Understanding the Cholesky decomposition I'm attempting to understand the Cholesky decomposition via the following site: http://en.wikipedia.org/wiki/Cholesky_decomposition
If I have a matrix, say 
$$A = \begin{bmatrix} 
2  & -1 & 0\\
-1 & 2 & -1\\
0  & -1 & 2\end{bmatrix},$$ then I'd like to use the Cholesky Algorithm to find a matrix $D$ such that $A = LDL^{T}$.
I left blank the parts of this example that I do not understand how to find.
The Cholesky Algorithm  for this example follows:
Skipping a few steps since we don't need to derive how they get $L$, we know that $L = L_{1}L_{2}L_{3}$ since $n=3$.
For $i=1$, 
$$L_1 = \begin{bmatrix} 
I_{1-1}  & 0 & 0\\
0 & \sqrt{a_{1,1}} & 0\\
0 & b_1/\sqrt{a_{1,1}} & I_{3-1}\end{bmatrix} = 
\begin{bmatrix} 
0  & 0 & 0 & 0\\
0 & \sqrt{2} & 0 & 0\\
0 & b_1/\sqrt{2} & 1 & 0\\
0 & 0 &             0 & 1\end{bmatrix}$$
For $i=2$, 
$$L_ = \begin{bmatrix} 
I_{2-1}  & 0 & 0\\
0 & \sqrt{a_{2,2}} & 0\\
0 & b_2/\sqrt{a_{2,2}} & I_{3-2}\end{bmatrix} = 
\begin{bmatrix} 
1  & 0 & 0 \\
0 & \sqrt{2} & 0\\
0 & b_2/\sqrt{2} & 1\end{bmatrix}$$
For $i=n = 3$, 
$$L_3 = \begin{bmatrix} 
I_{3-1}  & 0 & 0\\
0 & \sqrt{a_{3,3}} & 0\\
0 & b_3/\sqrt{a_{3,3}} & I_{3-3}\end{bmatrix} = 
\begin{bmatrix} 
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & \sqrt{2} & 0\\
0 & 0 & b_3/\sqrt{2} & 0\end{bmatrix}$$.
Thus, 
\begin{align}
L &:= L_1\cdot L_2\cdot L_3\\\notag
 &= \begin{bmatrix} 
 0  & 0 & 0 & 0\\
 0 & \sqrt{2} & 0 & 0\\
 0 & b_1/\sqrt{2} & 1 & 0\\
 0 & 0 &             0 & 1\end{bmatrix} \cdot 
 \begin{bmatrix} 
 1  & 0 & 0 \\
 0 & \sqrt{2} & 0\\
 0 & b_2/\sqrt{2} & 1\end{bmatrix} \cdot 
 \begin{bmatrix} 
 1 & 0 & 0 & 0\\
 0 & 1 & 0 & 0\\
 0 & 0 & \sqrt{2} & 0\\
 0 & 0 & b_3/\sqrt{2} & 0\end{bmatrix}\\\notag
 &= ?
\end{align}
I guess $b_i$ and $b_i^*$ come from $B$.  But how do I find the non-Hermitian matrix $B$ to finish this process?
 A: Wikipedia’s probably not the ideal place to learn about the Cholesky decomposition or of the various algorithms that can be used to generate the decomposition.  I’d recommend reading a textbook on numerical linear algebra or general book on numerical mathematics.  That being said, I’ve gone ahead and fixed up your steps in the outer-product Cholesky algorithm example that you’ve chosen to follow.  
For the record, I’m using the notation from the following snapshot of the Wikipedia article on the Cholesky decomposition: 
https://en.wikipedia.org/w/index.php?title=Cholesky_decomposition&oldid=711015429
Before I run through the algorithm, however, note that $I_i$ denotes the $i \times i$ identity matrix. $I_0$, in particular, refers to a $0 \times 0$ matrix, so when it occurs in an algorithm step it gets omitted from the matrix block partition.  Also, all matrices $A^{(i)}$ and $L_i$ should have the same size as the original matrix $A$.  Last, all elements $a_{ii}$, $b_i$, and $B$ are pulled directly from matrix $A^{(i)}$ using the partitioning described at the Wikipedia article.  
$A = A^{\left( 1 \right)}  = \left( {\begin{array}{*{20}c}
   2 & { - 1} & 0  \\
   { - 1} & 2 & { - 1}  \\
   0 & { - 1} & 2 \end{array}} \right)$, so $a_{11} = 2$, $b_1^T = \left( {\begin{array}{*{20}c} { - 1} & 0\end{array}} \right)$, and $B^{(1)}  = \left( {\begin{array}{*{20}c}
   2 & { - 1}  \\
   { - 1} & 2 \end{array}} \right)$.
Therefore, $L_1  = \left( {\begin{array}{*{20}c}
   {\sqrt 2 } & 0 & 0  \\
   {\frac{{ - 1}}{{\sqrt 2 }}} & 1 & 0  \\
   0 & 0 & 1\end{array}} \right)$ and $A^{\left( 2 \right)}  = \left( {\begin{array}{*{20}c}
   1 & 0 & 0  \\
   0 & {\frac{3}{2}} & { - 1}  \\
   0 & { - 1} & 2\end{array}} \right)$.
Continuing on, we see that $a_{22} = \frac{3}{2}$, $b_2 = -1$, and $B^{(2)} = 2$, so
$L_2  = \left( {\begin{array}{*{20}c}
   1 & 0 & 0  \\
   0 & {\sqrt {\frac{3}{2}} } & 0  \\
   0 & { - \sqrt {\frac{2}{3}} } & 1\end{array}} \right)$ and $A^{\left( 3 \right)}  = \left( {\begin{array}{*{20}c}
   1 & 0 & 0  \\
   0 & 1 & 0  \\
   0 & 0 & {\frac{4}{3}}\end{array}} \right)$.
From $A^{\left( 3 \right)}$ we immediately write down $L_3  = \left( {\begin{array}{*{20}c}
   1 & 0 & 0  \\
   0 & 1 & 0  \\
   0 & 0 & {\sqrt {\frac{4}{3}} }\end{array}} \right)$.
Therefore, $L = L_1 L_2 L_3  = \left( {\begin{array}{*{20}c}
   {\sqrt 2 } & 0 & 0  \\
   {\frac{{ - 1}}{{\sqrt 2 }}} & {\sqrt {\frac{3}{2}} } & 0  \\
   0 & { - \sqrt {\frac{2}{3}} } & {\sqrt {\frac{4}{3}} }\end{array}} \right)$ which you can confirm satisfies $LL^T  = A$.
