Part 1: Suppose $A=\sum_{i=1}^n\sigma_i v_i v_i^\top$, where $\sigma\ge 0$ is a non-negative real number (singular value) and $v_i\in\mathbb{R}^n$ is a vector. Suppose $A$ is positive semidefinite. Is that true that $B:=\sum_{i=1}^n\eta_i v_i v_i^\top$ (where $\eta_i = p_i \times \sigma_i, p_i>0 $) is also positive semidefinite? In other words, we just multiply the singular values by a positive number.
Part 2: Now let $A=\sum_{i=1}^n\sigma_i v_i w_i^\top$ where $\sigma\ge 0$ is a non-negative real number and $v_i\in\mathbb{R}^n,w_i\in\mathbb{R}^n$ are vectors. Again, suppose $A$ is positive semidefinite. Is that true that $B:=\sum_{i=1}^n\eta_i v_i w_i^\top$ (where $\eta_i = p_i \times \sigma_i, p_i>0 $) is also positive semidefinite?
Any comment or response is greatly appreciated.
Sure: evaluate $x^T Bx$ in terms of what you've got, and see that it is $\ge0$.