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.)
[Math] Cholesky decomposition with unit diagonal
cholesky decompositionlinear algebramatricesmatrix decompositionnumerical linear algebra
Related Solutions
What you described is a very indirect way of computing the QR decomposition (aka the Gram Schmidt process). Let us write the QR decomposition as $V = QR$. Then $V'V = R'Q'QR = R'R = LL'$. Therefore to recover the orthonormal $Q$, we can do $Q = VR^{-1}$, which is the last result. You're better off computing $V = QR$ directly.
A classical way is to use the induction. Let $A\in\mathbb{R}^{n\times n}$ be positive definite. It is trivial for $n=1$, just take the square root. Assume that a Cholesky factorization exists for positive definite matrices of dimension $n-1$ and partition $A$ as $$ A = \begin{bmatrix} \tilde{A}&a\\a^T&\alpha \end{bmatrix}, $$ where $\tilde{A}\in\mathbb{R}^{(n-1)\times(n-1)}$. Since a principal submatrix of a positive definite matrix is positive definite, $\tilde{A}$ has a Cholesky factorization $\tilde{A}=\tilde{L}\tilde{L}^T$. Consider $$\tag{1} L_1^{-1}AL_1^{-T} := \begin{bmatrix} \tilde{L}^{-1}&0\\ 0&1 \end{bmatrix} \begin{bmatrix} \tilde{A}&a\\ a^T&\alpha \end{bmatrix} \begin{bmatrix} \tilde{L}^{-T}&0\\ 0&1 \end{bmatrix} = \begin{bmatrix} I&b\\ b^T&\alpha \end{bmatrix} =:B, \quad b:=\tilde{L}^{-1}a. $$ Next we eliminate $b$ by $$\tag{2} L_2^{-1}BL_2^{-T} := \begin{bmatrix} I&0\\-b^T&1 \end{bmatrix} \begin{bmatrix} I&b\\ b^T&\alpha \end{bmatrix} \begin{bmatrix} I&-b\\0&1 \end{bmatrix} = \begin{bmatrix} I&0\\0&\alpha-b^Tb \end{bmatrix} = \begin{bmatrix} I&0\\0&\alpha-a^TA^{-1}a \end{bmatrix}. $$ The diagonal matrix on the right-hand side of (2) is a result of congruence transformations applied to $A$, so the right-hand side of (2) is positive definite and $0<\alpha-a^TA^{-1}a=\lambda^2$ for some real $\lambda$. Set $$ L_3:=\begin{bmatrix}I&0\\0&\lambda\end{bmatrix} $$ so $L_2^{-1}BL_2^{-T}=L_3L_3^T$. From (1) we have $$ L_2^{-1}L_1^{-1}AL_1^{-T}L_2^{-T}=L_3L_3^T, $$ so $$ A=LL^T, \quad L:=L_1L_2L_3 = \begin{bmatrix} \tilde{L}&0\\ 0&1 \end{bmatrix} \begin{bmatrix} I&0\\b^T&1 \end{bmatrix} \begin{bmatrix} I&0\\ 0&\lambda \end{bmatrix} = \begin{bmatrix} \tilde{L}&0\\ b^T&\lambda \end{bmatrix} = \begin{bmatrix} \tilde{L}&0\\ a^T\tilde{L}^{-T}&\lambda \end{bmatrix} $$ is a Cholesky factorization of $A$.
There is a source of non-uniqueness of the factorization in the choice of the sign of $\lambda$. As soon as one requires the signs of the diagonal terms of the Cholesky factors to be fixed (e.g., positive), the factorization is unique.
A simple way to confirm this can be made as follows. Assume $$ A=LL^T=MM^T $$ are two Cholesky factors of $A$. This gives $$\tag{3} I=L^{-1}MM^TL^{-T}=(L^{-1}M)(L^{-1}M)^T $$ and $$\tag{4} (L^{-1}M)=(L^{-1}M)^{-T}. $$ The left-hand and right-hand sides of (4) are, respectively, lower and upper triangular matrices which means that $D:=L^{-1}M$ is both lower and upper triangular and hence a diagonal matrix. From (3) we have $I=D^2$ so $D$ is a diagonal matrix with $\pm 1$ diagonal entries and $M=LD$ meaning that two Cholesky factors of $A$ differ by the signs of their columns.
Best Answer
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.