Cholesky decomposition how to prove

Question

From calculation of the Cholesky decomposition of covariance matrix, prove whether a symmetric matrix being positive or not can be determined from signs of principal diagonal minors.

Answer 1

The Cholesky decomposition $$ C=LL^\top\,,\quad\text{ where }\quad L=\begin{pmatrix}l_{11}&0&\dots&\dots&0\\l_{21}&l_{22}&0&\dots&0\\\vdots&\vdots&\ddots&\ddots&\vdots\\ l_{n-1,1}&l_{n-1,2}&\dots&\dots&0\\ l_{n1}&l_{n2}&\dots&\dots&l_{nn}\end{pmatrix} $$ of a symmetric real matrix $C$ exists if and only if $C$ is positive semi-definite.

IMO, the properly formulated question is therefore: Given that $C$ is positive semi-definite how can we use $L$ to determine wheter $C$ is strictly positive definite ?

Using $\det(AB)=\det A\det B$ where $A$ is any left upper submatrix of $L$ (aka principal minor) and $B=A^\top$ we find that the principal minors of $C$ are $$ \det C_1=l_{11}^2\,,\quad\det C_2=(l_{11}l_{22})^2\,,\dots, \det C=(l_{11}\dots l_{nn})^2\,. $$ By Sylvester's criterion if $C$ were only symmetric the strict positivity of these determinants would be a necessary and sufficient criterion for $C$ to be strictly positive definite.

This is obviously the case if and only if none of the $l_{ii}$ is zero.

• Note that only the full determinant $$ \det C=(l_{11}\dots l_{nn})^2 $$ needs to be strictly greater zero. In other words, Sylvester's criterion is not needed in full generality since we assume that the eigenvalues of $C$ are nonnegative from the beginning.