eigenvalue Eigenvalue 2002-01-19 23:40:28 2006-06-09 15:08:13 Definition eigenvalue spectral value \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \PMlinkescapeword{clearly} \PMlinkescapeword{information} \PMlinkescapeword{scheme} \PMlinkescapeword{connection} Let $V$ be a vector space over a field $k$, and let $A$ be an endomorphism of $V$ (meaning a linear mapping of $V$ into itself). A scalar $\lambda\in k$ is said to be an \emph{eigenvalue} of $A$ if there is a nonzero $x \in V$ for which \begin{equation} Ax = \lambda x\;. \end{equation} Geometrically, one thinks of a vector whose direction is unchanged by the action of $A$, but whose magnitude is multiplied by $\lambda$. If $V$ is finite dimensional, elementary linear algebra shows that there are several equivalent definitions of an eigenvalue: \noindent (2) The linear mapping $$B=\lambda I - A$$ i.e. $B:x\mapsto \lambda x-Ax$, has no inverse. \noindent (3) $B$ is not injective. \noindent (4) $B$ is not surjective. \noindent (5) $\det(B)=0$, i.e. $\det(\lambda I-A)=0$. But if $V$ is of infinite dimension, (5) has no meaning and the conditions (2) and (4) are not equivalent to (1). A scalar $\lambda$ satisfying (2) (called a \emph{spectral value} of $A$) need not be an eigenvalue. Consider for example the complex vector space $V$ of all sequences $(x_n)_{n=1}^{\infty}$ of complex numbers with the obvious operations, and the map $A: V \to V$ given by $$ A(x_1, x_2, x_3, \dots) = (0, x_1, x_2, x_3, \dots) \;. $$ Zero is a spectral value of $A$, but clearly not an eigenvalue. Now suppose again that $V$ is of finite dimension, say $n$. The function $$\chi(\lambda)=\det(B)$$ is a polynomial of degree $n$ over $k$ in the variable $\lambda$, called the \emph{characteristic polynomial} of the endomorphism $A$. (Note that some writers define the characteristic polynomial as $\det(A-\lambda I)$ rather than $\det(\lambda I-A)$, but the two have the same zeros.) If $k$ is $\C$ or any other algebraically closed field, or if $k=\R$ and $n$ is odd, then $\chi$ has at least one zero, meaning that $A$ has at least one eigenvalue. In no case does $A$ have more than $n$ eigenvalues. Although we didn't need to do so here, one can compute the coefficients of $\chi$ by introducing a basis of $V$ and the corresponding matrix for $B$. Unfortunately, computing $n \times n$ determinants and finding roots of polynomials of degree $n$ are computationally messy procedures for even moderately large $n$, so for most practical purposes variations on this naive scheme are needed. See the eigenvalue problem for more information. If $k=\C$ but the coefficients of $\chi$ are real (and in particular if $V$ has a basis for which the matrix of $A$ has only real entries), then the non-real eigenvalues of $A$ appear in conjugate pairs. For example, if $n=2$ and, for some basis, $A$ has the matrix $$ A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $$ then $\chi(\lambda)=\lambda^2+1$, with the two zeros $\pm i$. Eigenvalues are of relatively little importance in connection with an infinite-dimensional vector space, unless that space is endowed with some additional structure, typically that of a Banach space or Hilbert space. But in those cases the notion is of great value in physics, engineering, and mathematics proper. Look for ``spectral theory'' for more on that subject.