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command---while reading file lecture_12.aux +You've used 0 entries, + 0 wiz_defined-function locations, + 83 strings with 494 characters, +and the built_in function-call counts, 0 in all, are: += -- 0 +> -- 0 +< -- 0 ++ -- 0 +- -- 0 +* -- 0 +:= -- 0 +add.period$ -- 0 +call.type$ -- 0 +change.case$ -- 0 +chr.to.int$ -- 0 +cite$ -- 0 +duplicate$ -- 0 +empty$ -- 0 +format.name$ -- 0 +if$ -- 0 +int.to.chr$ -- 0 +int.to.str$ -- 0 +missing$ -- 0 +newline$ -- 0 +num.names$ -- 0 +pop$ -- 0 +preamble$ -- 0 +purify$ -- 0 +quote$ -- 0 +skip$ -- 0 +stack$ -- 0 +substring$ -- 0 +swap$ -- 0 +text.length$ -- 0 +text.prefix$ -- 0 +top$ -- 0 +type$ -- 0 +warning$ -- 0 +while$ -- 0 +width$ -- 0 +write$ -- 0 +(There were 3 error messages) diff --git a/lecture_12/lecture_12.ipynb b/lecture_12/lecture_12.ipynb index dc44a21..41f1959 100644 --- a/lecture_12/lecture_12.ipynb +++ b/lecture_12/lecture_12.ipynb @@ -22,6 +22,97 @@ "setdefaults" ] }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A =\n", + "\n", + " 0.447394 0.357071 0.720915 0.499926\n", + " 0.648313 0.323276 0.521677 0.288345\n", + " 0.084982 0.581513 0.466420 0.142342\n", + " 0.576580 0.658089 0.916987 0.923165\n", + "\n", + "L =\n", + "\n", + " 1.00000 0.00000 0.00000 0.00000\n", + " 0.13108 1.00000 0.00000 0.00000\n", + " 0.69009 0.24851 1.00000 0.00000\n", + " 0.88935 0.68736 0.68488 1.00000\n", + "\n", + "U =\n", + "\n", + " 0.64831 0.32328 0.52168 0.28834\n", + " 0.00000 0.53914 0.39804 0.10455\n", + " 0.00000 0.00000 0.26199 0.27496\n", + " 0.00000 0.00000 0.00000 0.40655\n", + "\n", + "P =\n", + "\n", + "Permutation Matrix\n", + "\n", + " 0 1 0 0\n", + " 0 0 1 0\n", + " 1 0 0 0\n", + " 0 0 0 1\n", + "\n", + "ans = 1\n" + ] + } + ], + "source": [ + "A=rand(4,4)\n", + "\n", + "[L,U,P]=lu(A)\n", + "\n", + "det(L)" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ans =\n", + "\n", + " 4 4\n", + "\n", + "ans = 23.586\n", + "ans = 35.826\n", + "ans = 14.869\n", + "C =\n", + "\n", + " 5.98549 4.28555 4.35707 4.31359\n", + " 0.00000 3.63950 1.35005 1.45342\n", + " 0.00000 0.00000 3.62851 1.50580\n", + " 0.00000 0.00000 0.00000 3.21911\n", + "\n" + ] + } + ], + "source": [ + "A=rand(4,100)';\n", + "A=A'*A;\n", + "size(A)\n", + "min(min(A))\n", + "max(max(A))\n", + "cond(A)\n", + "C=chol(A)" + ] + }, { "cell_type": "markdown", "metadata": {}, @@ -90,7 +181,7 @@ "1\\end{array}\\right]$\n", "\n", "$A^{-1}=\\frac{1}{2*3-1*1}\\left[ \\begin{array}{cc}\n", - "3 & 1 \\\\\n", + "3 & -1 \\\\\n", "-1 & 2 \\end{array} \\right]=\n", "\\left[ \\begin{array}{cc}\n", "3/5 & -1/5 \\\\\n", @@ -99,7 +190,7 @@ }, { "cell_type": "code", - "execution_count": 2, + "execution_count": 45, "metadata": { "collapsed": false }, @@ -203,7 +294,7 @@ "\n", "$A^{-1}=\\left[ \\begin{array}{cccc}\n", "| & | & & | \\\\\n", - "a_{1} & a_{2} & \\cdots & a_{3} \\\\\n", + "a_{1} & a_{2} & \\cdots & a_{n} \\\\\n", "| & | & & | \\end{array} \\right]$\n", "\n", "\n", @@ -236,9 +327,26 @@ "0 & -1 & 1 \\end{array} \\right]$\n" ] }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#### Note on solving for $A^{-1}$ column 1\n", + "\n", + "$Aa_1=I(:,1)$\n", + "\n", + "$LUa_1=I(:,1)$\n", + "\n", + "$(LUa_1-I(:,1))=0$\n", + "\n", + "$L(Ua_1-d_1)=0$\n", + "\n", + "$I(:,1)=Ld_1$" + ] + }, { "cell_type": "code", - "execution_count": 21, + "execution_count": 56, "metadata": { "collapsed": false }, @@ -278,7 +386,7 @@ }, { "cell_type": "code", - "execution_count": 22, + "execution_count": 57, "metadata": { "collapsed": false }, @@ -317,13 +425,23 @@ "1 \\\\ \n", "0 \\\\ \n", "\\vdots \\\\\n", - "0 \\end{array} \\right]$\n", - "$;~Ua_{1}=d_{1}$" + "0 \\end{array} \\right]=\n", + "\\left[\\begin{array}{ccc} \n", + "1 & 0 & 0 \\\\ \n", + "-1/2 & 1 & 0 \\\\\n", + "0 & -2/3 & 1 \\end{array} \\right]\\left[\\begin{array}{c} \n", + "d1(1) \\\\ \n", + "d1(2) \\\\ \n", + "d1(3)\\end{array} \\right]=\\left[\\begin{array}{c} \n", + "1 \\\\ \n", + "0 \\\\ \n", + "0 \\end{array} \\right]\n", + ";~Ua_{1}=d_{1}$" ] }, { "cell_type": "code", - "execution_count": 29, + "execution_count": 58, "metadata": { "collapsed": false }, @@ -350,7 +468,7 @@ }, { "cell_type": "code", - "execution_count": 30, + "execution_count": 59, "metadata": { "collapsed": false }, @@ -377,7 +495,7 @@ }, { "cell_type": "code", - "execution_count": 28, + "execution_count": 60, "metadata": { "collapsed": false }, @@ -404,7 +522,7 @@ }, { "cell_type": "code", - "execution_count": 31, + "execution_count": 61, "metadata": { "collapsed": false }, @@ -431,7 +549,7 @@ }, { "cell_type": "code", - "execution_count": 37, + "execution_count": 62, "metadata": { "collapsed": false }, @@ -458,7 +576,7 @@ }, { "cell_type": "code", - "execution_count": 38, + "execution_count": 63, "metadata": { "collapsed": false }, @@ -492,7 +610,7 @@ }, { "cell_type": "code", - "execution_count": 40, + "execution_count": 69, "metadata": { "collapsed": false }, @@ -507,18 +625,25 @@ " 1.00000 2.00000 2.00000\n", " 1.00000 2.00000 3.00000\n", "\n", - "ans =\n", + "I_app =\n", "\n", " 1.00000 0.00000 0.00000\n", " 0.00000 1.00000 -0.00000\n", " -0.00000 -0.00000 1.00000\n", - "\n" + "\n", + "ans = -4.4409e-16\n", + "ans = 2.2204e-16\n", + "ans = 0.0039062\n" ] } ], "source": [ "invA=[a1,a2,a3]\n", - "A*invA" + "I_app=A*invA\n", + "I_app(2,3)\n", + "eps\n", + "\n", + "2^-8" ] }, { @@ -530,7 +655,7 @@ }, { "cell_type": "code", - "execution_count": 44, + "execution_count": 70, "metadata": { "collapsed": false }, @@ -577,7 +702,7 @@ }, { "cell_type": "code", - "execution_count": 61, + "execution_count": 71, "metadata": { "collapsed": false }, @@ -703,7 +828,7 @@ "\t\n", "\n", "\n", - "\t\n", + "\t\n", "\t\n", "\tbackslash\n", "\n", @@ -712,7 +837,7 @@ "\t\n", "\n", "\n", - "\t\n", + "\t\n", "\t\n", "\tmultiplication\n", "\n", @@ -721,7 +846,7 @@ "\t\n", "\n", "\n", - "\t\n", + "\t\n", "\t\n", "\n", "\n", @@ -813,7 +938,7 @@ }, { "cell_type": "code", - "execution_count": 1, + "execution_count": 