%\documentclass[12pt]{article} \documentclass{article} \usepackage{amsmath,amsthm,amssymb} \usepackage[english]{babel} \usepackage[utf8]{inputenc} \usepackage{graphicx} \usepackage{listings} \usepackage{xcolor} \usepackage{fancyhdr} \pagestyle{fancy} \fancyhf{} \rhead{David Dobor} \lhead{Recitation 1, Linear Algebra } \rfoot{Page \thepage} \definecolor{mygreen}{rgb}{0,0.6,0} \definecolor{mygray}{rgb}{0.5,0.5,0.5} \definecolor{mymauve}{rgb}{0.58,0,0.82} \begin{document} \lstset{ % backgroundcolor=\color{white}, % choose the background color; you must add \usepackage{color} or \usepackage{xcolor} basicstyle=\footnotesize, % the size of the fonts that are used for the code breakatwhitespace=false, % sets if automatic breaks should only happen at whitespace breaklines=true, % sets automatic line breaking captionpos=b, % sets the caption-position to bottom commentstyle=\color{mygreen}, % comment style deletekeywords={...}, % if you want to delete keywords from the given language escapeinside={\%*}{*)}, % if you want to add LaTeX within your code extendedchars=true, % lets you use non-ASCII characters; for 8-bits encodings only, does not work with UTF-8 %frame=single, % adds a frame around the code keepspaces=true, % keeps spaces in text, useful for keeping indentation of code (possibly needs columns=flexible) keywordstyle=\color{blue}, % keyword style language=Matlab, % the language of the code morekeywords={*,...}, % if you want to add more keywords to the set numbers=none, % where to put the line-numbers; possible values are (none, left, right) numbersep=5pt, % how far the line-numbers are from the code numberstyle=\tiny\color{mygray}, % the style that is used for the line-numbers rulecolor=\color{black}, % if not set, the frame-color may be changed on line-breaks within not-black text (e.g. comments (green here)) showspaces=false, % show spaces everywhere adding particular underscores; it overrides 'showstringspaces' showstringspaces=false, % underline spaces within strings only showtabs=false, % show tabs within strings adding particular underscores %stepnumber=2, % the step between two line-numbers. If it's 1, each line will be numbered stringstyle=\color{mymauve}, % string literal style tabsize=2, % sets default tabsize to 2 spaces title=\lstname % show the filename of files included with \lstinputlisting; also try caption instead of title } \begin{center} \section*{\texttt{Solutions to Homework 1}} \end{center} \subsection*{Exercise 1} Adding page 5 as asked (pages 3 and 5 link to each other) gives the new link matrix: \begin{align*} A = \begin{bmatrix} 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\[0.3em] \frac{1}{3} & 0 & 0 & 0 & 0 \\[0.3em] \frac{1}{3} & \frac{1}{2} & 0 & \frac{1}{2} & 1 \\[0.3em] \frac{1}{3} & \frac{1}{2} & 0 & 0 & 0 \\[0.3em] 0 & 0 & \frac{1}{2} & 0 & 0 \\[0.3em] \end{bmatrix} \end{align*} As before, we are looking for an eigenvector $\mathbf{x} = [ x_1, x_2, x_3, x_4, x_5]^T$ corresponding to the eigenvalue 1 (which exists because $\mathbf{A}$ is column-stochastic). This $\mathbf{x}$ will be our ranking. Solve $\mathbf{A} \mathbf{x} = \mathbf{x}$, i.e. $(\mathbf{A} - \mathbf{I}) \mathbf{x} = 0$, for $\mathbf{x}$ : \begin{alignat}{5} -x_1 & & +\frac{1}{2}x_3 & +\frac{1}{2}x_4 & & =0 \\ \frac{1}{3}x_1 & -x_2 & & & &=0 \\ \frac{1}{3}x_1 & +\frac{1}{2}x_2 & -x_3 & +\frac{1}{2}x_4 & +x_5 &=0\\ \frac{1}{3}x_1 & +\frac{1}{2}x_2 & & - x_4 & &=0\\ & &\frac{1}{2}x_3 & & -x_5 &=0 \end{alignat} Solve `by hand' first, then confirm solution with \texttt{MATLAB}. Set $x_3 = 2$ and $x_5 = 1$ to satisfy equation (5). Then subtract (4) from (3) to obtain $x_4 = 2/3$, plug $x_3$ and $x_4$ into equation (1) to get $x_1 = 4/3$ . Finally plug $x_1$ into equation (2) to get $x_2 = 4/9$. \bigskip So we've got one eigenvector: $\mathbf{x} = [ 4/3, 4/9, 2, 2/3, 1]^T$ corresponding to $\lambda = 1$; there are many more solutions - as many as you'd like, all collinear to $\mathbf{x}$. (We can agree to pick the one that ensures the components of $\mathbf{x}$ sum to $1$, i.e. scale $\mathbf{x}$ by dividing each component by the sum of $\mathbf{x}$'s components.) \pagebreak Here's some \texttt{MATLAB} that does the same: {\ttfamily \begin{lstlisting} A = [ 0 0 1/2 1/2 0 ; 1/3 0 0 0 0 ; 1/3 1/2 0 1/2 1 ; 1/3 1/2 0 0 0 ; 0 0 1/2 0 0 ]; [V D] = eig(A1); disp('the eigenvector associated with eigenvalue 1:') V(:,1) x = V(:,1)/sum(V(:,1)) figure(1) bar(x) \end{lstlisting} } \begin{center} \includegraphics[scale=0.8]{Exercise1rank} \end{center} Thus the new ranking is $$ x = [\ 0.2449, \ 0.0816, \ 0.3673,\ 0.1224, \ 0.1837 \ ] $$ and page 3 is now ranked the highest. \pagebreak \subsection*{Exercise 7} Let $\mathbf{A}$ and $\mathbf{S}$ be column stochastic matrices, i.e. $\sum_i A_{ij} = 1$ and $\sum_i S_{ij} = 1$ for every $j$-th column of $\mathbf{A}$ or $\mathbf{S}$. Let $0 \leq m \leq 1$ and let $\mathbf{M} = m\mathbf{A} + (1-m)\mathbf{S}$. Consider the sum of the elements in $\mathbf{M}$'s $j$-th column: \begin{align*} \sum_i M_{ij} &= \sum_i (m A_{ij} + (1 - m) S_{ij}) \\ &= \sum_i m A_{ij} + \sum_i (1 - m) S_{ij} \\ &= m \sum_i A_{ij} + (1 - m) \sum_i S_{ij} \\ &= m + (1 - m) \\ &= 1 \end{align*} This being true for any $j$, $\mathbf{M}$ is column-stochastic. \bigskip \subsection*{Exercise 11} Here the link matrix $\mathbf{A}$ is the same as in problem 1. Set $ m = 0.15$ and write down the matrix $\mathbf{M} = 0.85 \mathbf{A} + 0.15 \mathbf{S}$ where $\mathbf{S}$ is the $5 \times 5$ matrix each element of which is 1/5.\\ {\ttfamily \begin{lstlisting} A = [ 0 0 1/2 1/2 0 ; 1/3 0 0 0 0 ; 1/3 1/2 0 1/2 1 ; 1/3 1/2 0 0 0 ; 0 0 1/2 0 0 ]; S = 1/5 * ones(5); m = 0.15; M = (1 - m)*A + m*S [V D] = eig(M) % diag(D) ranks = V(:,1) / sum(V(:,1)) bar(ranks) \end{lstlisting} } Here the output is: $$ ranks = [\ 0.2371, \ 0.0972, \ 0.3489, \ 0.1385, \ 0.1783 \ ] $$ Slightly different scores but the relative ranking of the pages is the same. \begin{center} \includegraphics[scale=0.7]{Exercise11rank} \end{center} \end{document} \end{document} This is never printed