Could you give a counter example of a symetric matrix for which the Cholesky factorization does not exist?
Also why does any eigenvalue solver have to be iterative?
2026-03-25 14:28:53.1774448933
Cholesky factorization counterexample
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If $M$ has a Cholesky decomposition $M=L^TL$, then $M$ must be symmetric positive-semidefinite, since $$x^TMx = (Lx)^T(Lx) = \|Lx\|^2\geq0.$$ Since every symmetric positive-semidefinite matrix has a Cholesky decomposition, this is a complete characterization.
As copper.hat points out in his comment, finding the eigenvalues of a matrix is equivalent to finding the roots of its characteristic polynomial. For matrices larger than $4\times 4$, by the Abel-Ruffini Theorem there does not exist a general formula for these roots in terms of addition, multiplication, and rational powers. That does not mean that a direct eigenvalue algorithm cannot exist -- but such an algorithm would have to be "sufficiently complex" and is not yet known.