72, "metadata": { "collapsed": false }, @@ -864,7 +989,7 @@ }, { "cell_type": "code", - "execution_count": 12, + "execution_count": 75, "metadata": { "collapsed": false }, @@ -908,7 +1033,7 @@ }, { "cell_type": "code", - "execution_count": 11, + "execution_count": 74, "metadata": { "collapsed": false }, diff --git a/lecture_12/lecture_12.log b/lecture_12/lecture_12.log new file mode 100644 index 0000000..c4a5ed3 --- /dev/null +++ b/lecture_12/lecture_12.log @@ -0,0 +1,841 @@ +This is pdfTeX, Version 3.14159265-2.6-1.40.16 (TeX Live 2015/Debian) (preloaded format=pdflatex 2017.1.11) 28 FEB 2017 14:24 +entering extended mode + restricted \write18 enabled. + %&-line parsing enabled. +**lecture_12.tex +(./lecture_12.tex +LaTeX2e <2016/02/01> +Babel <3.9q> and hyphenation 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statistics: + 177 PDF objects out of 1000 (max. 8388607) + 140 compressed objects within 2 object streams + 34 named destinations out of 1000 (max. 500000) + 77 words of extra memory for PDF output out of 10000 (max. 10000000) + diff --git a/lecture_12/lecture_12.md b/lecture_12/lecture_12.md index 9befcbc..82f9bea 100644 --- a/lecture_12/lecture_12.md +++ b/lecture_12/lecture_12.md @@ -9,6 +9,75 @@ setdefaults ``` + +```octave +A=rand(4,4) + +[L,U,P]=lu(A) + +det(L) +``` + + A = + + 0.447394 0.357071 0.720915 0.499926 + 0.648313 0.323276 0.521677 0.288345 + 0.084982 0.581513 0.466420 0.142342 + 0.576580 0.658089 0.916987 0.923165 + + L = + + 1.00000 0.00000 0.00000 0.00000 + 0.13108 1.00000 0.00000 0.00000 + 0.69009 0.24851 1.00000 0.00000 + 0.88935 0.68736 0.68488 1.00000 + + U = + + 0.64831 0.32328 0.52168 0.28834 + 0.00000 0.53914 0.39804 0.10455 + 0.00000 0.00000 0.26199 0.27496 + 0.00000 0.00000 0.00000 0.40655 + + P = + + Permutation Matrix + + 0 1 0 0 + 0 0 1 0 + 1 0 0 0 + 0 0 0 1 + + ans = 1 + + + +```octave +A=rand(4,100)'; +A=A'*A; +size(A) +min(min(A)) +max(max(A)) +cond(A) +C=chol(A) +``` + + ans = + + 4 4 + + ans = 23.586 + ans = 35.826 + ans = 14.869 + C = + + 5.98549 4.28555 4.35707 4.31359 + 0.00000 3.63950 1.35005 1.45342 + 0.00000 0.00000 3.62851 1.50580 + 0.00000 0.00000 0.00000 3.21911 + + + ## My question from last class ![q1](det_L.png) @@ -68,7 +137,7 @@ x_{2} \end{array}\right]= 1\end{array}\right]$ $A^{-1}=\frac{1}{2*3-1*1}\left[ \begin{array}{cc} -3 & 1 \\ +3 & -1 \\ -1 & 2 \end{array} \right]= \left[ \begin{array}{cc} 3/5 & -1/5 \\ @@ -156,7 +225,7 @@ Which we can solve for each $a_{n}$ with LU-decomposition, knowing the lower and $A^{-1}=\left[ \begin{array}{cccc} | & | & & | \\ -a_{1} & a_{2} & \cdots & a_{3} \\ +a_{1} & a_{2} & \cdots & a_{n} \\ | & | & & | \end{array} \right]$ @@ -189,6 +258,18 @@ $A=\left[ \begin{array}{ccc} 0 & -1 & 1 \end{array} \right]$ +#### Note on solving for $A^{-1}$ column 1 + +$Aa_1=I(:,1)$ + +$LUa_1=I(:,1)$ + +$(LUa_1-I(:,1))=0$ + +$L(Ua_1-d_1)=0$ + +$I(:,1)=Ld_1$ + ```octave A=[2,-1,0;-1,2,-1;0,-1,1] @@ -245,8 +326,18 @@ $Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ -0 \end{array} \right]$ -$;~Ua_{1}=d_{1}$ +0 \end{array} \right]= +\left[\begin{array}{ccc} +1 & 0 & 0 \\ +-1/2 & 1 & 0 \\ +0 & -2/3 & 1 \end{array} \right]\left[\begin{array}{c} +d1(1) \\ +d1(2) \\ +d1(3)\end{array} \right]=\left[\begin{array}{c} +1 \\ +0 \\ +0 \end{array} \right] +;~Ua_{1}=d_{1}$ ```octave @@ -349,7 +440,11 @@ Final solution for $A^{-1}$ is $[a_{1}~a_{2}~a_{3}]$ ```octave invA=[a1,a2,a3] -A*invA +I_app=A*invA +I_app(2,3) +eps + +2^-8 ``` invA = @@ -358,12 +453,15 @@ A*invA 1.00000 2.00000 2.00000 1.00000 2.00000 3.00000 - ans = + I_app = 1.00000 0.00000 0.00000 0.00000 1.00000 -0.00000 -0.00000 -0.00000 1.00000 + ans = -4.4409e-16 + ans = 2.2204e-16 + ans = 0.0039062 Now the solution of $x$ to $Ax=y$ is $x=A^{-1}y$ @@ -430,7 +528,7 @@ legend('inversion','backslash','multiplication','Location','NorthWest') ``` -![svg](lecture_12_files/lecture_12_21_0.svg) +![svg](lecture_12_files/lecture_12_24_0.svg) ## Condition of a matrix diff --git a/lecture_12/lecture_12.out b/lecture_12/lecture_12.out new file mode 100644 index 0000000..e7b5af8 --- /dev/null +++ b/lecture_12/lecture_12.out @@ -0,0 +1,7 @@ +\BOOKMARK [2][-]{subsection.0.1}{My question from last class}{}% 1 +\BOOKMARK [2][-]{subsection.0.2}{Your questions from last class}{}% 2 +\BOOKMARK [1][-]{section.1}{Matrix Inverse and Condition}{}% 3 +\BOOKMARK [2][-]{subsection.1.1}{Condition of a matrix}{section.1}% 4 +\BOOKMARK [3][-]{subsubsection.1.1.1}{just checked in to see what condition my condition was in}{subsection.1.1}% 5 +\BOOKMARK [3][-]{subsubsection.1.1.2}{Matrix norms}{subsection.1.1}% 6 +\BOOKMARK [3][-]{subsubsection.1.1.3}{Condition of Matrix}{subsection.1.1}% 7 diff --git a/lecture_12/lecture_12.pdf b/lecture_12/lecture_12.pdf index ed0f569..e0b4e6c 100644 Binary files a/lecture_12/lecture_12.pdf and b/lecture_12/lecture_12.pdf differ diff --git a/lecture_12/lecture_12.tex b/lecture_12/lecture_12.tex new file mode 100644 index 0000000..1265a06 --- /dev/null +++ b/lecture_12/lecture_12.tex @@ -0,0 +1,1015 @@ + +% Default to the notebook output style + + + + +% Inherit from the specified cell style. + + + + + +\documentclass[11pt]{article} + + + + \usepackage[T1]{fontenc} + % Nicer default font (+ math font) than Computer Modern for most use cases + \usepackage{mathpazo} + + % Basic figure setup, for now with no caption control since it's done + % automatically by Pandoc (which extracts ![](path) syntax from Markdown). + \usepackage{graphicx} + % We will generate all images so they have a width \maxwidth. This means + % that they will get their normal width if they fit onto the page, but + % are scaled down if they would overflow the margins. + \makeatletter + \def\maxwidth{\ifdim\Gin@nat@width>\linewidth\linewidth + \else\Gin@nat@width\fi} + \makeatother + \let\Oldincludegraphics\includegraphics + % Set max figure width to be 80% of text width, for now hardcoded. + \renewcommand{\includegraphics}[1]{\Oldincludegraphics[width=.8\maxwidth]{#1}} + % Ensure that by