Suppose that $A \geq B \geq 0$ where $A$ and $B$ are two symmetric n by n matrices. $A \geq B$ stands for $A-B$ is positive semi-definite. Then, is it able to show that $det(A)\geq det(B)$.
I don't find any such statement. So I conjecture this could be wrong, but how to construct a simple counterexample?
By a continuity argument, you may assume that $A\ge B>0$. Then $0\le A^{-1/2}BA^{-1/2}\le I$ and all eigenvalues of $A^{-1/2}BA^{-1/2}$ lie between $0$ and $1$. Hence $\det(A^{-1/2}BA^{-1/2})\le1$ and $\det(B)\le\det(A)$.
More fundamentally, by Courant-Fischer minimax principle, when $A\ge B$, we have $\lambda_i(A)\ge\lambda_i(B)$ for each $i$, where $\lambda_i(\cdot)$ denotes the $i$-th largest eigenvalue of a Hermitian matrix. The result thus follows.
