Added notes of lecture 6
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% vim: set ts=2 sw=2 et tw=80:
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\documentclass[12pt,a4paper]{article}
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\usepackage[utf8]{inputenc} \usepackage[margin=2cm]{geometry}
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\usepackage{amstext}
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\usepackage{amsmath}
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\usepackage{array}
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\usepackage[utf8]{inputenc}
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\usepackage[margin=2cm]{geometry}
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\usepackage{amstext}
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\usepackage{array}
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\newcommand{\lra}{\Leftrightarrow}
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\newcolumntype{L}{>{$}l<{$}}
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\DeclareMathOperator*{\argmax}{arg\,max}
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\DeclareMathOperator*{\argmin}{arg\,min}
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\title{Lecture notes 6 -- Introduction to Computational Science}
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\author{Micheal Multerer \\ Copied by: Claudio Maggioni}
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\begin{document}
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\maketitle
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\section*{2.5: Partial pivoting}
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Obviously, Algorithm 2.8 fils if one of the pivots becomes zero. In this case, we need to choose a different pivot element.
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\paragraph{Simple approach:} Column pivoting choose $|a^{(i)}_{k,i}|=\max_{i \leq l \leq n}|a^{(i)}_{l,i}|$
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In order to move the pivot element and the corresponding row we switch row k and row i by a \textit{primitive matrix}:
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\[
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\bar{P}_i := \begin{bmatrix}
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1_1 \\
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& \ddots \\
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&& 1_{i-1} \\
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&&& 0 & \hdots & 1_k \\
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&&&\vdots &1_{i+1} \\
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&&& 1_i && 0 \\
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&&&&&& \ddots \\
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&&&&&&& 1_n \\
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\end{bmatrix}
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\]
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The following rules apply:
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\begin{enumerate}
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\item Multiplication by $P_i$ from the left $\Rightarrow$ interchange rows i and k
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\item Multiplication by $P_i$ from the right $\Rightarrow$ interchange columns i and k
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\item $(\bar{P}_i)^2 = \bar{P}_i \cdot \bar{P}_i = \bar{I}$
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\end{enumerate}
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Then performing an LU decompisition with pivoting can be
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written in matrix notation as:
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$$A_{i+1} = L_i \cdot \bar{P}_i \cdot A_i$$
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\paragraph{Note:} For $j < i$, it holds $\bar{P}_i \bar{L}_j = \widetilde{L}_j \bar{P}_i$ where $\widetilde{L}$ is the same matrix as $\bar{L}_j$ except that $[\widetilde{l}_j]_i$ and $[\widetilde{L}_j]_k$ are interchanged:
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\[
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\bar{L}_i := \begin{bmatrix}
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1\\
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&\ddots\\
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&&1\\
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&&&\hat{l}^{(j)}_2\\
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&&&\hat{l}^{(j)}_i&&1\\
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&&&\hat{l}^{(j)}_k&&&\ddots\\
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&&&\hat{l}^{(j)}_n&&&&1\\
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\end{bmatrix}
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\text{\hspace{1cm}}
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\widetilde{L}_i := \begin{bmatrix}
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1\\
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&\ddots\\
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&&1\\
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&&&\hat{l}^{(j)}_2\\
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&&&\hat{l}^{(j)}_k&&1\\
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&&&\hat{l}^{(j)}_i&&&\ddots\\
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&&&\hat{l}^{(j)}_n&&&&1\\
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\end{bmatrix}
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\]
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Resolving (*) then yields:
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$$\bar{A}_{i+1} = \bar{L}_i \bar{P}_i \bar{L}_{i-1} \bar{P}_{i-1} \ldots \bar{L}_1 \bar{P}_1 A$$
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or
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$$\bar{L}_{n-1} \bar{P}_{n-1} \bar{L}_{n-2} \bar{P}_{n-2} \ldots \bar{L}_1 \bar{P}_1 \bar{A} = \bar{U}$$
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Now we can exploit $P_2 \bar{L}_1 \bar{P}_1 = \widetilde{L}_1 \bar{P}_2 \bar{P}_1$ and so on. This yields:
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$$\bar{P}\bar{A} = \bar{L}\bar{U}$$
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with
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$$\bar{P} = \bar{P}_{n-1} \bar{P}_{n-2} \ldots \bar{P}_1$$
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and
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$$\widetilde{L} = \widetilde{L}^{-1}_1 \ldots \widetilde{L}^{-1}_{n-1}$$
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and $$ \widetilde{L}_{n-1} = \bar{L}_{n-1} $$$$
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\widetilde{L}_{n-2} = \bar{P}_{n-1} \bar{L}_{n-2} \bar{P}_{n-1} $$$$
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\vdots $$$$
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\widetilde{L}_1 = \bar{P}_{n-1} \bar{P}_{n-2} \ldots \bar{P}_{2} \bar{L}_1 \bar{P}_2 \ldots \bar{P}_{n-2} \bar{P}_{n-1} $$
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\paragraph{Note:} If $\bar{A} \in R^{n \times n}$ is non-singular, the pivoted LU decomposition $\bar{P}\bar{A} = \bar{L}\bar{U}$ always exists.
