Cholesky factorization

This is a special case for LU factorization ($U=L^T$).
$A=L{L}^T$
where

• $A$ $m\times n$ is SPD

2x2 example

$$L=\left[\begin{array}{cc}x& 0\\ y& z\end{array}\right]$$ $$L^T=\left[\begin{array}{cc}x& y\\ 0& z\end{array}\right]$$

so

$$A=L{L}^T=\left[\begin{array}{cc}{x}^2& xy\\ xy& y^2+z^2\end{array}\right]$$

$\square$

Costs

Only lower triangular part of $A$ is accessed, so only $\frac{1}{2}$ the storage is required.
Only $\frac{n^3}{6}$ multiplications and additions required, so $\frac{1}{2}$ the work of LU factorization
No need to pivoting