The real Cholesky recurrence computes the positive-diagonal factor

Induct on $j$. At $j=1$, testing $e_1$ in [A1] gives $p_1=M_{11}>0$. For $j>1$, suppose columns $1,\dots ,j-1$ have been computed with positive diagonal and let $T=R_{\{1,\dots ,j-1\},\{1,\dots ,j-1\}}$ and $A=M_{\{1,\dots ,j-1\},\{1,\dots ,j-1\}}$; the induction hypothesis gives $A=T{T}^T$. The block $A$ is positive definite because extending any nonzero $x\in {\mathbb{R}}^{j-1}$ by zeros gives a nonzero vector to which [A1] applies. Its kernel is therefore zero; [F2] makes $A$ invertible. The lower-triangular matrix $T$ is invertible as well, since solving $Tx=0$ row by row with its positive diagonal forces every coordinate of $x$ to be zero. Put $m=(M_{1j},\dots ,M_{j-1,j}{)}^T$ and $u=(R_{j1},\dots ,R_{j,j-1}{)}^T$. The already computed recurrence at each column $i<j$ gives $M_{ij}={\sum }_{k<i}{R}_{ik}{R}_{jk}+R_{ii}{R}_{ji}$, so $Tu=m$. Since $A=T{T}^T$, this implies $m^T{A}^{-1}m=u^{T}u$. The leading $j\times j$ block $P$ of $M$ is positive definite by the same zero-extension argument. For the nonzero vector $(-A^{-1}m,1)$, its quadratic form under $P$ is $M_{jj}-m^T{A}^{-1}m=M_{jj}-u^{T}u=p_j>0$. Thus $R_{jj}=\sqrt{p_j}$ is positive and every division defining $R_{ij}$ is by a positive number. The recurrence is well-defined through column $j$, and the same formulas give the factorization identity on the leading $j\times j$ block.

For every $j$, the pivot equation gives $M_{jj}={\sum }_{k<j}{R}_{jk}^2+R_{jj}^2=(R{R}^T{)}_{jj}$. For every $i>j$, rearranging the recurrence gives $M_{ij}={\sum }_{k<j}{R}_{ik}{R}_{jk}+R_{ij}{R}_{jj}=(R{R}^T{)}_{ij}$. These identities fill the diagonal and lower triangle; symmetry of $M$ and $R{R}^T$ fills the upper triangle. The initialized matrix remains lower triangular, and each diagonal entry is positive by step 1.1, so it is the required Cholesky factor.