default, figures have no caption (until we provide a + % proper Figure object with a Caption API and a way to capture that + % in the conversion process - todo). + \usepackage{caption} + \DeclareCaptionLabelFormat{nolabel}{} + \captionsetup{labelformat=nolabel} + + \usepackage{adjustbox} % Used to constrain images to a maximum size + \usepackage{xcolor} % Allow colors to be defined + \usepackage{enumerate} % Needed for markdown enumerations to work + \usepackage{geometry} % Used to adjust the document margins + \usepackage{amsmath} % Equations + \usepackage{amssymb} % Equations + \usepackage{textcomp} % defines textquotesingle + % Hack from http://tex.stackexchange.com/a/47451/13684: + \AtBeginDocument{% + \def\PYZsq{\textquotesingle}% Upright quotes in Pygmentized code + } + \usepackage{upquote} % Upright quotes for verbatim code + \usepackage{eurosym} % defines \euro + \usepackage[mathletters]{ucs} % Extended unicode (utf-8) support + \usepackage[utf8x]{inputenc} % Allow utf-8 characters in the tex document + \usepackage{fancyvrb} % verbatim replacement that allows latex + \usepackage{grffile} % extends the file name processing of package graphics + % to support a larger range + % The hyperref package gives us a pdf with properly built + % internal navigation ('pdf bookmarks' for the table of contents, + % internal cross-reference links, web links for URLs, etc.) + \usepackage{hyperref} + \usepackage{longtable} % longtable support required by pandoc >1.10 + \usepackage{booktabs} % table support for pandoc > 1.12.2 + \usepackage[inline]{enumitem} % IRkernel/repr support (it uses the enumerate* environment) + \usepackage[normalem]{ulem} % ulem is needed to support strikethroughs (\sout) + % normalem makes italics be italics, not underlines + + + + + % Colors for the hyperref package + \definecolor{urlcolor}{rgb}{0,.145,.698} + \definecolor{linkcolor}{rgb}{.71,0.21,0.01} + \definecolor{citecolor}{rgb}{.12,.54,.11} + + % ANSI colors + \definecolor{ansi-black}{HTML}{3E424D} + \definecolor{ansi-black-intense}{HTML}{282C36} + \definecolor{ansi-red}{HTML}{E75C58} + \definecolor{ansi-red-intense}{HTML}{B22B31} + \definecolor{ansi-green}{HTML}{00A250} + \definecolor{ansi-green-intense}{HTML}{007427} + \definecolor{ansi-yellow}{HTML}{DDB62B} + \definecolor{ansi-yellow-intense}{HTML}{B27D12} + \definecolor{ansi-blue}{HTML}{208FFB} + \definecolor{ansi-blue-intense}{HTML}{0065CA} + \definecolor{ansi-magenta}{HTML}{D160C4} + \definecolor{ansi-magenta-intense}{HTML}{A03196} + \definecolor{ansi-cyan}{HTML}{60C6C8} + \definecolor{ansi-cyan-intense}{HTML}{258F8F} + \definecolor{ansi-white}{HTML}{C5C1B4} + \definecolor{ansi-white-intense}{HTML}{A1A6B2} + + % commands and environments needed by pandoc snippets + % extracted from the output of `pandoc -s` + \providecommand{\tightlist}{% + \setlength{\itemsep}{0pt}\setlength{\parskip}{0pt}} + \DefineVerbatimEnvironment{Highlighting}{Verbatim}{commandchars=\\\{\}} + % Add ',fontsize=\small' for more characters per line + \newenvironment{Shaded}{}{} + \newcommand{\KeywordTok}[1]{\textcolor[rgb]{0.00,0.44,0.13}{\textbf{{#1}}}} + \newcommand{\DataTypeTok}[1]{\textcolor[rgb]{0.56,0.13,0.00}{{#1}}} + \newcommand{\DecValTok}[1]{\textcolor[rgb]{0.25,0.63,0.44}{{#1}}} + \newcommand{\BaseNTok}[1]{\textcolor[rgb]{0.25,0.63,0.44}{{#1}}} + \newcommand{\FloatTok}[1]{\textcolor[rgb]{0.25,0.63,0.44}{{#1}}} + \newcommand{\CharTok}[1]{\textcolor[rgb]{0.25,0.44,0.63}{{#1}}} + 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PY@tok@c\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}} +\expandafter\def\csname PY@tok@mf\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@err\endcsname{\def\PY@bc##1{\setlength{\fboxsep}{0pt}\fcolorbox[rgb]{1.00,0.00,0.00}{1,1,1}{\strut ##1}}} +\expandafter\def\csname PY@tok@mb\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@ss\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.10,0.09,0.49}{##1}}} +\expandafter\def\csname PY@tok@sr\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.53}{##1}}} +\expandafter\def\csname PY@tok@mo\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@kd\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@mi\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.40,0.40,0.40}{##1}}} +\expandafter\def\csname PY@tok@kn\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@cpf\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.25,0.50,0.50}{##1}}} +\expandafter\def\csname PY@tok@kr\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@s\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@kp\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@w\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.73,0.73}{##1}}} +\expandafter\def\csname PY@tok@kt\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.69,0.00,0.25}{##1}}} +\expandafter\def\csname PY@tok@sc\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@sb\endcsname{\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} +\expandafter\def\csname PY@tok@k\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.00,0.50,0.00}{##1}}} +\expandafter\def\csname PY@tok@se\endcsname{\let\PY@bf=\textbf\def\PY@tc##1{\textcolor[rgb]{0.73,0.40,0.13}{##1}}} +\expandafter\def\csname PY@tok@sd\endcsname{\let\PY@it=\textit\def\PY@tc##1{\textcolor[rgb]{0.73,0.13,0.13}{##1}}} + +\def\PYZbs{\char`\\} +\def\PYZus{\char`\_} +\def\PYZob{\char`\{} +\def\PYZcb{\char`\}} +\def\PYZca{\char`\^} +\def\PYZam{\char`\&} +\def\PYZlt{\char`\<} +\def\PYZgt{\char`\>} +\def\PYZsh{\char`\#} +\def\PYZpc{\char`\%} +\def\PYZdl{\char`\$} +\def\PYZhy{\char`\-} +\def\PYZsq{\char`\'} +\def\PYZdq{\char`\"} +\def\PYZti{\char`\~} +% for compatibility with earlier versions +\def\PYZat{@} +\def\PYZlb{[} +\def\PYZrb{]} +\makeatother + + + % Exact colors from NB + \definecolor{incolor}{rgb}{0.0, 0.0, 0.5} + \definecolor{outcolor}{rgb}{0.545, 0.0, 0.0} + + + + + % Prevent overflowing lines due to hard-to-break entities + \sloppy + % Setup hyperref package + \hypersetup{ + breaklinks=true, % so long urls are correctly broken across lines + colorlinks=true, + urlcolor=urlcolor, + linkcolor=linkcolor, + citecolor=citecolor, + } + % Slightly bigger margins than the latex defaults + + \geometry{verbose,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in} + + + + \begin{document} + + + \maketitle + + + + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}27}]:} \PY{c}{\PYZpc{}plot \PYZhy{}\PYZhy{}format svg} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}28}]:} \PY{n}{setdefaults} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}29}]:} \PY{n}{A}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{)} + + \PY{p}{[}\PY{n}{L}\PY{p}{,}\PY{n}{U}\PY{p}{,}\PY{n}{P}\PY{p}{]}\PY{p}{=}\PY{n+nb}{lu}\PY{p}{(}\PY{n}{A}\PY{p}{)} + + \PY{n+nb}{det}\PY{p}{(}\PY{n}{L}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 0.447394 0.357071 0.720915 0.499926 + 0.648313 0.323276 0.521677 0.288345 + 0.084982 0.581513 0.466420 0.142342 + 0.576580 0.658089 0.916987 0.923165 + +L = + + 1.00000 0.00000 0.00000 0.00000 + 0.13108 1.00000 0.00000 0.00000 + 0.69009 0.24851 1.00000 0.00000 + 0.88935 0.68736 0.68488 1.00000 + +U = + + 0.64831 0.32328 0.52168 0.28834 + 0.00000 0.53914 0.39804 0.10455 + 0.00000 0.00000 0.26199 0.27496 + 0.00000 0.00000 0.00000 0.40655 + +P = + +Permutation Matrix + + 0 1 0 0 + 0 0 1 0 + 1 0 0 0 + 0 0 0 1 + +ans = 1 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}44}]:} \PY{n}{A}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{100}\PY{p}{)}\PY{o}{\PYZsq{}}\PY{p}{;} + \PY{n}{A}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZsq{}}\PY{o}{*}\PY{n}{A}\PY{p}{;} + \PY{n+nb}{size}\PY{p}{(}\PY{n}{A}\PY{p}{)} + \PY{n+nb}{min}\PY{p}{(}\PY{n+nb}{min}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{)} + \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{max}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{)} + \PY{n+nb}{cond}\PY{p}{(}\PY{n}{A}\PY{p}{)} + \PY{n}{C}\PY{p}{=}\PY{n+nb}{chol}\PY{p}{(}\PY{n}{A}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ans = + + 4 4 + +ans = 23.586 +ans = 35.826 +ans = 14.869 +C = + + 5.98549 4.28555 4.35707 4.31359 + 0.00000 3.63950 1.35005 1.45342 + 0.00000 0.00000 3.62851 1.50580 + 0.00000 0.00000 0.00000 3.21911 + + + \end{Verbatim} + + \subsection{My question from last +class}\label{my-question-from-last-class} + +\begin{figure}[htbp] +\centering +\includegraphics{det_L.png} +\caption{q1} +\end{figure} + +\begin{figure}[htbp] +\centering +\includegraphics{chol_pre.png} +\caption{q2} +\end{figure} + +\subsection{Your questions from last +class}\label{your-questions-from-last-class} + +\begin{enumerate} +\def\labelenumi{\arabic{enumi}.} +\item + Will the exam be more theoretical or problem based? +\item + Writing code is difficult +\item + What format can we expect for the midterm? +\item + Could we go over some example questions for the exam? +\item + Will the use of GitHub be tested on the Midterm exam? Or is it more + focused on linear algebra techniques/what was covered in the lectures? +\item + This is not my strong suit, getting a bit overwhelmed with matrix + multiplication. +\item + I forgot how much I learned in linear algebra. +\item + What's the most exciting project you've ever worked on with + Matlab/Octave? +\end{enumerate} + + \section{Matrix Inverse and +Condition}\label{matrix-inverse-and-condition} + +Considering the same solution set: + +\(y=Ax\) + +If we know that \(A^{-1}A=I\), then + +\(A^{-1}y=A^{-1}Ax=x\) + +so + +\(x=A^{-1}y\) + +Where, \(A^{-1}\) is the inverse of matrix \(A\). + +\(2x_{1}+x_{2}=1\) + +\(x_{1}+3x_{2}=1\) + +\(Ax=y\) + +\(\left[ \begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array} \right] \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right]= \left[\begin{array}{c} 1 \\ 1\end{array}\right]\) + +\(A^{-1}=\frac{1}{2*3-1*1}\left[ \begin{array}{cc} 3 & -1 \\ -1 & 2 \end{array} \right]= \left[ \begin{array}{cc} 3/5 & -1/5 \\ -1/5 & 2/5 \end{array} \right]\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}45}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]} + \PY{n}{invA}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{5}\PY{o}{*}\PY{p}{[}\PY{l+m+mi}{3}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{]} + + \PY{n}{A}\PY{o}{*}\PY{n}{invA} + \PY{n}{invA}\PY{o}{*}\PY{n}{A} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 2 1 + 1 3 + +invA = + + 0.60000 -0.20000 + -0.20000 0.40000 + +ans = + + 1.00000 0.00000 + 0.00000 1.00000 + +ans = + + 1.00000 0.00000 + 0.00000 1.00000 + + + \end{Verbatim} + + How did we know the inverse of A? + +for 2$\times$2 matrices, it is always: + +$A=\left[ \begin{array}{cc} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array} \right]$ + +$A^{-1}=\frac{1}{det(A)}\left[ \begin{array}{cc} A_{22} & -A_{12} \\ -A_{21} & A_{11} \end{array} \right]$ + + $AA^{-1}=\frac{1}{A_{11}A_{22}-A_{21}A_{12}}\left[ \begin{array}{cc} A_{11}A_{22}-A_{21}A_{12} & -A_{11}A_{12}+A_{12}A_{11} \\ A_{21}A_{22}-A_{22}A_{21} & -A_{21}A_{12}+A_{22}A_{11} \end{array} \right] =\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$ + + What about bigger matrices? + +We can use the LU-decomposition + +\(A=LU\) + +\(A^{-1}=(LU)^{-1}=U^{-1}L^{-1}\) + +if we divide \(A^{-1}\) into n-column vectors, \(a_{n}\), then + +\(Aa_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right]\) +\(Aa_{2}=\left[\begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array} \right]\) +\(Aa_{n}=\left[\begin{array}{c} 0 \\ 0 \\ \vdots \\ 1 \end{array} \right]\) + +Which we can solve for each \(a_{n}\) with LU-decomposition, knowing the +lower and upper triangular decompositions, then + +\(A^{-1}=\left[ \begin{array}{cccc} | & | & & | \\ a_{1} & a_{2} & \cdots & a_{n} \\ | & | & & | \end{array} \right]\) + +\(Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right]\) +\(;~Ua_{1}=d_{1}\) + +\(Ld_{2}=\left[\begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array} \right]\) +\(;~Ua_{2}=d_{2}\) + +\(Ld_{n}=\left[\begin{array}{c} 