TITLE: How to calculate the cost of Cholesky decomposition?
QUESTION [7 upvotes]: The cost of Cholesky decomposition is $n^3/3$ flops (A is a $n \times n$ matrix). Could anyone show me some steps to get this number? Thank you very much.

REPLY [2 votes]: I found Section 1.6 ("Serial complexity of the algorithm") of the following webpage to be useful for this topic: https://algowiki-project.org/en/Cholesky_decomposition
Edit: Here's the info from that page.

1.6 Serial complexity of the algorithm
The following number of operations should be performed to decompose a matrix of order $n$ using a serial version of the Cholesky algorithm:

$n$ square roots
$\frac{n(n-1)}{2}$ divisions
$\frac{n^3-n}{6}$ multiplications and $\frac{n^3-n}{6}$ additions (subtractions): the main amount of computational work.

In the accumulation mode, the multiplication and subtraction operations should be made in double precision (or by using the corresponding function, like the DPROD function in Fortran), which increases the overall computation time of the Cholesky algorithm.
Thus, a serial version of the Cholesky algorithm is of cubic complexity.