OperatornamekerL0IfAndOnlyIfLIsInjective 2004-10-17 11:02:20 2006-10-27 12:14:54 Theorem kernel of a linear mapping % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{mathrsfs} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions % % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\sR}[0]{\mathbb{R}} \newcommand{\sC}[0]{\mathbb{C}} \newcommand{\sN}[0]{\mathbb{N}} \newcommand{\sZ}[0]{\mathbb{Z}} \usepackage{bbm} \newcommand{\Z}{\mathbbmss{Z}} \newcommand{\C}{\mathbbmss{C}} \newcommand{\R}{\mathbbmss{R}} \newcommand{\Q}{\mathbbmss{Q}} \newcommand*{\norm}[1]{\lVert #1 \rVert} \newcommand*{\abs}[1]{| #1 |} \newtheorem*{thm}{Theorem} \newtheorem{defn}{Definition} \newtheorem{prop}{Proposition} \newtheorem{lemma}{Lemma} \newtheorem{cor}{Corollary} \begin{thm} A linear map between vector spaces is injective if and only if its kernel is $\{0\}$. \end{thm} \begin{proof} Let $L: V \to W$ be a linear map. Suppose $L$ is injective, and $L(v)=0$ for some vector $v\in V$. Also, $L(0) =0$ because $L$ is linear. Then $L(v)=L(0)$, so $v=0$. On the other hand, suppose $\operatorname{ker}L=\{0\}$, and $L(v)=L(v')$ for vectors $v,v'\in V$. Hence $L(v-v') = L(v)-L(v') = 0$ because $L$ is linear. Therefore, $v-v'$ is in $\operatorname{ker}L=\{0\}$, which means that $v-v'$ must be $0$. \end{proof}