0 \\ 1 \\ \vdots \\ n \end{array} \right]\) +\(;~Ua_{n}=d_{n}\) + +Consider the following matrix: + +\(A=\left[ \begin{array}{ccc} 2 & -1 & 0\\ -1 & 2 & -1\\ 0 & -1 & 1 \end{array} \right]\) + + \paragraph{\texorpdfstring{Note on solving for \(A^{-1}\) column +1}{Note on solving for A\^{}\{-1\} column 1}}\label{note-on-solving-for-a-1-column-1} + +\(Aa_1=I(:,1)\) + +\(LUa_1=I(:,1)\) + +\((LUa_1-I(:,1))=0\) + +\(L(Ua_1-d_1)=0\) + +\(I(:,1)=Ld_1\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}56}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{0}\PY{p}{;}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{,}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{]} + \PY{n}{U}\PY{p}{=}\PY{n}{A}\PY{p}{;} + \PY{n}{L}\PY{p}{=}\PY{n+nb}{eye}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;} + \PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{p}{:}\PY{p}{)} + \PY{n}{L}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{/}\PY{n}{A}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 2 -1 0 + -1 2 -1 + 0 -1 1 + +U = + + 2.00000 -1.00000 0.00000 + 0.00000 1.50000 -1.00000 + 0.00000 -1.00000 1.00000 + +L = + + 1.00000 0.00000 0.00000 + -0.50000 1.00000 0.00000 + 0.00000 0.00000 1.00000 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}57}]:} \PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)} + \PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{p}{:}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{p}{:}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +L = + + 1.00000 0.00000 0.00000 + -0.50000 1.00000 0.00000 + 0.00000 -0.66667 1.00000 + +U = + + 2.00000 -1.00000 0.00000 + 0.00000 1.50000 -1.00000 + 0.00000 0.00000 0.33333 + + + \end{Verbatim} + + Now solve for \(d_1\) then \(a_1\), \(d_2\) then \(a_2\), and \(d_3\) +then \(a_{3}\) + +\(Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right]= \left[\begin{array}{ccc} 1 & 0 & 0 \\ -1/2 & 1 & 0 \\ 0 & -2/3 & 1 \end{array} \right]\left[\begin{array}{c} d1(1) \\ d1(2) \\ d1(3)\end{array} \right]=\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] ;~Ua_{1}=d_{1}\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}58}]:} \PY{n}{d1}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{d1}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} + 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+\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +a1 = + + 1.00000 + 1.00000 + 1.00000 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}60}]:} \PY{n}{d2}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{p}{;} + \PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +d2 = + + 0.00000 + 1.00000 + 0.66667 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}61}]:} \PY{n}{a2}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;} + \PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}\PY{p}{;} + \PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d2}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a2}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +a2 = + + 1.0000 + 2.0000 + 2.0000 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}62}]:} \PY{n}{d3}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{p}{;} + \PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{0}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{L}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +d3 = + + 0 + 0 + 1 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}63}]:} \PY{n}{a3}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;} + \PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)}\PY{p}{;} + \PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{l+m+mi}{1}\PY{o}{/}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{*}\PY{p}{(}\PY{n}{d3}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{*}\PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{)}\PY{o}{\PYZhy{}}\PY{n}{U}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{o}{*}\PY{n}{a3}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +a3 = + + 1.00000 + 2.00000 + 3.00000 + + + \end{Verbatim} + + Final solution for \(A^{-1}\) is \([a_{1}~a_{2}~a_{3}]\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}69}]:} \PY{n}{invA}\PY{p}{=}\PY{p}{[}\PY{n}{a1}\PY{p}{,}\PY{n}{a2}\PY{p}{,}\PY{n}{a3}\PY{p}{]} + \PY{n}{I\PYZus{}app}\PY{p}{=}\PY{n}{A}\PY{o}{*}\PY{n}{invA} + \PY{n}{I\PYZus{}app}\PY{p}{(}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)} + \PY{n+nb}{eps} + + \PY{l+m+mi}{2}\PYZca{}\PY{o}{\PYZhy{}}\PY{l+m+mi}{8} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +invA = + + 1.00000 1.00000 1.00000 + 1.00000 2.00000 2.00000 + 1.00000 2.00000 3.00000 + +I\_app = + + 1.00000 0.00000 0.00000 + 0.00000 1.00000 -0.00000 + -0.00000 -0.00000 1.00000 + +ans = -4.4409e-16 +ans = 2.2204e-16 +ans = 0.0039062 + + \end{Verbatim} + + Now the solution of \(x\) to \(Ax=y\) is \(x=A^{-1}y\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}70}]:} \PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{2}\PY{p}{;}\PY{l+m+mi}{3}\PY{p}{]} + \PY{n}{x}\PY{p}{=}\PY{n}{invA}\PY{o}{*}\PY{n}{y} + \PY{n}{xbs}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{y} + \PY{n}{x}\PY{o}{\PYZhy{}}\PY{n}{xbs} + \PY{n+nb}{eps} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +y = + + 1 + 2 + 3 + +x = + + 6.0000 + 11.0000 + 14.0000 + +xbs = + + 6.0000 + 11.0000 + 14.0000 + +ans = + + -3.5527e-15 + -8.8818e-15 + -1.0658e-14 + +ans = 2.2204e-16 