Let $A \in \mathbb{R}^{n \times n}$ be a positive semi-definite $A \succcurlyeq 0$, and with positive diagonal elements ($A_{i,i} > 0$ for all $i$). Let $A$ have at least one eigenvalue equal to $0$.
Let $b, c \in \mathbb{R}^n$ be element-wise non-negative vectors ($b, c \geq 0$), with $c^\top b > 0$.
Prove that the matrix $$ M := \left[ \begin{matrix} A & b \\ c^\top A & c^\top b\end{matrix} \right] \in \mathbb{R}^{(n+1) \times (n+1)} $$ has non-negative eigenvalues.
Comments. Since $A$ has at least one $0$ eigenvalue, $M$ has determinant $0$, hence at least one $0$ eigenvalue. The trace of $M$ is positive, thus the sum of its eigenvalues is positive. In addition, the traces of all the square sub-matrices is positive. I am not sure whether this fact can be exploited. If $A$ were positive definite, then one could use the Schur determinant formula to prove the claim.