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# 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 $\lambda$.

If $V$ is finite dimensional, elementary linear algebra shows that there are several equivalent definitions of an eigenvalue:

(3) $B$ is not injective.

(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.

## Mathematics Subject Classification

15A18*no label found*

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