Is a matrix whose quadrants obey eigenvalue restrictions positive semidefinite?

Question

Suppose I have some Hermitian matrix $M$. Let $M = \begin{pmatrix} A & B\\B^* & D \end{pmatrix}$. Let $d = \dim(A) = \dim(B)$. Does the following hold?

If $\lambda_i(A)=\lambda_i(D) \geq \lambda_i(B) \text{ for all } i\in [d]$ Then: $A$ is positive semidefinite.

Answer 1

No it does not imply that $M$ is positive semidefinite. Counter example:

Let $M = \begin{pmatrix} 2 &0 &3 &0 \\ 0& 4 &0& 0 \\ 3 &0 &2& 0 \\ 0 &0 &0 &4 \end{pmatrix}$ Then $M$ has a $-1$ eigenvalue.

Perhaps this holds only if the eigenvalues that have some over lap in their eigenvectors are bigger...