# 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 $$B>0$$ 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?
• Hi, thanks for the hint. I'm still not sure how to find det($C$) and trace($AC$). – Ilan Aizelman WS Apr 23 at 14:00
• I do. But I don't understand how $C$ is constructed. Same about $A$. – Ilan Aizelman WS Apr 23 at 17:22
• @user1551 nice hint, $det(C)$ is indeed 1, but how do you calculate trace of $AC$ ,you dont know what is trace of $B$ after all... also how do you construct a sequence from that $C$? – user2323232 Apr 23 at 18:45
• @user2323232 The trace of $B$ is precisely the trace of $A$. – user1551 Apr 23 at 19:18