Why does the Cholesky factorization requires the matrix A to be positive definite? What happens when we factorize non-positive definite matrix?
Let's assume that we have a matrix A' that is not positive definite (so at least one leading principal minor is negative). Can one prove that there is no L such as A' = LL*? If not, wouldn't the positive definite criteria remove some of the matrices that could be potentially decomposed?
We could also put this question in the form of a demonstration for the next statement: For any square matrix L, the product LL* is a positive definite matrix.
Best Answer
Suppose a matrix $A$ factors as $A = L^* L$. Then \begin{align} x^* Ax &= x^* L^* L x \\ &= (Lx)^* (Lx) \\ &= \| Lx \|^2 \\ &\geq 0. \end{align} This shows that $A$ is positive semidefinite.
If we further assume that $L$ is square and triangular with positive real diagonal entries, then $L$ is invertible, so $Lx = 0 \iff x = 0$. In this case, we see that $A$ is positive definite.