If $A \in C^{nxn}$ , $A \ge 0 $ and A is sing., there exists a sequence of matrices $C_k$, that $C_k \ge 0$,$|C_k| = 1$ and trace $AC_k \le 1/k$ Question: Show that if $A \in C^{n \times n}$ , $ A \ge 0 $ and A is singular, then there exists a sequence of matrices $C_k$, $k = 1,2,...$ such that $C_k \ge 0$, det $C_k = 1$ and trace $AC_k \le 1/k$

My approach: I have no idea how to even approach this exercise, any help is highly appreciated.
Hint. If $A=B\oplus0_{(n-r)\times(n-r)}$ where $B&gt;0$ and $r=\operatorname{rank}(A)$, consider $C=\left(\varepsilon I_r\right)\oplus\left(\varepsilon^{-r/(n-r)}I_{n-r}\right)$ for some positive scalar $\varepsilon$. What are $\det(C)$ and $\operatorname{trace}(AC)$ in this case?
Hint. If $A=B\oplus0_{(n-r)\times(n-r)}$ where $B&gt;0$ and $r=\operatorname{rank}(A)$, consider $C=\left(\varepsilon I_r\right)\oplus\left(\varepsilon^{-r/(n-r)}I_{n-r}\right)$ for some positive scalar $\varepsilon$. What are $\det(C)$ and $\operatorname{trace}(AC)$ in this case?