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Let $A$ be real-valued (strictly) positive definite (P.D.) so that it has a unique Cholesky decomposition of the form $A = LL^T$ where $L$ is lower triangular. What PD matrices $A$ have a Cholesky decomposition where the diagonal entries of $L$ are all ones? (A necessary and sufficient characterization would be nice.)

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Let $A\in\mathbb{R}^{n\times n}$ be symmetric and positive definite (SPD) and let $A=R^TR$ be its Cholesky decomposition such that $R$ is upper triangular with positive diagonal entries. Then $R$ is a unit upper triangular matrix (with diagonal entries equal to $1$) if and only if $\det(B)=1$ for each leading principal submatrix (LPS) $B$ of $A$.

This can be proved by induction. It is trivial for $n=1$ so assume that it is true for $n-1$ and consider a partitioning of an $n\times n$ SPD matrix $A$ in the form $$ A=\pmatrix{B&c\\c^T&\delta}, $$ where $B$ is $(n-1)\times(n-1)$. Assume a conforming partitioning of the Cholesky factor $R$ in the form $$ R=\pmatrix{S&t\\0&\mu}. $$ You can easily check that $B=S^TS$ is the Cholesky factorization of $B$.

Let all leading principal submatrices of $A$ have unit determinant. By the induction hypothesis, since $B$ is a LPS of $A$, $S$ has unit diagonal. But $\det(A)=1$ ($A$ is also its own LPS) $$ 1=\det(A)=\det(S)^2\mu^2=\mu^2. $$ Since we fix the Cholesky factor to have positive diagonal, we have hence $\mu=1$.

The other direction of the equivalence can be shown similarly.

Example The matrix $$ A=\pmatrix{1&2&3\\2&5&7\\3&7&11} $$ satisfies the conditions on the LPS: $$ \det(1)=\det\pmatrix{1&2\\2&5}=\det\pmatrix{1&2&3\\2&5&7\\3&7&11}=1. $$ Its Cholesky factor $$ R=\pmatrix{1&2&3\\0&1&1\\0&0&1} $$ has indeed a unit diagonal.

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