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\\\\
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We can easily add column priority in Algorithm 2.8:
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\paragraph{Algorithm 2.9}(Outer product LU decomposition with column pivoting) \\
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input: matrix $\bar{A} = [a_{i,j}]^n_{i,j = 1} \in R^{n\times n}$ \\
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output: pivoted LU decomposition $ \bar{L}\bar{U} = \bar{P}\bar{A}$
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\begin{enumerate}
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\item Set $\bar{A}_1 = \bar{A}, \bar{p} = [1,2,\ldots,n]$
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\item For $i = 1,2,\ldots,n$ \begin{itemize}
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\item compute: k = $\argmax_{1 \leq j \leq n} | a^{(i)}_{p_j,i} |$ \% find pivot
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\item swap: $ p_i \gets\to p_k$
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\item $\bar{l}_i := \bar{a}^{(i)}_{:,i} / a^{(i)}_{p_i,p_i}$
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\item $\bar{u}_i := a_{p_i,:}$
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\item compute: $\bar{A}_{i+1} = \bar{A}_i - \bar{l}_i \cdot \bar{u}_i$
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\end{itemize}
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\item set $\bar{P} := [\bar{l}_{p_1},\bar{l}_{p_2},\ldots, \bar{l}_{p_n}]^T$ \% $\bar{e}_i$ is i-th unit vector
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\item set $bar{L} = bar{P}[\bar{l}_1,\bar{l}_2,\cdots,\bar{l}_n]$
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\end{enumerate}
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\paragraph{Example 2.10} \textit{(omitted)}
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\section*{2.6: Cholesky decomposition}
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If $\bar{A}$ is symmetric and positive definite, i.e. all eigenvalues of $\bar{A}$ are bigger than zero or equivalently $bar{x}^T\bar{A}\bar{x} > 0$ for all $\bar{x} \neq 0$, we can compute a symmetric decomposition of $\bar{A}$.
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\paragraph{Note:} if $\bar{A}$ is symmetric and positive definite, then the \textit{Schur complement} $\bar{S}:=\bar{A}_{2:n,2:n} - (\bar{a}_{2:n},1/a_{1,1})\bar{a}^T_{2:n,1}$, is symmetric and positive definite as well. In particular, it holds $s_{i,i} > 0$ and $a_{i,i} > 0$ !
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\paragraph{Definition 2.11} A decomposition $\bar{A} = \bar{L}\bar{L}^T$ with a lower triangular matrix $\bar{L}$ with positive diagonal elements is called \textit{Cholesky decomposition of $\bar{A}$}.
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\paragraph{Note:} A Cholesky decomposition exists, if $\bar{A}$ is symmetric and positive definite.
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\paragraph{Algorithm 2.12} (outer product of Cholesky decomposition) \\
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input: matrix $\bar{A}$ symmetric and positive definite \\
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output: Cholesky decomposition $\bar{A} = \bar{L}\bar{L}^T =
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[\bar{l}_1,\bar{l}_2,\ldots,\bar{l}_n] [\bar{l}_1,\bar{l}_2,\ldots,\bar{l}_n]^T$
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\begin{enumerate}
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\item set: $\bar{A}_1 := \bar{A}$
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\item for $i = 1,2,\ldots,n$ \begin{itemize}
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\item set: $\bar{l}_i := a^{(i)}_{:,i} / \sqrt{a^{(i)}_{i,i}}$
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\item set: $\bar{A}_{i+1} := \bar{A}_{i} - \bar{l}_i\bar{l}^T_i$
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\end{itemize}
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\item set: $\bar{L} = [\bar{l}_1,\bar{l}_2,\ldots,\bar{l}_n]$
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\end{enumerate}
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The computational cost is $\frac{1}{6}n^{3}+O(n^{2})$ and thus only half the cost of LU decomposition.
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\paragraph{Example 2.13} \textit{(omitted)}
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\end{document}
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