Proof of Every Invertible Real Matrix Is an Orthogonal Matrix Times a Positive Diagonal Matrix Times an Orthogonal Matrix

The matrix $A^{\top}A$ is positive definite. Let $B=A^{\top}A$. By Rules for Symmetric and Positive Semidefinite Matrices: Closure, Self-Adjointness, Polarization, Congruence and Cones (vi), $B\in\mathcal{S}(n)$ and $B$ is positive semidefinite. By The Determinant is Multiplicative (i) and Rules for the Determinant: Rows, Scalar Multiples and the Transpose (viii),

$$\det B=\det(A^{\top})\det A=(\det A)^{2},$$

which is nonzero because $\det A\neq0$ (A Square Real Matrix Is Invertible if and only if Its Determinant Is Nonzero (i)); so $B$ is invertible, by the same theorem, and therefore positive definite (Positive Semidefinite and Positive Definite Matrices by Their Eigenvalues (ii)).

Diagonalising $A^{\top}A$. By the spectral theorem (The Spectral Theorem for Real Symmetric Matrices (i)) there are $V\in\mathrm{O}(n)$ and $\lambda\in\mathbb{R}^{n}$ with

$$A^{\top}A=V\operatorname{diag}(\lambda)V^{\top}.$$

Each $\lambda_{k}$ is an eigenvalue of $A^{\top}A$ (The Spectral Theorem for Real Symmetric Matrices (ii)), hence $\lambda_{k}>0$ (Positive Semidefinite and Positive Definite Matrices by Their Eigenvalues (ii)). Let $d\in\mathbb{R}^{n}$ have components $d_{k}=\sqrt{\lambda_{k}}>0$, the square roots, let $d'\in\mathbb{R}^{n}$ have components $d'_{k}=1/d_{k}$, and write $D=\operatorname{diag}(d)$. The rules for diagonal matrices (Eigenvalues: the Determinant Criterion, Diagonal Matrices, Orthogonal Similarity and Orthogonality of Eigenvectors of Symmetric Matrices (ii)) give

$$D^{\top}=D,\qquad D^{2}=\operatorname{diag}(\lambda),\qquad D\operatorname{diag}(d')=\operatorname{diag}(d')D=I_{n},$$

so $D$ is invertible with $D^{-1}=\operatorname{diag}(d')$ (Inverse Matrix and Invertible Real Square Matrix (i)), and $(D^{-1})^{\top}=D^{-1}$.

The orthogonal factor $U$. Let $U=AVD^{-1}$. Using $V^{\top}V=I_{n}$ (Orthogonal Real Matrices (i)),

$$U^{\top}U=D^{-1}V^{\top}(A^{\top}A)VD^{-1}=D^{-1}(V^{\top}V)\operatorname{diag}(\lambda)(V^{\top}V)D^{-1}=D^{-1}D^{2}D^{-1}=I_{n},$$

so $U\in\mathrm{O}(n)$ (Orthogonal Real Matrices (iii)). Since $VV^{\top}=I_{n}$ (Rules for Orthogonal Matrices: Inverse, Products, Isometry, Columns, Determinant, Orthonormal Extension and Expansion (i)),

$$U\operatorname{diag}(d)V^{\top}=AV(D^{-1}D)V^{\top}=A(VV^{\top})=A .$$

This proves (i).

The determinant. Let $U,V\in\mathrm{O}(n)$ and $d\in\mathbb{R}^{n}$ with positive components satisfy $A=U\operatorname{diag}(d)V^{\top}$. By Rules for Orthogonal Matrices: Inverse, Products, Isometry, Columns, Determinant, Orthonormal Extension and Expansion (vi), $\det U,\det V\in\{1,-1\}$, so $|\det U|=|\det V|=1$. The matrix $\operatorname{diag}(d)$ has entries $d_{i}\delta_{ij}$ (Diagonal Matrix with Prescribed Diagonal (i)), so it is triangular with diagonal entries $d_{k}$, and

$$\det\operatorname{diag}(d)=\prod_{k=1}^{n}d_{k}>0$$

by The Determinant of a Triangular Matrix is the Product of its Diagonal Entries (ii). Multiplicativity of the determinant (The Determinant is Multiplicative (i)) and $\det V^{\top}=\det V$ (Rules for the Determinant: Rows, Scalar Multiples and the Transpose (viii)) give

$$|\det A|=|\det U|\,\det\operatorname{diag}(d)\,|\det V|=\prod_{k=1}^{n}d_{k},$$

which proves (ii).