From wikipedia, given any matrix $A$, we can sometimes decompose $A = LU$ using Gaussian elimination. Other times, a permutation matrix is needed, giving $PA = LU$.
If $A$ is Hermitian positive-definite, I can show that IF no permutation matrix is needed, then Gaussian elimination gives $A=LU$ which I can eventually massage and get the Cholesky decomposition $A=LL^*$. However, it seems that Hermitian positive-definite matrices are special in that no permutaiton matrix is ever needed, and hence the Cholesky decomposition always exist. Why?