# How to prove the existence and uniqueness of Cholesky decomposition?

Given a real Hermitian positive-definite matrix $A$ is a decomposition of the form $A=L L^T$ where L is a lower triangular matrix with positive diagonal entries. I read some proofs about the existence of Cholesky decomposition. Most of them start from LDU decomposition. Then the proof shows that $U^T=L$ and $A=LDU=LD^{\frac{1}{2}} D^{\frac{1}{2}}L^T=CC^T$ where $C=L D^\frac{1}{2}$.

How can I prove the existence of Cholesky decomposition without any preassumption like LDU decomposition exists? Or how can I prove LDU decomposition exists? I know it may be easy. But I just cannot figure it out.

For uniqueness, I think it's not hard to prove.

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.