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}71}]:} \PY{n}{N}\PY{p}{=}\PY{l+m+mi}{100}\PY{p}{;} + \PY{n}{n}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{:}\PY{n}{N}\PY{p}{]}\PY{p}{;} + \PY{n}{t\PYZus{}inv}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{t\PYZus{}bs}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n}{t\PYZus{}mult}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{n}{N}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{k}{for} \PY{n}{i}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{:}\PY{n}{N} + \PY{n}{A}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{n}{i}\PY{p}{,}\PY{n}{i}\PY{p}{)}\PY{p}{;} + \PY{n+nb}{tic} + \PY{n}{invA}\PY{p}{=}\PY{n+nb}{inv}\PY{p}{(}\PY{n}{A}\PY{p}{)}\PY{p}{;} + \PY{n}{t\PYZus{}inv}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;} + \PY{n}{b}\PY{p}{=}\PY{n+nb}{rand}\PY{p}{(}\PY{n}{i}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{;} + \PY{n+nb}{tic}\PY{p}{;} + \PY{n}{x}\PY{p}{=}\PY{n}{A}\PY{o}{\PYZbs{}}\PY{n}{b}\PY{p}{;} + \PY{n}{t\PYZus{}bs}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;} + \PY{n+nb}{tic}\PY{p}{;} + \PY{n}{x}\PY{p}{=}\PY{n}{invA}\PY{o}{*}\PY{n}{b}\PY{p}{;} + \PY{n}{t\PYZus{}mult}\PY{p}{(}\PY{n}{i}\PY{p}{)}\PY{p}{=}\PY{n+nb}{toc}\PY{p}{;} + \PY{k}{end} + \PY{n+nb}{plot}\PY{p}{(}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}inv}\PY{p}{,}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}bs}\PY{p}{,}\PY{n}{n}\PY{p}{,}\PY{n}{t\PYZus{}mult}\PY{p}{)} + \PY{n+nb}{axis}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{0} \PY{l+m+mi}{100} \PY{l+m+mi}{0} \PY{l+m+mf}{0.002}\PY{p}{]}\PY{p}{)} + \PY{n+nb}{legend}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{inversion\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{backslash\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{multiplication\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{Location\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{NorthWest\PYZsq{}}\PY{p}{)} +\end{Verbatim} + + \begin{center} + \adjustimage{max size={0.9\linewidth}{0.9\paperheight}}{lecture_12_files/lecture_12_24_0.pdf} + \end{center} + { \hspace*{\fill} \\} + + \subsection{Condition of a matrix}\label{condition-of-a-matrix} + +\subsubsection{\texorpdfstring{\emph{just checked in to see what +condition my condition was +in}}{just checked in to see what condition my condition was in}}\label{just-checked-in-to-see-what-condition-my-condition-was-in} + +\subsubsection{Matrix norms}\label{matrix-norms} + +The Euclidean norm of a vector is measure of the magnitude (in 3D this +would be: \(|x|=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}\)) in general the +equation is: + +\(||x||_{e}=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}\) + +For a matrix, A, the same norm is called the Frobenius norm: + +\(||A||_{f}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{2}}\) + +In general we can calculate any \(p\)-norm where + +\(||A||_{p}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{p}}\) + +so the p=1, 1-norm is + +\(||A||_{1}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{1}}=\sum_{i=1}^{n}\sum_{i=1}^{m}|A_{i,j}|\) + +\(||A||_{\infty}=\sqrt{\sum_{i=1}^{n}\sum_{i=1}^{m}A_{i,j}^{\infty}}=\max_{1\le i \le n}\sum_{j=1}^{m}|A_{i,j}|\) + +\subsubsection{Condition of Matrix}\label{condition-of-matrix} + +The matrix condition is the product of + +\(Cond(A) = ||A||\cdot||A^{-1}||\) + +So each norm will have a different condition number, but the limit is +\(Cond(A)\ge 1\) + +An estimate of the rounding error is based on the condition of A: + +\(\frac{||\Delta x||}{x} \le Cond(A) \frac{||\Delta A||}{||A||}\) + +So if the coefficients of A have accuracy to \$10\^{}\{-t\} + +and the condition of A, \(Cond(A)=10^{c}\) + +then the solution for x can have rounding errors up to \(10^{c-t}\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}72}]:} \PY{n}{A}\PY{p}{=}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{;}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{4}\PY{p}{;}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{1}\PY{o}{/}\PY{l+m+mi}{5}\PY{p}{]} + \PY{p}{[}\PY{n}{L}\PY{p}{,}\PY{n}{U}\PY{p}{]}\PY{p}{=}\PY{n}{LU\PYZus{}naive}\PY{p}{(}\PY{n}{A}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +A = + + 1.00000 0.50000 0.33333 + 0.50000 0.33333 0.25000 + 0.33333 0.25000 0.20000 + +L = + + 1.00000 0.00000 0.00000 + 0.50000 1.00000 0.00000 + 0.33333 1.00000 1.00000 + +U = + + 1.00000 0.50000 0.33333 + 0.00000 0.08333 0.08333 + 0.00000 -0.00000 0.00556 + + + \end{Verbatim} + + Then, \(A^{-1}=(LU)^{-1}=U^{-1}L^{-1}\) + +\(Ld_{1}=\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]\), +\(Ux_{1}=d_{1}\) ... + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}75}]:} \PY{n}{invA}\PY{p}{=}\PY{n+nb}{zeros}\PY{p}{(}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{;} + \PY{n}{d1}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;} + \PY{n}{d2}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{;} + \PY{n}{d3}\PY{p}{=}\PY{n}{L}\PY{o}{\PYZbs{}}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{0}\PY{p}{;}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{;} + \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d1}\PY{p}{;} + \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d2}\PY{p}{;} + \PY{n}{invA}\PY{p}{(}\PY{p}{:}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}\PY{p}{=}\PY{n}{U}\PY{o}{\PYZbs{}}\PY{n}{d3} + \PY{n}{invA}\PY{o}{*}\PY{n}{A} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +invA = + + 9.0000 -36.0000 30.0000 + -36.0000 192.0000 -180.0000 + 30.0000 -180.0000 180.0000 + +ans = + + 1.0000e+00 3.5527e-15 2.9976e-15 + -1.3249e-14 1.0000e+00 -9.1038e-15 + 8.5117e-15 7.1054e-15 1.0000e+00 + + + \end{Verbatim} + + Find the condition of A, \(cond(A)\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}74}]:} \PY{c}{\PYZpc{} Frobenius norm} + \PY{n}{normf\PYZus{}A} \PY{p}{=} \PY{n+nb}{sqrt}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{o}{.\PYZca{}}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)} + \PY{n}{normf\PYZus{}invA} \PY{p}{=} \PY{n+nb}{sqrt}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{o}{.