According to the Cholesky factorization on Wikipedia, the computational complexity of it is $\frac{n^3}{3}$ FLOPs where $n$ is the size of the considered matrix $\mathbf{A}$.
There are various methods for calculating the Cholesky decomposition. The computational complexity of commonly used algorithms is $\mathcal{O}(n^3)$ in general.The algorithms described below all involve about $n^3/3$ FLOPs ($n^3/6$ multiplications and the same number of additions), where $n$ is the size of the matrix A. Hence, they have half the cost of the LU decomposition, which uses $2n^3/3$ FLOPs (see Trefethen and Bau 1997).
To more understand how this number comes out, I was using a $3\times3$ real symmetric matrix, which can be found from this lecture note, pp.12.17--12.18.
For convenience, let me assume the computational complexity (or the floating point operation, FLOP) for multiplication, addition/subtraction, division, and square-root is $1$.
The matrix $\mathbf{A}$ is defined as
$$ \mathbf{A} = \begin{bmatrix} 25 & 15 & -5\\15 & 18 & 0\\-5 & 0 & 11\end{bmatrix} = \begin{bmatrix} R_{11} & 0 & 0\\R_{12} & R_{22} & 0\\R_{13} & R_{23} & R_{33} \end{bmatrix} \begin{bmatrix} R_{11} & R_{12} & R_{13}\\ 0 & R_{22} & R_{23} \\ 0 & 0 & R_{33} \end{bmatrix}\ . $$
We need to compute the following to find $R_{11}$:
$$R_{11}^2 = 25 \rightarrow R_{11} = 5,$$
and the plus 5 is chosen for convenience.
From the equation above, we easily can see that one squareroot is required to compute $R_{11}$.
For $R_{12}$ and $R_{13}$, we might compute the following:
$$ R_{12} = 15R_{11}^{-1} = 3, $$
$$ R_{13} = -5R_{11}^{-1} = -1, $$
which requires one division, respectively.
In total, so far, a total of $3$ FLOPs is required for $R_{11}$, $R_{12}$, and $R_{13}$.
Let me write the rest elements below.
$$R_{22} = \sqrt{18-R_{12}^2} = 3,$$
$$R_{23} = \frac{0-R_{13}R_{12}}{R_{22}} = 1,$$
$$R_{33} = \sqrt{11 - R_{13}^2 - R_{23}^2} = 3,$$
Therefore, it is a total of 14 FLOPS for the Cholesky factorization, which can be found in the following table:
| Element | $\times$ | $+, -$ | $\div$ | $\sqrt{\cdot}$ |
|---|---|---|---|---|
| $R_{11}$ | $0$ | $0$ | $0$ | $1$ |
| $R_{12}$ | $0$ | $0$ | $1$ | $0$ |
| $R_{13}$ | $0$ | $0$ | $1$ | $0$ |
| $R_{22}$ | $1$ | $1$ | $0$ | $1$ |
| $R_{23}$ | $1$ | $1$ | $1$ | $0$ |
| $R_{33}$ | $2$ | $2$ | $0$ | $1$ |
| Total | $4$ | $4$ | $3$ | $3$ |
It should have been $9$ FLOPs according to $n^3/3$, but the FLOPs that were obtained were $14$.
Is there any miscomputation? Or there should be something that I am missing or misunderstanding?