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65F15-Eigenvector.tex
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65F15-Eigenvector.tex
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\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{Eigenvector}
\pmcreated{2013-03-22 12:11:55}
\pmmodified{2013-03-22 12:11:55}
\pmowner{mathcam}{2727}
\pmmodifier{mathcam}{2727}
\pmtitle{eigenvector}
\pmrecord{12}{31497}
\pmprivacy{1}
\pmauthor{mathcam}{2727}
\pmtype{Definition}
\pmcomment{trigger rebuild}
\pmclassification{msc}{65F15}
\pmclassification{msc}{65-00}
\pmclassification{msc}{15A18}
\pmclassification{msc}{15-00}
\pmrelated{SingularValueDecomposition}
\pmrelated{Eigenvalue}
\pmrelated{EigenvalueProblem}
\pmrelated{SimilarMatrix}
\pmrelated{DiagonalizationLinearAlgebra}
\pmdefines{scalar multiple}
\endmetadata
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
%\usepackage{psfrag}
%\usepackage{graphicx}
%%%%\usepackage{xypic}
\begin{document}
Let $A$ be an $n \times n$ square matrix and $x$ an $n\times 1$ column vector. Then a (right) \emph{eigenvector} of $A$ is a nonzero vector $x$ such that
$$ Ax = \lambda x $$
for some scalar $\lambda$, i.e. such that the image of $x$ under the transformation $A$ is a \emph{scalar \PMlinkescapetext{multiple}} of $x$. One can similarly define left eigenvectors in the case that $A$ acts on the right.
One can find eigenvectors by first finding eigenvalues, then for each eigenvalue $\lambda_i$, solving the system
$$ (A-\lambda_i I) x_i = 0 $$
to find a form which characterizes the eigenvector $x_i$ (any \PMlinkescapetext{multiple} of $x_i$ is also an eigenvector). Of course, this is not necessarily the best way to do it; for this, see singular value decomposition.
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\end{document}