\documentclass[12pt]{article} \usepackage{graphics} \date{} \textwidth=135mm \textheight=190mm \oddsidemargin=.15in \evensidemargin=.15in \topmargin=-1.5cm \parskip=11pt \def\qed{{\hfill\rule{1.2ex}{1.2ex}}} \newtheorem{Th}{Theorem} \newenvironment{Proof}{\begingroup\noindent{\bf Proof:}}{\qed\medskip\endgroup} \newenvironment{Proof of the theorem}{\begingroup\noindent{\bf Proof of the theorem:}}{\qed\medskip\endgroup} \newenvironment{Proof of corollary}{\begingroup\noindent{\bf Proof of corollary:}}{\qed\medskip\endgroup} \newtheorem{Cor}{Corollary} \def\Min{\mathop{\rm Min\,}\nolimits} \def\dim{\mathop{\rm dim}} \def\det{\mathop{\rm det}} \def\C{{\bf C}} \def\R{{\bf R}} \def\N{{\bf N}} \def\Z{{\bf Z}} \def\Span{\mathop{\hbox{\rm Span}}\nolimits} \title{Eigenspaces and the Fundamental Theorem of Algebra} \author{Norbert A'Campo} \begin{document} \maketitle \noindent The following well-known theorem will be proved first, and then the so-called Fundamental Theorem of Algebra will be deduced from it. \begin{Th} Let $V$ be a $\C$-vectorspace of finite dimension, $\dim V\geq 2$, and let $A:V\to V$ be a linear transformation. Then there exists a linear subspace $W$ in $V$ with $A(W)\subset W $ and $ 0< \dim W < \dim V $. \end{Th} \medskip\noindent{\bf Preliminaries:} {\bf 1.} The angular variation $\theta=\theta(x,y)\in(-\pi/3,\pi/3)$ is defined for $x\in\C,y\in\C$, with $|x-y|<\Min \{|x|,|y|\}$ by the equation $${y\over x} \cdot {|x| \over |y|}=\cos\theta+i \sin\theta=e^{i\theta}.$$ For $x,y,z\in\C$ with $|x-y|<\Min\{|x|,|y|\}$, $|y-z|<\Min\{|y|,|z|\}$ and $|x-z|<\Min\{|x|,|z|\}$ the angular variations obey the following additive rule: $\theta(x,z)=\theta(x,y)+\theta(y,z).$ \noindent{\bf 2.} For a continuous curve $\gamma:[0,1]\to\C-\{0\} $ the angular variation along $ \gamma$ is defined to be $$\theta(\gamma):=\theta_N(\gamma):=\sum^N_{n=1} \theta\left(\gamma\left({n-1 \over N}\right),\gamma\left({n\over N}\right)\right)\in\R,$$ where $N\in \N,N\not=0$, is chosen in such a way that for all $t,t'\in[0,1]$ with $|t-t'|\leq{1\over N}$ the inequality $|\gamma(t)-\gamma(t')|<\Min\{|\gamma(t)|,|\gamma(t')|\}$ holds. The angular variation $\theta(\gamma)$ does not depend on the actual choice of $N,$ since for two such choices $N$ and $N',$ it follows from the additive rule: $\theta_N(\gamma)=\theta_{NN'}(\gamma)=\theta_{N'}(\gamma).$ The angular variation $\theta(\gamma)=\theta_N(\gamma)$ satisfies: $${\gamma(0)\over\gamma(1)}\cdot e^{i\theta(\gamma)}= \prod^N_{n=1} {|\gamma({n-1\over N})|\over |\gamma({n\over N})|} \in\R_+.$$ \noindent{\bf 3.} For a closed continuous curve $\gamma:[0,1]\to\C-\{0\}$ it follows ${1\over2\pi}\theta(\gamma)\in\Z.$ To see this, observe: $e^{i\theta(\gamma)}={\gamma(o)\over\gamma(1)} \cdot e^{i\theta(\gamma)}\in\R_+.$ \noindent{\bf 4.} For a constant curve $\gamma:[0,1]\to\C-\{0\}$ we have $\theta(\gamma)=0$. To see this, compute with $N=1$. \noindent{\bf 5.} For the curve $\gamma:[0,1]\to\C-\{0\}, \gamma(t)=a e^{2\pi int},n\in\Z, a\in\C, a\not=0$, we have $\theta(\gamma)=2\pi n$. Compute with $N=6|n|+1$. \noindent{\bf 6.} For a continuous family $(\gamma_{s})_{s\in[0,1]}$ of closed, continuous curves $\gamma_{s}:[0,1]\to\C-\{0\}$ we have $\theta(\gamma_{0}) = \theta(\gamma_{1}).