<s>[INST] If $A$ and $B$ are linear transformations on a finite-dimensional inner product space, and if $\textbf{0} \leq A \leq B$, then det $A \leq$ det $B$. Exercise 12 from SEC. 82 of Finite-Dimensional Vector Spaces - 2nd Edition by Paul R. Halmos.

If $A$ and $B$ are linear transformations on a finite-dimensional inner product space, and if $\textbf{0} \le A \leq B$, then $\det A \leq \det B$.

Notation: $0 \le A$ means that $A$ is positive.

If $\det B = 0$, choose $x \in \ker B$, then $0 \le ((B-A)x,x) = (Bx, x) - (Ax, x) = -(Ax, x) \le 0$. Therefore $\ker B \subseteq \ker A$ and $0 = det A \le det B = 0$.
How to prove for the case when $B$ is invertible? Intuitively, I feel that since both $A$ and $B$ are positive their proper values are also positive and $B$'s proper values should be greater the $A$'s and same for determinants.
 [/INST] A verbose ("for dummies") version of Omnomnomnom's and Andreo's proof is as follows.
We have two cases: (a) $B$ is non-invertible, and (b) $B$ is invertible.
Case (a): since $B$ is not invertible, det $B = 0$. Also, the null-space of $B$ has some non-zero vector $x$. It follows from the positivity of $(B-A)$ that $\big\langle(B-A)x, x \big\rangle \geq 0$
$\implies \langle Bx-Ax, x\rangle \geq 0 \implies \langle Bx, x\rangle - \langle Ax, x\rangle \geq 0\\ \implies 0 - \langle Ax, x\rangle \geq 0 \implies \langle Ax, x\rangle = 0$ [$\because \langle Ax, x\rangle \geq 0$ due to the positivity of $A$]
$\implies \langle \sqrt A \sqrt Ax, x\rangle = 0$ [$\because$ positivity of $A$ implies that it has a positive square-root]
$\implies \langle \sqrt Ax, {\sqrt A}^*x \rangle = 0$ $\implies \langle \sqrt Ax, \sqrt Ax \rangle = 0$ [$\because \sqrt A$ is positive, it is self-adjoint too] $\implies \left \Vert \sqrt Ax \right \Vert^2 = 0 \implies \sqrt Ax = 0$ $\implies \sqrt A \sqrt Ax = 0 \implies Ax = 0$
$\implies$ (non-zero) $x$ is in the null-space of $A \implies A$ is not invertible $\implies$ det $A = 0$
$\implies$ det $A =$ det $B$ (since both det are $0$).
Case (b): in finite dimensions, the invertibility of $B \geq \textbf{0}$ implies that $B > \textbf{0}$ (positive-definite). Thus, $B$ has a positive-definite square root, say $B^\frac{1}{2}$ which is also invertible. If $B^{-\frac{1}{2}}$ is the inverse of $B^\frac{1}{2}$, then $B^{-\frac{1}{2}}$ is also positive-definite.
Now, because $B-A$ is given as positive, $C^*(B-A)C$ is positive for an arbitrary $C$, i.e, $C^*(B-A)C \geq \textbf{0}$. If we let $C = B^{-\frac{1}{2}}$, then we have $C^* = {B^{-\frac{1}{2}}}^* = B^{-\frac{1}{2}}$. Substituting $C$ and $C^*$ in the inequality on hand yields $\textbf{1}- B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \geq \textbf{0}$, i.e., $B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \leq \textbf{1}$. We observe that $B^{-\frac{1}{2}}AB^{-\frac{1}{2}}$ is positive transformation, and thus unitarily diagonalizable. This fact, together with the inequality $B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \leq \textbf{1}$, reveals that each Eigenvalue (diagonal entry in the diagonal form) of $B^{-\frac{1}{2}}AB^{-\frac{1}{2}}$ is $\leq 1$.  This implies that det $B^{-\frac{1}{2}}AB^{-\frac{1}{2}} \leq 1$. Further, because det $CD$ = det $C \cdot$ det $D$ holds for invertible transformations $C$ and $D$, it follows that det $B^{-\frac{1}{2}}\cdot$ det $A \cdot$ det $B^{-\frac{1}{2}} \leq 1$ where det $B^{-\frac{1}{2}} \neq 0$. This fact, together with the finding that det $B^{-\frac{1}{2}}\cdot$ det $B \cdot$ det $B^{-\frac{1}{2}} =$ det $B^{-\frac{1}{2}}BB^{-\frac{1}{2}} =$ det $B^{-\frac{1}{2}} B^\frac{1}{2} B^\frac{1}{2} B^{-\frac{1}{2}} = $ det $\textbf{1} = 1$, informs that det $A \leq$ det $B$.
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