Proof of Cholesky Factorisation of a Symmetric Positive Definite Real Matrix

By Symmetric, Positive Semidefinite, and Positive Definite Real Matrices, $G$ is symmetric and satisfies $0<c\cdot(Gc)$ for every nonzero $c$ in Euclidean space $\mathbb{R}^{n}$, with the dot product and the matrix-vector product. These are exactly the hypotheses on $G$ in Triangular Orthonormalization of a Positive Definite Gram Matrix, and the notion of lower triangular matrix used there agrees with the one used here.

By claim 1 of Triangular Orthonormalization of a Positive Definite Gram Matrix there is a lower triangular real $n\times n$ matrix $T$ with $0<T_{ii}$ for every $i\in[n]$ such that

$$T\,G\,T^{\top}=I_{n},$$

with the identity matrix $I_{n}$. By claim 2 of that lemma, $T$ is invertible and $L=T^{-1}$ is lower triangular with $0<L_{ii}$ for every $i\in[n]$.

By claim 6 of Elementary Properties of the Transpose of a Real Matrix, $T^{\top}$ is invertible with inverse $(T^{-1})^{\top}=L^{\top}$, so $T^{\top}L^{\top}=I_{n}$; and $L\,T=I_{n}$ by the definition of the inverse. Multiplying the displayed identity on the left by $L$ and on the right by $L^{\top}$, and regrouping by Associativity of the Matrix Product,

$$L\,I_{n}\,L^{\top}=L\,\bigl(T\,G\,T^{\top}\bigr)\,L^{\top}=(L\,T)\,G\,(T^{\top}L^{\top})=I_{n}\,G\,I_{n}.$$

By The Identity Matrix is a Two-Sided Multiplicative Identity the left-hand side is $L\,L^{\top}$ and the right-hand side is $G$. Hence $G=L\,L^{\top}$, and $L$ is a lower triangular matrix with positive diagonal entries as required.