approach for $a_k \ngeq 0$. Any additional reasonable assumptions can be made for this proof.
Thank you.