Consider the set $\mathcal{D}_n$ of $n$-dimensional positive semidefinite matrices. 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 $D(M)$ to be its diagonal part --- set all off diagonal term to 0 and leave the diagonal elements. Let $\Lambda(M)$ denote the diagonal matrix whose diagonal entries are the sorted eigenvalues of M.

Now suppose $M\in \mathcal{D}_n$ satisfies

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

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