Given an SPD matrix $A$ and Cholesky decomposition $A = LL^T$.
Prove that $A(i,j) = 0$ means that $L(i,j) = 0$ and that $A(i,j) \neq 0$ means that $L(i,j) \neq 0$
I've been messing around with the formulas that determine the elements of the L matrix for too long now. I thought about proving this through induction and through a contradiction but nothing worked. Any help is appreciated.
This is not true. Consider $$ A=LL^T=\pmatrix{ 1&1&-1&1\\ 1&2&0&1\\ -1&\color{red}{\mathbf0}&3&0\\ 1&\color{forestgreen}{\mathbf1}&0&3}\ \text{ where } \ L=\pmatrix{ 1&0&0&0\\ 1&1&0&0\\ -1&\color{red}{\mathbf1}&1&0\\ 1&\color{forestgreen}{\mathbf0}&1&1}. $$ It is true, however, if $A$ is both entrywise positive and positive definite and you are talking about the entries of $L$ in the lower triangular part.