Cholesky decomposition with unit diagonal 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.)
 A: 
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.
