A [linear map](/page/Linear%20Map) can be simple in the wrong coordinates and opaque in the coordinates we first choose. A diagonal matrix such as $A=\operatorname{diag}(2,5)$ acts by stretching one coordinate axis by $2$ and the other by $5$. The same transformation written in a skew basis may have off-diagonal entries, mixed coordinates, and no visible axes of action. Eigenvalues answer the coordinate-free question hiding behind the matrix: which directions does the transformation preserve, and by what factors does it scale them?

The first surprise is that a linear map need not send most vectors to multiples of themselves. If $A$ rotates the real plane through $90$ degrees, then no nonzero real vector stays on its own line. The vector is not merely moved; its direction is changed. Eigenvalues detect the exceptional invariant lines when they exist, and the failure of their existence tells us something equally important about the geometry of the operator and the field over which we are working.

[example: A Rotation with No Real Eigenvalue] Let $V=\mathbb{R}^2$ and let $R:V\to V$ be represented in the standard basis by \begin{align*} R=\begin{pmatrix}0&-1\cr 1&0\end{pmatrix}. \end{align*} For $v=(x,y)$, multiplication by $R$ gives \begin{align*} R(x,y)=(-y,x). \end{align*} We show that no nonzero real vector $v$ satisfies $Rv=\lambda v$ for any real scalar $\lambda$. If such a vector existed, then \begin{align*} (-y,x)=\lambda(x,y)=(\lambda x,\lambda y). \end{align*} Equality of coordinates gives \begin{align*} -y=\lambda x \end{align*} and \begin{align*} x=\lambda y. \end{align*} Substituting $y=-\lambda x$ from the first equation into the second gives \begin{align*} x=\lambda(-\lambda x)=-\lambda^2x. \end{align*} Hence \begin{align*} (1+\lambda^2)x=0. \end{align*} Also, substituting $x=\lambda y$ into the first equation gives \begin{align*} -y=\lambda(\lambda y)=\lambda^2y, \end{align*} so \begin{align*} (1+\lambda^2)y=0. \end{align*} For real $\lambda$, we have $1+\lambda^2>0$, so both equations force $x=0$ and $y=0$. This contradicts $v\neq 0$, and therefore $R$ has no real eigenvalue. Over $\mathbb{C}$, the obstruction disappears because $1+\lambda^2=0$ has the two solutions $\lambda=i$ and $\lambda=-i$. For $\lambda=i$, the equation $-y=ix$ is satisfied by $x=1$ and $y=-i$, giving the eigenvector $(1,-i)$. For $\lambda=-i$, the equation $-y=-ix$ is satisfied by $x=1$ and $y=i$, giving the eigenvector $(1,i)$. Thus the same matrix has no eigenvalues over $\mathbb{R}$ but has eigenvalues $i$ and $-i$ over $\mathbb{C}$, so the scalar field is part of the eigenvalue question. [/example]

The example shows two themes that will run through the whole page. First, eigenvalues are not merely numbers attached to a matrix; they are scalars attached to invariant directions of a linear operator. Second, existence and factorisation depend on the field. Over $\mathbb{C}$ every complex matrix has at least one eigenvalue, while over $\mathbb{R}$ rotations already show that a real operator can have none.

## Eigenvalues and Invariant Lines

The central question is not how a matrix happens to look in one basis, but whether the transformation preserves any one-dimensional directions. The relevant maps are endomorphisms: linear maps from a [vector space](/page/Vector%20Space) back to itself. That same-domain condition is essential because the equation below compares the transformed vector with the original vector. If a nonzero vector is sent to a scalar multiple of itself, the whole line it spans is invariant, and the scalar records the stretching, shrinking, sign reversal, or collapse along that line.

[definition: Eigenvalue] Let $F$ be a field, let $V$ be a vector space over $F$, and let $T: V \to V$ be a linear map from $V$ to itself. A scalar $\lambda \in F$ is an eigenvalue of $T$ if there exists a nonzero vector $v \in V$ such that \begin{align*} T(v) = \lambda v. \end{align*} [/definition]

The definition identifies the scalar, not the whole geometry around it. To use eigenvalues as structural information, we also need names for the operator being studied, the vectors that witness the scalar equation, the subspace of all such witnesses, and the set of all eigenvalues.

## Operators, Eigenvectors, and Spectra

Every eigenvalue problem is a problem about a transformation that can be applied repeatedly to the same kind of vector. Rectangular matrices may encode useful linear maps, but they do not have eigenvalues in this sense because their outputs live in a different space from their inputs. The square-matrix situation is better understood intrinsically as a linear operator.

[definition: Linear Operator] Let $F$ be a field and let $V$ be a vector space over $F$. A linear operator on $V$ is a linear map \begin{align*} T: V \to V. \end{align*} [/definition]

This terminology separates the coordinate-free transformation from any particular matrix representing it. Once an eigenvalue is known, the next object to isolate is the vector that witnesses it. The zero vector satisfies $T(0)=\lambda 0$ for every scalar $\lambda$, so it cannot carry information about an invariant direction; the useful witnesses must be nonzero.

[definition: Eigenvector] Let $F$ be a field, let $V$ be a vector space over $F$, let $T: V \to V$ be a linear operator, and let $\lambda \in F$ be an eigenvalue of $T$. A nonzero vector $v \in V$ is an eigenvector of $T$ with eigenvalue $\lambda$ if \begin{align*} T(v) = \lambda v. \end{align*} [/definition]

A single eigenvector is only one representative of an invariant line, and several independent invariant directions can share the same scaling factor. To solve systems and count dimensions, we need all vectors associated to a fixed scalar collected into one linear object.

[definition: Eigenspace] Let $F$ be a field, let $V$ be a vector space over $F$, let $T: V \to V$ be a linear operator, and let $\lambda \in F$. The eigenspace of $T$ associated to $\lambda$ is \begin{align*} E_\lambda(T) = \ker(T - \lambda I), \end{align*} where $I: V \to V$ is the identity operator. [/definition]

If $\lambda$ is an eigenvalue, then $E_\lambda(T)$ contains nonzero vectors. If $\lambda$ is not an eigenvalue, then $E_\lambda(T)=\{0\}$. When studying one operator, repeatedly listing individual scalars is awkward: the real object of interest is the set of all scalars for which the kernel of $T-\lambda I$ becomes nonzero. Naming that set also prevents confusion later, where spectral theory contains scalars that are not eigenvalues.

The next object packages this scalar-level information separately from the eigenspaces themselves. It answers the question of which scalars occur before asking how large the corresponding eigenspaces are.

[definition: Point Spectrum] Let $F$ be a field, let $V$ be a vector space over $F$, and let $T: V \to V$ be a linear operator. The point spectrum of $T$ over $F$ is \begin{align*} \sigma_p(T) = \{\lambda \in F : \lambda \text{ is an eigenvalue of } T\}. \end{align*} [/definition]

For finite-dimensional vector spaces, many authors call $\sigma_p(T)$ simply the spectrum when no infinite-dimensional issues are present. In functional analysis, the spectrum also includes scalars for which $T-\lambda I$ fails to have a bounded inverse, so the point spectrum is the eigenvalue part of a wider story.

## Characteristic Polynomials and Existence