A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique

Suppose $A=L{L}^{\ast }$ as in [L1]. Then $A^{\ast }=L{L}^{\ast }=A$, so $A$ is Hermitian. For any nonzero $x$, one has $x^{\ast }Ax=x^{\ast }L{L}^{\ast }x=(L^{\ast }x{)}^{\ast }(L^{\ast }x)=\mathrm{\parallel }{L}^{\ast }x{\mathrm{\parallel }}_2^2\mathrm{.}$ The diagonal of $L$ is positive, hence nonzero; [L2] and [L3] give $\det L\mathrm{\ne }0$, so $L$ is invertible and $L^{\ast }x\mathrm{\ne }0$ when $x\mathrm{\ne }0$. Therefore $x^{\ast }Ax>0$.

For the converse, use induction on $n$. The case $n=1$ is immediate: $A=[a]$ is Hermitian positive definite exactly when $a>0$, in which case $A=[\sqrt{a}][\sqrt{a}{]}^{\ast }$.

Assume $n>1$ and the converse for size $n-1$. Write $A=\left(\begin{array}{cc}a& r^{\ast }\\ r& C\end{array}\right)\mathrm{.}$ Positive definiteness of $A$ at $x=e_0$ gives $a>0$. Set $\mathrm{\ell }:=r\mathrm{/}\sqrt{a}$ and $S:=C-\mathrm{\ell }{\mathrm{\ell }}^{\ast }$. For nonzero $y\in {\mathbb{C}}^{n-1}$, apply positive definiteness to $x=\left(\genfrac{}{}{0px}{}{-a^{-1}{r}^{\ast }y}{y}\right)$: $0<x^{\ast }Ax=y^{\ast }Sy\mathrm{.}$ Thus $S$ is Hermitian positive definite. By the induction hypothesis, $S=\tilde{L}{\tilde{L}}^{\ast }$ with $\tilde{L}$ lower triangular and positive diagonal.

Define $L:=\left(\begin{array}{cc}\sqrt{a}& 0\\ \mathrm{\ell }& \tilde{L}\end{array}\right)\mathrm{.}$ Direct multiplication gives $L{L}^{\ast }=\left(\begin{array}{cc}a& r^{\ast }\\ r& \mathrm{\ell }{\mathrm{\ell }}^{\ast }+\tilde{L}{\tilde{L}}^{\ast }\end{array}\right)=\left(\begin{array}{cc}a& r^{\ast }\\ r& C\end{array}\right)=A\mathrm{.}$ The matrix $L$ is lower triangular and its diagonal is positive real, so this is a Cholesky factorisation.

For uniqueness, suppose $A=L_1{L}_1^{\ast }=L_2{L}_2^{\ast }$ are two such factorizations. Then $M:=L_2^{-1}{L}_1$ is lower triangular, and $M{M}^{\ast }=I$ because $L_{2}M{M}^{\ast }{L}_2^{\ast }=L_1{L}_1^{\ast }=L_2{L}_2^{\ast }$. A lower-triangular unitary matrix has zero strict-lower entries and diagonal entries of modulus $1$; since the diagonal of $M$ is the ratio of two positive real diagonals, each diagonal entry is also positive real, hence equals $1$. Therefore $M=I$ and $L_1=L_2$.

Step 1.1 proves that a Cholesky factorisation implies Hermitian positive definiteness. Steps 1.2-2.1 prove the converse by induction, and step 3.1 gives uniqueness.