Determinant inequality positive definite matrices

Question

If $A$ and $B$ are two positive definite matrices such that: $$A \geq B,$$ can I conclude that: $$\mathrm{det}(A) \geq \mathrm{det}(B).$$ (With the notation $A \geq B$, I mean that $A - B \geq 0$, i.e. $A-B$ is positive semi-definite).

Thanks.

Answer 1

Yes. $A\ge B$ is equivalent to $B^{-1/2}AB^{-1/2}A\ge I$. So, it implies that all eigenvalues of $B^{-1/2}AB^{-1/2}A$ are $\ge1$. It follows that $\det(B^{-1/2}AB^{-1/2}A)\ge1$ and $\det(A)\ge\det(B)$.

Actually, by using the variational characterisation of eigenvalues of Hermitian matrices, it can be shown that $\lambda_i(A)\ge\lambda_i(B)$ when $A\ge B$. From this, we may also infer that $\det(A)\ge\det(B)$ when $B\ge0$.