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?

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?

We are considering the set of $n$-dimensional Positive semidefinite Matrix, $\mathcal{D}_n$. A matrix $M\in \mathcal{D}_n$ is called $\epsilon$ diagonal in trace distance only 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, $\Lambda(D)$ denotes 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?

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?