# difference in determinants of Positive Definite Matrices

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.

## 1 Answer

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 $$\det A=\prod_k\lambda_k(A)\geq\prod_k\lambda_k(B)=\det B.$$