$A$ is normal if and only if $trace(A^\ast A)=\sum_{i=1}^n|\lambda_i|^2.$ where the $\lambda_i$ are the eigenvalues of $A$.
Show that $A\in\mathbb{C}_n$ is normal $\iff$ $tr(A*A) = \sum_{i = 1}^n|\lambda_i|^2$, where $\lambda_1,...,\lambda_n$ are the eigenvalues of $A$.