$ \begin{Proof} The function $(t,s)\in[0,1]\times[0,1] \mapsto\gamma_{s}(t)\in\C-\{0\}$ is uniformly continuous. Choose $N\in\N,N\not=0,$ in such a way that for all $(t,s),(t',s')\in[0,1]\times [0,1]$ with $|t-t'|\leq{1\over N}, |s-s'|\leq{1\over N}$ the inequality $|\gamma_{s}(t)-\gamma_{s'}(t')|<\Min\{|\gamma_{s}(t)|, |\gamma_{s'}(t')|\}$ holds. The function $$s\in[0,1]\mapsto{1\over 2\pi}\theta(\gamma_{s})= {1\over 2\pi} \sum^N_{n=1} \theta\left( \gamma_{s}\left({n-1\over N}\right),\gamma_{s}\left({n\over N}\right) \right)\in \R$$ is continuous, with values in $\Z$ by 3, and therefore constant. Hence $\theta(\gamma_{0})=\theta(\gamma_{1}).$ \end{Proof} \begin{Proof of the theorem} Choose a basis $f_{1}, ... ,f_{d}$, $d=\dim(V)$, for $V$ and consider the function $$ \phi: V\to\C, u\in V\mapsto \phi(u):= \det(B_{u}),$$ where $B_{u}: V\to V, u\in V,$ is the linear transformation given by $$B_{u}(f_{j})= A^{j-1}(u),\ 1\leq j\leq d.$$ We have $\phi(\lambda u)= \lambda^{\dim(V)} \phi(u),$ for $\lambda\in\C, u\in V$. For any $w\in W$ with $w\not=0$ and $\phi(w)=0$ the subspace $W=: \Span[w,A(w), ... ,A^{d-1}(w)]$ satisfies: $A(W)\subset W \quad \hbox{\rm and } \quad 0<\dim(W)<\dim(V).$ Therefore it suffices to prove the existence of a vector $w\in V$ with $w\not=0 $ and $\phi(w)=0.$ \begin{center} \scalebox{0.5}[0.5]{\includegraphics{ea.eps}} \newline {Fig. 1. $a:=e^{2\pi it}u,\,\,b:=(1-s)a+sv=(1-s)e^{2\pi it}u +sv$} \end{center} Choose in $V$ two linear independent vectors $u$ and $v$, and consider the family of curves $\gamma_{s} : [0,1] \to \C, s\in[0,1]$ given by: $$\gamma_{s}(t)= \phi\left(\left(1-s\right) e^{2\pi i t} u +s v\right).$$ We claim: there exists $(t,s)\in [0,1] \times [0,1]$ such that $\gamma_{s}(t) = 0 $, and the vector $w:= (1-s) e^{2\pi i t} u + s v\in V$ satisfies $w\not=0$ and $\phi(w)=0$. We prove the claim by contradiction. Assume $\gamma_{s}(t)\not=0$ for all $(t,s)\in [0,1] \times [0,1]$. Observe $$ \gamma_{0}(t)= e^{2\pi i t \dim(V) } \phi(u),\,\,\,\gamma_{1}(t) = \phi(v).$$ We obtain: $2\pi\dim(V) = \theta(\gamma_0) = \theta(\gamma_1) = 0,$ using the preliminaries 5, 6 and 4, which contradicts the hypothesis $\dim V \ge 2$. \end{Proof of the theorem} \begin{Cor}Let $V$ be a vectorspace of finite dimension, $\dim V\geq1$, and let $A:V\to V$ a linear transformation. Then there exists a linear subspace $L$ in $V$ with $A(L)\subset L$ and $\dim L=1$. Moreover, every $v\in L$, $v\not=0$, is an eigenvector, i.e. for some $\lambda\in \C$ the equation $A(v)=\lambda v$ holds. The matrix of $A$ is uppertriangular with respect to some basis of $V.$ \end{Cor} \begin{Proof} Consider a chain of maximal length of linear subspaces in V,$W_0\subset W_1\subset ... \subset W_r$ with $A(W_{j})\subset W_{j}$ and $\dim W_{j+1}<\dim W_{j}.$ Then $W_0=\{0\},W_r=V$ and $A$ induces on each quotient space $W_{j}/W_{j-1}$ a linear transformation, say $A_j,$ such that there is no subspace $W$ in $W_{j}/W_{j-1}$ with $A_j(W) \subset W$ and $0 < \dim(W) < \dim(W_{j}/W_{j-1}).$ So, by the theorem we have $\dim(W_{j}/W_{j-1})=1.$ Then $\dim W_{j}=j$ and the subspace $L:=W_{1}$ has the property of the corollary. A system $e_1,e_2, .. ,e_r$ such that $e_j \notin W_{j-1},1\leq j\leq r,$ is a basis of $V$ for which the matrix of $A$ is uppertriangular. \end{Proof} \begin{Cor} {\rm (``Fundamental Theorem of Algebra'')} Let $P(X)\in\C[X]$ be a polynomial of degree $\geq 1$. Then their exists $\lambda\in\C$ with $P(\lambda)=0$. \end{Cor} \begin{Proof} Let $P(X)=X^{n}+c_{n-1}X^{n-1}+ ... +c_{0}\in\C[X]$ be a polynomial of degree $n\geq 1$. Consider the linear transformation $A:\C^{n} \to \C^{n}$ acting on the standard basis $e_{1},e_{2}, \dots ,e_{n}$ of $\C^{n}$ as follows: $$A(e_{j})=e_{j+1},\ 1\leq j