I found a proof in an online course note which purports to show that two definitions of matrix norms are equivalent, however, I have some doubts regarding the proof, I would like a second pair of eyes     Claim: $\|A\| = \max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} =  \max\limits_{\|x\| = 1} \|Ax\|$     Proof:     $(\text{first show} \max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} \leq \max\limits_{\|x\| = 1} \|Ax\|)$       Suppose $x \neq 0$, then $\max\limits_{ x \neq 0}   \dfrac{\|Ax\|}{\|x\|} = \max\limits_{ x \neq 0}  \|A\dfrac{x}{\|x\|}\|$. Let $z = \dfrac{x}{\|x\|}$,    $\|z\| = 1$. Therefore, $\max\limits_{ x \neq 0}   \dfrac{\|Ax\|}{\|x\|} \leq  \max\limits_{ \|z\| =1} \|Az\| =  \max\limits_{ \|x\| =1} \|Ax\|$   -     $(\text{next show:}\max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} \geq \max\limits_{\|x\| = 1} \|Ax\|)$      Suppose that $\|x\| = 1$, therefore $\max\limits_{\|x\| = 1} \|Ax\| =  \max\limits_{\|x\| = 1}  \dfrac{\|Ax\|}{\|x\|} \leq \max\limits_{x \neq 0}  \dfrac{\|Ax\|}{\|x\|}$, where the inequality follows from maximization   over a superset of $\{x\in \mathbb{R}^n:\|x\|=1\}$.   -     Thus, $\max\limits_{ x \neq 0} \dfrac{\|Ax\|}{\|x\|} =   \max\limits_{\|x\| = 1} \|Ax\|$.   It seems that there is no justification for $\max\limits_{ x \neq 0}   \dfrac{\|Ax\|}{\|x\|} \leq  \max\limits_{ \|z\| =1} \|Az\|$. It doesn't make sense, because we are maximizing over a much larger set than the unit circle. Can someone see if there is indeed a problem with the proof and if there is some opportunity to fix it?
How do you prove that these four definitions of the operator norm are equivalent? $$\begin{align*} \lVert A\rVert_{\mathrm{op}} &amp;= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\ &amp;=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\ &amp;=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\ &amp;=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}. \end{align*}$$
Sharing Test code in Maven What is the difference between the three sentences below?   I have been married with her for three years. I was married with her for three years. I married her for three years. require_once () or die() not working I want to create a function that performs a function passed by parameter on a set of data. How do you pass a function as a parameter in C? I'm looking for a library/method to parse an html file with more html specific features than generic xml parsing libraries. In my Master thesis, I would like to have the title page without any page number and an acknowledgments section and the table of contents with roman numbers. The rest of the document should be numbered arabic.  I have tried this, to have roman numbers for the Table of contents and normal numbers for the rest.  \documentclass[12pt,ngerman,openany]{report} \usepackage{titlesec} \usepackage{titletoc} \usepackage{cite} \usepackage[a4paper]{geometry} %\usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage[ngerman]{babel} %\usepackage{ngerman}{babel} \usepackage[T1]{fontenc} \usepackage{graphicx} \usepackage{float} \selectlanguage{ngerman} \usepackage{amssymb} \usepackage{amsmath} \usepackage{url} \usepackage{epstopdf} \usepackage{scrhack} \usepackage{listings} \usepackage{caption} \titleformat{\chapter}{\rmfamily\huge}{\textbf\thechapter.}{25pt}{\huge\textbf} \defcaptionname{ngerman}{\lstlistingname}{Code} \defcaptionname{ngerman}{\lstlistlistingname }{Codeverzeichnis}  \lstloadlanguages{R}  \lstset{% general command to set parameter(s)    basicstyle=\sffamily\footnotesize, % print whole listing small    keywordstyle=\sffamily\footnotesize\bfseries, % ubold black keywords    identifierstyle=, % nothing happens    commentstyle=\sffamily\footnotesize\slshape, % green comments    stringstyle=\sffamily\footnotesize, % typewriter type for strings    showstringspaces=false, % no special string spaces    numbers=left,    numberstyle=\sffamily\footnotesize,    stepnumber=1,    numbersep=10pt,    showspaces=false,    showtabs=false,    frame=lines,    morecomment=[l]{\%},    float=htbp,    numberbychapter=true  }   \begin{document} \let\cleardoublepage\clearpage   \pagenumbering{roman}  \tableofcontents  \clearpage  \vspace*{75mm} \chapter{Einleitung} \section{Motivation} bla bla bla \section{Ziel}  bla bla \chapter{Stand der Technik zur Mustererkennung} \section{Die Zeitreihen} bla bla \subsection{Beschreibung} bla bla \subsection{Erklärung}  bla bla  \subsection{Vorhersage} bla bla \subsection{Kontrolle} bla bla \section{Feature Extraktion} bla bla \subsection{Autokorrelation} \subsection{Was ist Autokorrelation?