\PYZca{}}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)}\PY{p}{)} + + \PY{n}{cond\PYZus{}f\PYZus{}A} \PY{p}{=} \PY{n}{normf\PYZus{}A}\PY{o}{*}\PY{n}{normf\PYZus{}invA} + + \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{fro\PYZsq{}}\PY{p}{)} + + \PY{c}{\PYZpc{} p=1, column sum norm} + \PY{n}{norm1\PYZus{}A} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)} + \PY{n}{norm1\PYZus{}invA} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)}\PY{p}{)} + \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)} + + \PY{n}{cond\PYZus{}1\PYZus{}A}\PY{p}{=}\PY{n}{norm1\PYZus{}A}\PY{o}{*}\PY{n}{norm1\PYZus{}invA} + + \PY{c}{\PYZpc{} p=inf, row sum norm} + \PY{n}{norminf\PYZus{}A} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{)} + \PY{n}{norminf\PYZus{}invA} \PY{p}{=} \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{sum}\PY{p}{(}\PY{n}{invA}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)}\PY{p}{)} + \PY{n+nb}{norm}\PY{p}{(}\PY{n}{A}\PY{p}{,}\PY{n+nb}{inf}\PY{p}{)} + + \PY{n}{cond\PYZus{}inf\PYZus{}A}\PY{p}{=}\PY{n}{norminf\PYZus{}A}\PY{o}{*}\PY{n}{norminf\PYZus{}invA} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +normf\_A = 1.4136 +normf\_invA = 372.21 +cond\_f\_A = 526.16 +ans = 1.4136 +norm1\_A = 1.8333 +norm1\_invA = 30.000 +ans = 1.8333 +cond\_1\_A = 55.000 +norminf\_A = 1.8333 +norminf\_invA = 30.000 +ans = 1.8333 +cond\_inf\_A = 55.000 + + \end{Verbatim} + + Consider the problem again from the intro to Linear Algebra, 4 masses +are connected in series to 4 springs with spring constants \(K_{i}\). +What does a high condition number mean for this problem? + +\begin{figure}[htbp] +\centering +\includegraphics{../lecture_09/mass_springs.png} +\caption{Springs-masses} +\end{figure} + +The masses haves the following amounts, 1, 2, 3, and 4 kg for masses +1-4. Using a FBD for each mass: + +\(m_{1}g+k_{2}(x_{2}-x_{1})-k_{1}x_{1}=0\) + +\(m_{2}g+k_{3}(x_{3}-x_{2})-k_{2}(x_{2}-x_{1})=0\) + +\(m_{3}g+k_{4}(x_{4}-x_{3})-k_{3}(x_{3}-x_{2})=0\) + +\(m_{4}g-k_{4}(x_{4}-x_{3})=0\) + +in matrix form: + +\(\left[ \begin{array}{cccc} k_{1}+k_{2} & -k_{2} & 0 & 0 \\ -k_{2} & k_{2}+k_{3} & -k_{3} & 0 \\ 0 & -k_{3} & k_{3}+k_{4} & -k_{4} \\ 0 & 0 & -k_{4} & k_{4} \end{array} \right] \left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right]= \left[ \begin{array}{c} m_{1}g \\ m_{2}g \\ m_{3}g \\ m_{4}g \end{array} \right]\) + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}21}]:} \PY{n}{k1}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} \PY{c}{\PYZpc{} N/m} + \PY{n}{k2}\PY{p}{=}\PY{l+m+mi}{100000}\PY{p}{;} + \PY{n}{k3}\PY{p}{=}\PY{l+m+mi}{10}\PY{p}{;} + \PY{n}{k4}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} + \PY{n}{m1}\PY{p}{=}\PY{l+m+mi}{1}\PY{p}{;} \PY{c}{\PYZpc{} kg} + \PY{n}{m2}\PY{p}{=}\PY{l+m+mi}{2}\PY{p}{;} + \PY{n}{m3}\PY{p}{=}\PY{l+m+mi}{3}\PY{p}{;} + \PY{n}{m4}\PY{p}{=}\PY{l+m+mi}{4}\PY{p}{;} + \PY{n}{g}\PY{p}{=}\PY{l+m+mf}{9.81}\PY{p}{;} \PY{c}{\PYZpc{} m/s\PYZca{}2} + \PY{n}{K}\PY{p}{=}\PY{p}{[}\PY{n}{k1}\PY{o}{+}\PY{n}{k2} \PY{o}{\PYZhy{}}\PY{n}{k2} \PY{l+m+mi}{0} \PY{l+m+mi}{0}\PY{p}{;} \PY{o}{\PYZhy{}}\PY{n}{k2} \PY{n}{k2}\PY{o}{+}\PY{n}{k3} \PY{o}{\PYZhy{}}\PY{n}{k3} \PY{l+m+mi}{0}\PY{p}{;} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k3} \PY{n}{k3}\PY{o}{+}\PY{n}{k4} \PY{o}{\PYZhy{}}\PY{n}{k4}\PY{p}{;} \PY{l+m+mi}{0} \PY{l+m+mi}{0} \PY{o}{\PYZhy{}}\PY{n}{k4} \PY{n}{k4}\PY{p}{]} + \PY{n}{y}\PY{p}{=}\PY{p}{[}\PY{n}{m1}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m2}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m3}\PY{o}{*}\PY{n}{g}\PY{p}{;}\PY{n}{m4}\PY{o}{*}\PY{n}{g}\PY{p}{]} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +K = + + 100010 -100000 0 0 + -100000 100010 -10 0 + 0 -10 11 -1 + 0 0 -1 1 + +y = + + 9.8100 + 19.6200 + 29.4300 + 39.2400 + + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}25}]:} \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{n+nb}{inf}\PY{p}{)} + \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{)} + \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{fro\PYZsq{}}\PY{p}{)} + \PY{n+nb}{cond}\PY{p}{(}\PY{n}{K}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +ans = 3.2004e+05 +ans = 3.2004e+05 +ans = 2.5925e+05 +ans = 2.5293e+05 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor}26}]:} \PY{n+nb}{e}\PY{p}{=}\PY{n+nb}{eig}\PY{p}{(}\PY{n}{K}\PY{p}{)} + \PY{n+nb}{max}\PY{p}{(}\PY{n+nb}{e}\PY{p}{)}\PY{o}{/}\PY{n+nb}{min}\PY{p}{(}\PY{n+nb}{e}\PY{p}{)} +\end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +e = + + 7.9078e-01 + 3.5881e+00 + 1.7621e+01 + 2.0001e+05 + +ans = 2.5293e+05 + + \end{Verbatim} + + \begin{Verbatim}[commandchars=\\\{\}] +{\color{incolor}In [{\color{incolor} }]:} +\end{Verbatim} + + + % Add a bibliography block to the postdoc + + + + \end{document} diff --git a/lecture_12/lecture_12_files/lecture_12_21_0.png b/lecture_12/lecture_12_files/lecture_12_21_0.png new file mode 100644 index 0000000..2633731 Binary files /dev/null and b/lecture_12/lecture_12_files/lecture_12_21_0.png differ diff --git a/lecture_12/lecture_12_files/lecture_12_24_0.pdf b/lecture_12/lecture_12_files/lecture_12_24_0.pdf new file mode 100644 index 0000000..e2d1d9b Binary files /dev/null and b/lecture_12/lecture_12_files/lecture_12_24_0.pdf differ diff --git a/lecture_12/lecture_12_files/lecture_12_24_0.png b/lecture_12/lecture_12_files/lecture_12_24_0.png new file mode 100644 index 0000000..c8959a7 Binary files /dev/null and b/lecture_12/lecture_12_files/lecture_12_24_0.png differ diff --git a/lecture_12/lecture_12_files/lecture_12_24_0.svg b/lecture_12/lecture_12_files/lecture_12_24_0.svg new file mode 100644 index 0000000..4705883 --- /dev/null +++ b/lecture_12/lecture_12_files/lecture_12_24_0.svg @@ -0,0 +1,148 @@ + + +Gnuplot +Produced by GNUPLOT 5.0 patchlevel 3 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + 0 + + + + + 0.0005 + + + + + 0.001 + + + + + 0.0015 + + + + + 0.002 + + + + + 0 + + + + + 20 + + + + + 40 + + + + + 60 + + + + + 80 + + + + + 100 + + + + + + + + + + + + + inversion + + + + + inversion + + + + + + backslash + + + backslash + + + + + + multiplication + + + multiplication + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/linear_algebra/LU_suggested.html b/linear_algebra/LU_suggested.html new file mode 100644 index 0000000..f1998e1 --- /dev/null +++ b/linear_algebra/LU_suggested.html @@ -0,0 +1,39 @@ + + + + + + + + + + + +

