Is A-B positive semidefinite if A's eigenvalues are greater than B's

Question

I know that if matrices $A$ and $B$ are positive semi-definite and $A\succeq B$ ($A-B$ is positive semi-definite), then the $i$-th eigenvalue of $A$ is greater than or equal to the i-th eigenvalue of $B$. In other words: $$\text{If } A \succ 0, B \succ 0, A\succeq B \text{ then } \lambda_i(A) \ge \lambda_i(B), i=1,...,n$$

My question: Is the opposite of this true? As in, if I know $A$ and $B$ are positive semi-definite and the $i$-th eigenvalue of $A$ is greater than or equal to the $i$-th eigenvalue of $B$, can I conclude that $A\succeq B$? If so, how? Again in other words: $$\text{If } A \succ 0, B \succ 0, \lambda_i(A) \ge \lambda_i(B), i=1,...,n \text{ then } A\succeq B?$$

Answer 1

Certainly not. E.g. $\pmatrix{2\\ &1}-\pmatrix{1\\ &2}$ is indefinite.