} bla bla   \subsection{Autokorrelation und Zeitreihen} bla bla  \subsubsection{Zeitreihe-Diagramm}  bla bla \subsubsection{Lagged Streudiagramm} bla bla  \subsubsection{Autocorrelation-Function (Korrelogramm)} bla bla \subsection{Spektralanalyse} \subsection{Was ist Spektral-Analyse} bla bla \end{document} What is a NullReferenceException, and how do I fix it? Extra space between lines (gb4e) I am using gb4e.sty to provide inter-linear language glossing. This works well, but I want to get wider spacing between lines when sentences have 2 or more lines like this;   Here is my source code:  \documentclass{article} \usepackage{times} \usepackage{gb4e} \noautomath \let\eachwordone\it \setlength{\glossglue}{10pt plus 2pt minus 1pt}  \begin{document} \begin{exe} \ex \gll mukasi monosugoka jii atama=no an wanrjoku=no an samurai=ga utan=bee=cjo\\ old.days very good head=NOM exist strength=NOM exist samurai=NOM existed=HSY=ASS\\ \trans `Once upon a time, there was a samurai with a smart brain and strength.' \end{exe} \end{document}   Does anyone have a solution to this problem? I am currently using Ubuntu 10.04. I know there is a 10.10 release, but can I upgrade directly to 11.04? Could you walk me through the steps please? Why does Bash's source not need the execution bit? Carmichael number square free Show that if $n$ is a , then $n$ is a square-free.  I did this: Let $n= (p^t)(m)$  where $t &gt;1$. Then by modular property,  $$b^p= b \mod n , \,\, b^m= b \mod n$$ Above two equations are also true in$\mod p$, $\mod p^t$. But in$\mod m$, just $b^n=b \mod m$.  And I tried to use CRT But I couldn't.  I think that I've chosen wrong way CSH - How to check if input is NOT number Need to apply it on csh.  to check if the inputted $2 is not number nor the word "all"  if ($#argv == 2 &amp;&amp; ($2 != all &amp;&amp; $2 != **any number**)) then    echo "wrong parameter" MySQL Results as comma separated list I need to run a query like:  SELECT p.id, p.name,         (SELECT name            FROM sites s           WHERE s.id = p.site_id) AS site_list   FROM publications p   But I'd like the sub-select to return a comma separated list, instead of a column of data.  Is this even possible, and if so, how? Variation of the Kempner series – convergence of series $\sum\frac{1}{n}$ where $9$ is not a digit of $1/n$. It is easy to argue that the  converges:  $$  \sum\limits_{\substack{n \text{ s.t. 9 is}\\\text{ not a digit} \\\text{ of } n}}  \frac{1}{n} &lt; \infty$$  Let $E \subset \Bbb N_{&gt;0}$  the subset of the positive integers  such that $9$ is not a digit of the decimal expansion of $1/n$ (the decimal expansion is not allowed to have a trailing infinite sequence of "$9$"s. For instance $0.24999...$ is not allowed).  Here are the  that don't belong to $E$ : $11,13,17,19,21,23,29,31,34,38,41,…$ (not known by the OEIS, by the way).  My question is:     Does the series    $$ \sum\limits_{n \in E} \frac{1}{n} \tag 1$$ converge?   My attempt is :   Let $1/n = 0,a_1 a_2 \dots a_k \overline{b_1 b_2 \dots b_m}$ with $n \in E$. Since $1/n$ has no digit "9", we have at most $9^{k+m}$ possibilities for the $a_i$'s and $b_j$'s. Moreover, $1/n ≥ 0,00...0\overline{00...01}≥1/10^{k+m}$. But then I can only bound my series $(1)$ from below, by some real number. So, this is not a clue for the divergence of the series. Apparently, the numbers of the form $n=10k+1$ don't belong to $E$. Maybe we can find sufficiently many numbers that have $9$ in the decimal representation of their reciprocals, so that $(1)$ could converge...   Any comment will be appreciated ! Unit testing void methods? What is the best way to unit test a method that doesn't return anything? Specifically in c#.  