Linear Algebra Review

+

(Gauss Elimination) Suggested problems

+

No due date

+
    +

  1. Determine the lower (L) and upper (U) triangular matrices with LU-decomposition for the following matrices:

    +
      +

    1. \(A=\left[ \begin{array}{cc} 1 & 3 \\ 2 & 1 \end{array} \right]\)

      2. +

    2. \(A=\left[ \begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array} \right]\)

      3. +

    3. \(A=\left[ \begin{array}{cc} 1 & 1 \\ 2 & -2 \end{array} \right]\)

      4. +

    4. \(A=\left[ \begin{array}{ccc} 1 & 3 & 1 \\ -4 & -9 & 2 \\ 0 & 3 & 6\end{array} \right]\)

      5. +

    5. \(A=\left[ \begin{array}{ccc} 1 & 3 & 1 \\ -4 & -9 & 2 \\ 0 & 3 & 6\end{array} \right]\)

      6. +

    6. \(A=\left[ \begin{array}{ccc} 1 & 3 & -5 \\ 1 & 4 & -8 \\ -3 & -7 & 9\end{array} \right]\)

      7. +

    7. \(A=\left[ \begin{array}{ccc} 1 & 2 & -1 \\ 2 & 2 & 2 \\ 1 & -1 & 2\end{array} \right]\)

      8. +
    2. +

  2. Calculate the determinant of A from 1a-g.

    3. +

  3. Determine the Cholesky factorization, C, of the following matrices, where

    +

    \(C_{ii}=\sqrt{a_{ii}-\sum_{k=1}^{i-1}C_{ki}^{2}}\)

    +

    \(C_{ij}=\frac{a_{ij}-\sum_{k=1}^{i-1}C_{ki}C_{kj}}{C_{ii}}\).

    +
      +

    1. A=\(\left[ \begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array} \right]\)

      2. +

    2. A=\(\left[ \begin{array}{cc} 10 & 5 \\ 5 & 20 \end{array} \right]\)

      3. +

    3. A=\(\left[ \begin{array}{ccc} 10 & -10 & 20 \\ -10 & 20 & 10 \\ 20 & 10 & 30 \end{array} \right]\)

      4. +

    4. A=\(\left[ \begin{array}{cccc} 21 & -1 & 0 & 0 \\ -1 & 21 & -1 & 0 \\ 0 & -1 & 21 & -1 \\ 0 & 0 & -1 & 1 \end{array} \right]\)

      5. +
    4. +

  4. Verify that \(C^{T}C=A\) for 3a-d

    5. +
+ + diff --git a/linear_algebra/LU_suggested.md b/linear_algebra/LU_suggested.md new file mode 100644 index 0000000..f809813 --- /dev/null +++ b/linear_algebra/LU_suggested.md @@ -0,0 +1,68 @@ +# Linear Algebra Review +## (Gauss Elimination) Suggested problems +### No due date + +1. Determine the lower (L) and upper (U) triangular matrices with LU-decomposition for the +following matrices: + + a. $A=\left[ \begin{array}{cc} + 1 & 3 \\ + 2 & 1 \end{array} \right]$ + + a. $A=\left[ \begin{array}{cc} + 1 & 1 \\ + 2 & 3 \end{array} \right]$ + + a. $A=\left[ \begin{array}{cc} + 1 & 1 \\ + 2 & -2 \end{array} \right]$ + + b. $A=\left[ \begin{array}{ccc} + 1 & 3 & 1 \\ + -4 & -9 & 2 \\ + 0 & 3 & 6\end{array} \right]$ + + c. $A=\left[ \begin{array}{ccc} + 1 & 3 & 1 \\ + -4 & -9 & 2 \\ + 0 & 3 & 6\end{array} \right]$ + + d. $A=\left[ \begin{array}{ccc} + 1 & 3 & -5 \\ + 1 & 4 & -8 \\ + -3 & -7 & 9\end{array} \right]$ + + d. $A=\left[ \begin{array}{ccc} + 1 & 2 & -1 \\ + 2 & 2 & 2 \\ + 1 & -1 & 2\end{array} \right]$ + +2. Calculate the determinant of A from 1a-g. + +3. Determine the Cholesky factorization, C, of the following matrices, where + + $C_{ii}=\sqrt{a_{ii}-\sum_{k=1}^{i-1}C_{ki}^{2}}$ + + $C_{ij}=\frac{a_{ij}-\sum_{k=1}^{i-1}C_{ki}C_{kj}}{C_{ii}}$. + + a. A=$\left[ \begin{array}{cc} + 3 & 2 \\ + 2 & 1 \end{array} \right]$ + + a. A=$\left[ \begin{array}{cc} + 10 & 5 \\ + 5 & 20 \end{array} \right]$ + + a. A=$\left[ \begin{array}{ccc} + 10 & -10 & 20 \\ + -10 & 20 & 10 \\ + 20 & 10 & 30 \end{array} \right]$ + + a. A=$\left[ \begin{array}{cccc} + 21 & -1 & 0 & 0 \\ + -1 & 21 & -1 & 0 \\ + 0 & -1 & 21 & -1 \\ + 0 & 0 & -1 & 1 \end{array} \right]$ + +4. Verify that $C^{T}C=A$ for 3a-d +