Question 10.5: Compute the Cholesky factorization for the symmetric matrix ...

For the first row (i = 1), Eq. (10.15) is employed to compute

u_{ii} = \sqrt{a_{ii} − \sum\limits_{k=1}^{i−1}{u_{ki}^2}}                                                                            (10.15)
u_{11} = \sqrt{a_{11}} = \sqrt{6} = 2.44949
Then, Eq. (10.16) can be used to determine

u_{ij} = \frac {a_{ij} − \sum\limits_{k=1}^{i−1}{u_{ki}u_{kj}}}{u_{ii}}                            for j = i + 1, . . . , n             (10.16)
u_{12} =\frac{a_{12}}{u_{11}} = \frac{15}{2.44949} = 6.123724
u_{13} =\frac{a_{13}}{u_{11}} = \frac{55}{2.44949} = 22.45366
For the second row (i = 2):
u_{22} = \sqrt{a_{22} − u_{12}^2} = \sqrt{55 – (6.123724)^2} = 4.1833
u_{23} =\frac{{a_{23} −u_{12}u_{13}}}{u_{22}} = \frac{225 − 6.123724(22.45366)}{4.1833} = 20.9165
For the third row (i = 3):
u_{33} = \sqrt{a_{33} − u_{13}^2 − u_{23}^2} = \sqrt{979 − (22.45366)^2 − (20.9165)^2} = 6.110101
Thus, the Cholesky factorization yields
[U] =\begin{bmatrix} 2.44949 & 6.123724 & 22.45366 \\  & 4.1833 & 20.9165 \\  &  & 6.110101 \end{bmatrix}
The validity of this factorization can be verified by substituting it and its transpose into Eq. (10.14) to see if their product yields the original matrix [A]. This is left for an exercise.
[A] = [U]^T [U] \quad(10.14)