Considering the set of $n$-dimensional positive semidefinite matrices, $\mathcal{D}_n$. A matrix $M\in \mathcal{D}_n$ is called $\epsilon$ diagonal in trace distance if there is a diagonal matrix $D\in \mathcal{D}_n$ such that

$$|M-D|_{tr}\leq \epsilon.$$

For any $M\in \mathcal{D}_n$ , we can define $D(M)$ be its diagonal matrix---set all off diagonal term 0 and leave the diaognal element. Let $\Lambda(M)$ denote the diagonal matrix whose diagonal entries are the sorted eigenvalues of M.

Now suppose $M\in \mathcal{D}_n$ enjoys the following

$$|D(M)-\Lambda(M)|_{tr}\leq \epsilon,$$

Can we conclude $M$ is $2\epsilon$ diagonal?