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).
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5Does this answer your question? If $A$ and $B$ are linear transformations on finite-dimensional inner product vector space, and if $0 \le A \le B$, then $det A \le det B$. – user8675309 Feb 08 '20 at 19:03
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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 $B\ge0$.
user1551
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Just to elaborate slightly, if the eigenvalues of $A$ are ordered $\lambda_1 \ge ...$ then $\lambda_k = \min_{\dim S = n=k+1} \max_{|x|=1, x \in S} x^* Ax$. It follows that $\lambda_k(A) \ge \lambda_k(B)$. – copper.hat Feb 08 '20 at 21:08