If $A$ and $B$ are respectively positive-definite and positive-semidefinite matrices, is $A^T B A - B \ge 0$?

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I would like to prove that $A^T B A - B \ge 0$ for given $A > 0$ and $B \ge 0$.

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No.

Consider the 1 by 1 matrices $A$ with $0< A < 1$, and $B \gt 0$. Then $A^TBA - B < 0$