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Eigenspaces and the Fundamental Theorem of Algebra

Norbert A'Campo

The following well-known theorem will be proved first, and then the so-called Fundamental Theorem of Algebra will be deduced from it.

Theorem 1   Let $V$ be a ${\bf C}$-vectorspace of finite dimension, $\mathop{\rm dim}V\geq 2$, 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< \mathop{\rm dim}W < \mathop{\rm dim}V $.


Preliminaries: 1. The angular variation $\theta=\theta(x,y)\in(-\pi/3,\pi/3)$ is defined for $x\in{\bf C},y\in{\bf C}$, with $\vert x-y\vert<\mathop{\rm Min\,}\nolimits \{\vert x\vert,\vert y\vert\}$ by the equation

\begin{displaymath}{y\over x} \cdot {\vert x\vert \over \vert y\vert}=\cos\theta+i \sin\theta=e^{i\theta}.\end{displaymath}

For $x,y,z\in{\bf C}$ with $\vert x-y\vert<\mathop{\rm Min\,}\nolimits \{\vert x\vert,\vert y\vert\}$, $\vert y-z\vert<\mathop{\rm Min\,}\nolimits \{\vert y\vert,\vert z\vert\}$ and $\vert x-z\vert<\mathop{\rm Min\,}\nolimits \{\vert x\vert,\vert z\vert\}$ the angular variations obey the following additive rule: $\theta(x,z)=\theta(x,y)+\theta(y,z).$

2. For a continuous curve $\gamma:[0,1]\to{\bf C}-\{0\} $ the angular variation along $ \gamma$ is defined to be

\begin{displaymath}\theta(\gamma):=\theta_N(\gamma):=\sum^N_{n=1} \theta\left(\g...
...\over N}\right),\gamma\left({n\over N}\right)\right)\in{\bf R},\end{displaymath}

where $N\in {\bf N},N\not=0$, is chosen in such a way that for all $t,t'\in[0,1]$ with $\vert t-t'\vert\leq{1\over N}$ the inequality $\vert\gamma(t)-\gamma(t')\vert<\mathop{\rm Min\,}\nolimits \{\vert\gamma(t)\vert,\vert\gamma(t')\vert\}$ 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:

\begin{displaymath}{\gamma(0)\over\gamma(1)}\cdot e^{i\theta(\gamma)}=
\prod^N_{...
...\over N})\vert\over \vert\gamma({n\over N})\vert}
\in{\bf R}_+.\end{displaymath}

3. For a closed continuous curve $\gamma:[0,1]\to{\bf C}-\{0\} $ it follows ${1\over2\pi}\theta(\gamma)\in{\bf Z}.$ To see this, observe: $e^{i\theta(\gamma)}={\gamma(o)\over\gamma(1)} \cdot e^{i\theta(\gamma)}\in{\bf R}_+.$

4. For a constant curve $\gamma:[0,1]\to{\bf C}-\{0\} $ we have $\theta(\gamma)=0$. To see this, compute with $N=1$.

5. For the curve $\gamma:[0,1]\to{\bf C}-\{0\}, \gamma(t)=a e^{2\pi int},n\in{\bf Z}, a\in{\bf C}, a\not=0$, we have $\theta(\gamma)=2\pi n$. Compute with $N=6\vert n\vert+1$.

6. For a continuous family $(\gamma_{s})_{s\in[0,1]}$ of closed, continuous curves $\gamma_{s}:[0,1]\to{\bf C}-\{0\}$ we have $\theta(\gamma_{0}) = \theta(\gamma_{1}).$

Proof:The function $(t,s)\in[0,1]\times[0,1] \mapsto\gamma_{s}(t)\in{\bf C}-\{0\}$ is uniformly continuous. Choose $N\in{\bf N},N\not=0,$ in such a way that for all $(t,s),(t',s')\in[0,1]\times
[0,1]$ with $\vert t-t'\vert\leq{1\over N}, \vert s-s'\vert\leq{1\over N}$ the inequality $\vert\gamma_{s}(t)-\gamma_{s'}(t')\vert<\mathop{\rm Min\,}\nolimits \{\vert\gamma_{s}(t)\vert,
\vert\gamma_{s'}(t')\vert\}$ holds. The function

\begin{displaymath}s\in[0,1]\mapsto{1\over 2\pi}\theta(\gamma_{s})=
{1\over 2\pi...
... N}\right),\gamma_{s}\left({n\over N}\right)
\right)\in {\bf R}\end{displaymath}

is continuous, with values in ${\bf Z}$ by 3, and therefore constant. Hence $\theta(\gamma_{0}) = \theta(\gamma_{1}).$


