## 每週問題 March 6, 2017

Let $A$ be an $n\times n$ nonzero Hermitian matrix. Prove that

$\displaystyle \hbox{rank}A\ge\frac{(\hbox{trace}A)^2}{\hbox{trace}(A^2)}$.

$\displaystyle (\hbox{trace}A)^2=(\lambda_1+\cdots+\lambda_r)^2\le r(\lambda_1^2+\cdots+\lambda_r^2)=r\,\hbox{trace}(A^2)$