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lets say, after Cholesky factorization, at some point in my solution I get this:

$L_{11}*L_{11}^T=A$, $L_{11}*L_{21}^T=u$, $L_{21}*L_{11}=u^T$, and $L_{21}*L_{21}^T=a$

So, $a$ is a scalar, $u$ and $u^T$ are vectors, and $A=L*L^T$ the complexity of $A$'s is given. Now i need to express $L_{11}, L_{11}^T, L_{21}, L_{21}^T$ with $A (or L,L^T), u, and u^T$

Basically the formulas you see are first Cholesky factorization of $B_{(n+1)(n+1)}$ = $ \begin{bmatrix} A & u \\ u^T & 1 \\ \end{bmatrix}$

The formulas are found from here: $B_{(n+1)(n+1)}$ = $ \begin{bmatrix} A & u \\ u^T & 1 \\ \end{bmatrix}$ = $\begin{bmatrix} L_{11} & 0 \\ L_{21} & l_{22} \\ \end{bmatrix} $ $\begin{bmatrix} L_{11}^T & L_{21}^T \\ 0 & l_{22} \\ \end{bmatrix} $

So, i need to do this Cholesky decompostion on B, after decomposition i have to find the complexity of it (given that A has a complexity N)

I hope this is more clear, Thanks

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I dont understand " L is a given as it is some square matrix with some complexity N" –  Inquest Oct 28 '12 at 19:48
    
I cleared the problem for you :) –  ASROMA Oct 28 '12 at 20:44
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