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Let $A$ and $B$ be positive definite matrices (psd) of the same size, such that $A>B$ (i.e. $A-B$ is also psd).

I wonder if $det(A)>det(B)$?

I have tried to find a counter example, but couldn't find.

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Since $A$ and $B$ are symmetric and real, the Min-Max Theorem applies: $$ \lambda_k(A)=\min_{\dim M=k}\max\{\langle Ax,x\rangle:\ x\in M, \|x\|=1\}, $$ where $\lambda_k(A)$ denotes the $k^{\rm th}$ eigenvalue of $A$ in nonincreasing order. As $\langle Ax,x\rangle>\langle Bx,x\rangle$ for all $x$, it follows that $$\lambda_k(A)\geq\lambda_k(B),\ \ \ k=1,\ldots,n.$$ Then $$ \det A=\prod_k\lambda_k(A)\geq\prod_k\lambda_k(B)=\det B. $$

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