Suppose $0\neq X_n\in \mathbb{M}_{k(n)}(\mathbb{C})$,if $\lim_{n \to \infty}tr(X_n^*X_n)=0$,can we conclude that $\lim_{n \to \infty}\|X_n^*X_n\|=0$,where $tr$ is the standard trace on complex matrix,$\|·\|$ is the norm of matrix.
My thought: the eigenvalues of $X_n^*X_n$ is non-negative,$ tr(X_n^*X_n) $ is equal to the sum of non-negative eigenvalues,so each non-negative eigenvalue is small enough,the norm of $X_n^*X_n$ is equal to the product of eigenvalues,so it tends to zero.Is my idea correct?
Yes. The matrices $X_n^*X_n$ are positive, so their eigenvalues are nonnegative. The trace is sum of the eigenvalues and the norm (if you are using the operator norm, you don't say) is the greatest eigenvalue. So you always have $$ \|X_n^*X_n\|\leq\operatorname{Tr}(X_n^*X_n). $$ If you allow an arbitrary norm, the answer is no. For instance, take the norm(s) to be $$ \|A\|_n=\sum_{j=1}^n\left(\sum_{k=1}^n|A_{kj}|^2\right)^{1/2}.$$ Now put $X_n=n^{-5/4}E_n$, where $E_n$ is the $n\times n$ matrix with all entries equal to $1$. It is easy to check that $E_n^2=nE_n$, and that $E_n$ is positive. Now $$ \operatorname{Tr}(X_n^*X_n)=n^{-5/2}\operatorname{Tr}(E_n^2)=n^{-1/2}\to0. $$ Meanwhile, $|X_n^*X_n|=X_n^*X_n$, and \begin{align} \|X_n^*X_n\|_n&=n^{-5/2}\sum_{j=1}^n\left(\sum_{k=1}^nn^2\right)^{1/2} =n^{-5/2}nn^{3/2}=1. \end{align}