Why does the Cholesky decomposition requires a positive definite matrix? 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.
 A: 
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.

Semidefinite.
Let's go with the definition:
$$x^* L L^* x = (L^*x)^* (L^*x) = y^* y, \quad y := L^*x.$$
By the definition of the Euclidean scalar product, $y^* y = \langle y, y \rangle \ge 0$, with the equality if and only if $0 = y = L^* x$.
So, $x^* L L^* x \ge 0$ for all $x$, hence $L L^*$ is positive semidefinite. Furthermore, it is positive definite if
$$L^*x = 0 \quad \Leftrightarrow \quad x = 0,$$
i.e., if $L$ is of a full row rank.
A: Just consider the case of numbers $\Bbb M_1(\Bbb R)$. If we have a negative number $a<0$, then its Cholesky decomposition doesn't exist:
$$ll^\ast  = |l|^2 = a<0.$$
A: 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.
