# eigenvalue

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 eigenvalue     of $A$ if there is a nonzero $x\in V$ for which

 $Ax=\lambda x\;.$ (1)

Geometrically, one thinks of a vector whose direction is unchanged by the action of $A$, but whose magnitude is multiplied by $\lambda$.

(2) The linear mapping

 $B=\lambda I-A$

(4) $B$ is not surjective.

(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 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 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 $\mathbb{C}$ or any other algebraically closed field, or if $k=\mathbb{R}$ 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 $k=\mathbb{C}$ 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.

 Title eigenvalue Canonical name Eigenvalue Date of creation 2013-03-22 12:11:52 Last modified on 2013-03-22 12:11:52 Owner Koro (127) Last modified by Koro (127) Numerical id 15 Author Koro (127) Entry type Definition Classification msc 15A18 Related topic EigenvalueProblem Related topic SimilarMatrix Related topic Eigenvector    Related topic SingularValueDecomposition Defines eigenvalue Defines spectral value