Proof of the theorem:Choose a basis $f_{1}, ... ,f_{d}$, $d=\mathop{\rm dim}(V)$, for $V$ and consider the function

\begin{displaymath}\phi: V\to{\bf C}, u\in V\mapsto \phi(u):= \mathop{\rm det}(B_{u}),\end{displaymath}

where $B_{u}: V\to V, u\in V,$ is the linear transformation given by

\begin{displaymath}B_{u}(f_{j})= A^{j-1}(u),\ 1\leq j\leq d.\end{displaymath}

We have $\phi(\lambda u)= \lambda^{\mathop{\rm dim}(V)} \phi(u),$ for $\lambda\in{\bf C}, u\in V$. For any $w\in W$ with $w\not=0$ and $\phi(w)=0$ the subspace $W=: \mathop{\hbox{\rm Span}}\nolimits [w,A(w), ... ,A^{d-1}(w)]$ satisfies: $A(W)\subset W \quad \hbox{\rm and } \quad 0<\mathop{\rm dim}(W)<\mathop{\rm dim}(V).$ Therefore it suffices to prove the existence of a vector $w\in V$ with $w\not=0$ and $\phi(w)=0.$

\scalebox{0.5}[0.5]{\includegraphics{ea.eps}}
Fig. 1. $a:=e^{2\pi it}u,\,\,b:=(1-s)a+sv=(1-s)e^{2\pi it}u +sv$

Choose in $V$ two linear independent vectors $u$ and $v$, and consider the family of curves $\gamma_{s} : [0,1] \to {\bf C}, s\in[0,1]$ given by:

\begin{displaymath}\gamma_{s}(t)= \phi\left(\left(1-s\right) e^{2\pi i t} u +s v\right).\end{displaymath}

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

\begin{displaymath}\gamma_{0}(t)= e^{2\pi i t \mathop{\rm dim}(V) } \phi(u),\,\,\,\gamma_{1}(t) = \phi(v).\end{displaymath}

We obtain: $2\pi\mathop{\rm dim}(V) = \theta(\gamma_0) = \theta(\gamma_1) = 0,$ using the preliminaries 5, 6 and 4, which contradicts the hypothesis $\mathop{\rm dim}V \ge 2$.


Corollary 1   Let $V$ be a vectorspace of finite dimension, $\mathop{\rm dim}V\geq1$, and let $A:V\to V$ a linear transformation. Then there exists a linear subspace $L$ in $V$ with $A(L)\subset L$ and $\mathop{\rm dim}L=1$. Moreover, every $v\in L$, $v\not=0$, is an eigenvector, i.e. for some $\lambda\in {\bf C}$ the equation $A(v)=\lambda v$ holds. The matrix of $A$ is uppertriangular with respect to some basis of $V.$

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 $\mathop{\rm dim}W_{j+1}<\mathop{\rm dim}W_{j}.$ 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 $0 < \mathop{\rm dim}(W) < \mathop{\rm dim}(W_{j}/W_{j-1}).$ So, by the theorem we have $\mathop{\rm dim}(W_{j}/W_{j-1})=1.$ Then $\mathop{\rm dim}W_{j}=j$ 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.


Corollary 2   (``Fundamental Theorem of Algebra'') Let $P(X)\in{\bf C}[X]$ be a polynomial of degree $\geq 1$. Then their exists $\lambda\in {\bf C}$ with $P(\lambda)=0$.

Proof:Let $P(X)=X^{n}+c_{n-1}X^{n-1}+ ... +c_{0}\in{\bf C}[X]$ be a polynomial of degree $n\geq 1$. Consider the linear transformation $A:{\bf C}^{n} \to {\bf C}^{n}$ acting on the standard basis $e_{1},e_{2}, \dots ,e_{n}$ of ${\bf C}^{n}$ as follows:

\begin{displaymath}A(e_{j})=e_{j+1},\ 1\leq j <n,\,\,\,\,A(e_{n})=-\sum_{j=1}^{j=n}c_{j-1}e_{j}.\end{displaymath}

Clearly $P(A)(e_{1})=0$ and $P(A)(e_{j})=P(A)(A^{j-1}(e_{1}))=A^{j-1}(P(A)(e_{1}))=0$, $1<j\leq n$. We conclude that $P(A)=0$. Thus for an eigenvector $v\in {\bf C}^{n}$, with eigenvalue $\lambda\in {\bf C}$, we get $0=P(A)(v)=P(\lambda)v$, and hence $P(\lambda)=0$.


Remark. The Fundamental Theorem of Algebra was proved by C. Gauss [G]. As is done in the textbooks on Linear Algebra, one can use it also to proof directly the existence of eigenvectors for endomorphism of finite dimensional vector spaces over ${\bf C}$. A first appearance of eigenvectors for symetric matrices is in the work of Cauchy on inertials elipsoids (see the textbook of P.M. Cohn [C]). The Jordan Normal Form Theorem is formulated and proved by C. Jordan [J] for general fields $k$. If any polynomial $\chi(t) \in k[t]$ of positiv degree factors in $k[t]$ in polynomials of degree one, the Normal Form Theorem proofs the existence of invariant flags for endomorphisms of finite dimensional vector spaces over the field $k$.

[C] P.M. Cohn, Algebra, Wiley, New York, 1982.

[J] Camille Jordan, Traité des substitutions et équations algébriques, Gauthiers-Villars, Paris, 1957.

[G] C. Gauss, .




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