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I am studying the lecture 23 in Numerical Linear Algebra book and I cannot follow the part that explains the Cholesky Factorization's algorithm. Specifically it is written:

When Cholesky factorization is implemented, only half of the matrix being operated on needs to be represented explicitly. This simplification allows half of the arithmetic to be avoided. A formal statement of the algorithm (only one of many possibilities) is given below. The input matrix $A$ represents the super-diagonal half of the $m\times m$ hermitian positive definite matrix to be factored. The output matrix $R$ represents the upper triangular factor for which $A=R^{*}R$. Each outer iteration correspond to single elementary factorization: the upper-triangular part of the sub-matrix $R^{*}_{k:m,k:m}$ represents the super diagonal part of the hermitian matrix being factored at step $k$.

$\underline{\text{Algorithm 23.1: Chelosky Factorization}}$

$R=A$

for $k = 1$ to $m$

$\quad$for $j = k+1$ to $m$

$\quad\quad\quad R_{j,j:m} = R_{j,j:m} - R_{k,j:m}\bar{R}_{kj}/R_{kk}$

$\quad R_{k,k:m} = \frac {R_{k,k:m}} { \sqrt{R_{k,k}}}$

Since I did not completely understand the explanation, I traced the algorithm with an example. At the end, the algorithm did not make the sub-diagonal entries of $A$ zero and only super diagonals were correct. But from what I understood the output , $R$ , should be upper triangular.

Does the explanation before the algorithm means: we need to zero out all sub diagonal entries of $A$ before entering the first for loop?

Any insight would be appreciated. Thank you.

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    $\begingroup$ Yes. If you want a "fancy" $R$, you can, e.g., append $R_{k+1:m,k}=0$ at the end of the outer loop. $\endgroup$ – Algebraic Pavel Oct 7 '16 at 8:17

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