Proof of The Determinant of a Positive Definite Matrix Is Positive

Routine facts about real numbers are used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity).

The determinant of $\Sigma$ is nonzero. A positive definite matrix is invertible (Invertibility of Positive Definite Matrices (i)), and an invertible matrix has nonzero determinant (A Square Real Matrix Is Invertible if and only if Its Determinant Is Nonzero (i)); hence $\det\Sigma\neq0$.

Factors of $\Sigma$. Let $A\in\mathcal{M}_{n}(\mathbb{R})$ satisfy $AA^{\top}=\Sigma$. Since the determinant is multiplicative (The Determinant is Multiplicative (i)) and unchanged by transposition (Rules for the Determinant: Rows, Scalar Multiples and the Transpose (viii)),

$$\det\Sigma=\det\bigl(AA^{\top}\bigr)=\det A\cdot\det A^{\top}=(\det A)^{2}.$$

As $\det\Sigma\neq0$ by The determinant of $\Sigma$ is nonzero, also $\det A\neq0$, so $A$ is invertible (A Square Real Matrix Is Invertible if and only if Its Determinant Is Nonzero (i)). Finally, the square root of a square is the absolute value (Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits (ii)), so

$$\sqrt{\det\Sigma}=\sqrt{(\det A)^{2}}=|\det A|,$$

which proves (ii).

Positivity of $\det\Sigma$. The matrix $\Sigma$ is positive semidefinite (Rules for Symmetric and Positive Semidefinite Matrices: Closure, Self-Adjointness, Polarization, Congruence and Cones (x)), so it has a positive semidefinite square root $S=\Sigma^{1/2}$; this square matrix is symmetric with $SS=\Sigma$, hence $SS^{\top}=SS=\Sigma$. The computation of Factors of $\Sigma$, applied to $A=S$, gives $\det\Sigma=(\det S)^{2}$ with $\det S\neq0$. The square of a nonzero real number is positive, so $\det\Sigma>0$, which is (i).