I am trying to find the Cholesky factorization $AA^T$ of the below covariance matrix $C$, to decompose a gaussian vector into independent standard normal random variables. However, the entry $a_{(3,3)}$ in the resultant matrix $A$ after my calculations seems off. I checked my work a couple of times over, and can't spot where the error is.
I would like to ask for some help, to check my work.
Problem 2.40. (IID Decomposition.) Consider the Gaussian vector $X=(X_{1},X_{2},X_{3})$ with mean $0$ and covariance matrix
\begin{align} C =\left[\begin{array}{ccc} 2 & 1 & 1\\ 1 & 2 & 1\\ 1 & 1 & 2 \end{array}\right] \end{align}
Find $Z=(Z_{1},Z_{2},Z_{3})$ and a matrix $A$ such that $X=AZ$.
My Solution.
Let $Z_{1}=\frac{X_{1}}{\sqrt{2}}$. Thus, $X_{1}=\sqrt{2}Z_{1}$.
Let $Z_{2}'$ be defined as:
\begin{align} Z_{2}' &=X_{2}-\mathbb{E}\left[X_{2}Z_{1}\right]Z_{1}\\ &=X_{2}-\frac{1}{\sqrt{2}}\mathbb{E}\left[X_{2}X_{1}\right]\frac{X_{1}}{\sqrt{2}}\\ &=-\frac{1}{2}X_{1}+X_{2} \end{align}
Now,
\begin{align} \mathbb{E}\left[\left(Z_{2}'\right)^{2}\right] &=\mathbb{E}\left[X_{2}^{2}-X_{1}X_{2}+\frac{1}{4}X_{1}^{2}\right]\\ &=2-1+\frac{1}{4}\cdot 2\\ &=\frac{3}{2} \end{align}
So,
\begin{align} Z_{2} &=\frac{Z_{2}'}{\sqrt{\mathbb{E}\left[\left(Z_{2}'\right)^{2}\right]}}\\ &=\sqrt{\frac{2}{3}}\left(-\frac{1}{2}X_{1}+X_{2}\right)\\ &=-\frac{1}{\sqrt{6}}X_{1}+\sqrt{\frac{2}{3}}X_{2} \end{align}
Thus,
\begin{align} Z_{2} &=-\frac{1}{\sqrt{3}}Z_{1}+\frac{\sqrt{2}}{\sqrt{3}}X_{2} \\ \sqrt{2}X_{2} &=Z_{1}+\sqrt{3}Z_{2}\\ X_{2} &=\frac{1}{\sqrt{2}}Z_{1}+\sqrt{\frac{3}{2}}Z_{2} \end{align}
Let $Z_{3}'$ be defined as:
$$Z_{3}' =X_{3}-\mathbb{E}[X_{3}Z_{1}]Z_{1}-\mathbb{E}[X_{3}Z_{2}]Z_{2}$$
We have,
\begin{align} \mathbb{E}[X_{3}Z_{1}]&=\frac{1}{\sqrt{2}}\mathbb{E}[X_{1}X_{3}]\\ &=\frac{1}{\sqrt{2}} \end{align}
And
\begin{align} \mathbb{E}[X_{3}Z_{2}] &=\mathbb{E}\left[X_{3}\left(-\frac{1}{\sqrt{6}}X_{1}+\sqrt{\frac{2}{3}}X_{2}\right)\right]\\ &=-\frac{1}{\sqrt{6}}\mathbb{E}[X_{1}X_{3}]+\frac{\sqrt{2}}{\sqrt{3}}\mathbb{E}[X_{2}X_{3}]\\ &=-\frac{1}{\sqrt{6}}+\frac{\sqrt{2}}{\sqrt{3}}\\ &=\frac{-\sqrt{3}+2\sqrt{3}}{3\sqrt{2}}\\ &=\frac{\sqrt{3}}{3\sqrt{2}}=\frac{1}{\sqrt{6}} \end{align}
Consequently,
$$Z_{3}' =X_{3}-\frac{1}{\sqrt{2}}Z_{1}-\frac{1}{\sqrt{6}}Z_{2}$$
Now,
\begin{align} \mathbb{E}[(Z_{3}')^{2}] &=\mathbb{E}\left[X_{3}^{2}+\frac{1}{2}Z_{1}^{2}+\frac{1}{6}Z_{2}^{2}-\sqrt{2}X_{3}Z_{1}-\sqrt{\frac{2}{3}}X_{3}Z_{2}+\frac{1}{2\sqrt{3}}Z_{1}Z_{2}\right]\\ &=2+\frac{1}{2}\cdot1+\frac{1}{6}\cdot1-\sqrt{2}\cdot\frac{1}{\sqrt{2}}-\sqrt{\frac{2}{3}}\cdot\frac{1}{\sqrt{6}}+\frac{1}{2\sqrt{3}}\cdot0\\ &=\frac{5}{3}-1-\frac{1}{3}=\frac{1}{3} \end{align}
Thus,
\begin{align} Z_{3} &=\sqrt{3}Z_{3}'\\ &=\sqrt{3}X_{3}-\sqrt{\frac{3}{2}}Z_{1}-\frac{1}{\sqrt{2}}Z_{2} \end{align}
And hence,
\begin{align} \sqrt{3}X_{3} &=\sqrt{\frac{3}{2}}Z_{1}+\frac{1}{\sqrt{2}}Z_{2}+Z_{3}\\ X_{3} &=\frac{1}{\sqrt{2}}Z_{1}+\frac{1}{\sqrt{6}}Z_{2}+\frac{1}{\sqrt{3}}Z_{3} \end{align}
Thus,
\begin{align} A =\left[\begin{array}{ccc} \sqrt{2} & 0 & 0\\ \frac{1}{\sqrt{2}} & \sqrt{\frac{3}{2}} & 0\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{array}\right] \end{align}