# Does $\lim_{N\rightarrow\infty}\frac{tr(A'A)}{N}=0$ imply $\lim_{N\rightarrow\infty}\frac{tr(A)}{N}=0$?

$A$ is an $N\times N$ matrix with bounded row and column norms. Does $\lim_{N\rightarrow\infty}\frac{tr(A'A)}{N}=0$ imply $\lim_{N\rightarrow\infty}\frac{tr(A)}{N}=0$?

I know this is true for finite $N$, but how to prove it when $N\rightarrow\infty$? If this statement is false, can someone give an example?

$\langle A,B\rangle=\operatorname{tr}(B^\ast A)$ is an inner product on $M_n(\mathbb C)$. By Cauchy-Schwarz inequality, $$\frac{\left|\operatorname{tr}(A)\right|}{n} =\frac{\left|\langle A,I\rangle\right|}{n} \le\frac{\sqrt{\langle I,I\rangle\, \langle A,A\rangle}}{n} =\frac{\sqrt{n\operatorname{tr}(A^\ast A)}}{n} =\sqrt{\frac{\operatorname{tr}(A^\ast A)}{n}}.$$