What I am really trying to test is a method that takes a log file and parses it for specific strings. The strings are then inserted into a database. Nothing that hasn't been done before but being VERY new to TDD I am wondering if it is possible to test this or is it something that doesn't really get tested. Why do we use this definition of "algebraic integer"? A number is an &quot;algebraic integer&quot; if it is the root to a monic polynomial with integer coefficients. Artin says (Algebra, p. 411):  The concept of algebraic integer was one of the most important discoveries of number theory. It is not easy to explain quickly why it is the right definition to use, but roughly speaking, we can think of the leading coefficient of the primitive irreducible polynomials $f(x)$ as a &quot;denominator.&quot; If $\alpha$ is the root of an integer polynomial $f(x)=dx^n+a_{n-1}x^{n-1}...$ then $d\alpha$ is an algebraic integer, because it is a root of the monic integer polynomial $x^n + a_{n-1}x^{n-1} + ... + d^{n-1}a_0$. Thus we can &quot;clear the denominator&quot; in any algebraic number by multiplying it with a suitable integer to get an algebraic integer.  When I first learned of algebraic integers, I looked online and saw some hints that maybe they were used to prove the Abel-Ruffini theorem. So I put off questioning their usage for a while; I now think I understand one proof of this theorem (the one at the end of Artin's Algebra) and it has nothing to do with algebraic integers (that I can tell). So basically: why is it important if a number is an algebraic integer? I think I understand what he's saying about the relationship between roots of integer polynomials and algebraic integers, but I fail to see why this is &quot;one of the most important discoveries of number theory.&quot; The proof of shrinking coefficients using ridge regression through "spectral decomposition" I have understood how ridge regression shrinks coefficients towards zero geometrically. Moreover I know how to prove that in the special "Orthonormal Case," but I am confused how that works in the general case via "Spectral decomposition." Can anybody please help me to prove this:  Let $p$ be greater than or equal to $1$. Show that for the space $\ell_p=\{(u_n):\sum_{n=1}^\infty |u_n|^p&lt;\infty\}$ of all $p$-summable sequences (with norm $||u||_p=\sqrt[p]{\sum_{n=1}^\infty |u_n|^p}\ )$, there is an inner product $&lt;\_\,|\,\_&gt; $ s.t. $||u||^2=&lt;u\,|\,u&gt;$ if and only if $p=2$. Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc. Let $P(z)=a_0+a_1z+\cdots+a_nz^n$ be a polynomial whose coefficients satisfy $$0&lt;a_0&lt;a_1&lt;\cdots&lt;a_n.$$  I want to show that the roots of $P$ live in unit disc. The obvious idea is to use Rouche's theorem, but that doesn't quite work here, at least with the choice $f(z)=a_nz^n, g(z)=$ (the rest).  Any ideas? ref to subsection number only is there a way to reference the subsection part only? Example:  \documentclass{article} \begin{document}  \section{This is my section} For further information see subsection \ref{subsecthree} within this section. \subsection{One} \subsection{Two} \subsection{Three}\label{subsecthree}  \end{document}   Usually gives (without the stars):  1 This is my section For further information see subsection **1.3** within this section.   But I want:  1 This is my section For further information see subsection **3** within this section.   Is that possible?  [Edit] To make it more clear: I am looking for a command that works like \ref, but prints (the arabic number of) the value of the subsection counter at the position of the \label.  ======================= == Solution in LaTeX ==  There is another solution below. This is a different solution I came up with  (I cannot post an answer to my own question, so I put it here):  \documentclass{article}  \newcommand{\labelsubseccounter}[1]{     \renewcommand\thesubsection{\arabic{subsection}}     \addtocounter{subsection}{-1}     \refstepcounter{subsection}     \label{#1}     \renewcommand\thesubsection{\thesection.\arabic{subsection}} }  \begin{document}  \section{This is my section} For further information see subsection \ref{subsecthree} within this section,  which is called \ref{onepointthree}, not \ref{onepointfour}. \subsection{One} \subsection{Two} \subsection{Three}\label{onepointthree}\labelsubseccounter{subsecthree} \subsection{Four}\label{onepointfour}  \end{document}   (I would like to insert the compiled code, but don't know how, sorry)