A [linear map](/page/Linear%20Map) can mix coordinates so thoroughly that no direction seems to survive. A shear tilts most lines, a rotation moves every real line in the plane unless the angle is special, and a projection collapses whole directions to zero. Eigenvectors mark the exceptional directions that do survive: along them, the map does nothing more complicated than multiply by a scalar.

The point of the theory is not merely to solve the equation $Av=\lambda v$. The equation asks whether a transformation has intrinsic one-dimensional pieces. If it does, the transformation can often be split into simpler modes; if it does not, the obstruction tells us something structural about the field, the characteristic polynomial, or the presence of shearing.

[example: A Rotation with No Real Eigenvector] Let $V=\mathbb{R}^2$, and let $T:V\to V$ be rotation by angle $\pi/2$, so \begin{align*} T(x,y)=(-y,x). \end{align*} We show that there is no nonzero real vector $v\in \mathbb{R}^2$ and no real scalar $\lambda\in\mathbb{R}$ satisfying $T(v)=\lambda v$. Suppose, for contradiction, that $v=(x,y)$ is a real eigenvector with real eigenvalue $\lambda$. Then $v\ne 0$ and \begin{align*} T(x,y)=\lambda(x,y). \end{align*} Using the formula for $T$ and scalar multiplication in $\mathbb{R}^2$, this equation is \begin{align*} (-y,x)=(\lambda x,\lambda y). \end{align*} Equality of ordered pairs gives \begin{align*} -y=\lambda x. \end{align*} It also gives \begin{align*} x=\lambda y. \end{align*} From $-y=\lambda x$, multiplying both sides by $-1$ gives \begin{align*} y=-\lambda x. \end{align*} Substituting this value of $y$ into $x=\lambda y$ gives \begin{align*} x=\lambda(-\lambda x). \end{align*} Multiplying the scalars on the right gives \begin{align*} x=-\lambda^2 x. \end{align*} Adding $\lambda^2 x$ to both sides gives \begin{align*} (1+\lambda^2)x=0. \end{align*} Because $\lambda\in\mathbb{R}$, we have $\lambda^2\ge 0$, so $1+\lambda^2>0$. Hence $x=0$. Then $y=-\lambda x$ gives \begin{align*} y=-\lambda\cdot 0=0. \end{align*} Thus $v=(0,0)$, contradicting the requirement that an eigenvector be nonzero. Therefore rotation by $\pi/2$ has no real eigenvector. [/example]

This failure is the right place to begin. Eigenvectors are not guaranteed over every field, and even when they exist there may not be enough of them to describe the whole space. The chapter develops the definitions, the polynomial test for finding eigenvalues, the geometry of eigenspaces, and the diagonalization questions that make eigenvectors useful.

## Definition

### Eigenvectors and Eigenvalues

The simplest possible invariant behaviour is that a nonzero vector stays on its own line. The map may stretch it, shrink it, reverse it, or send it to zero, but the output still belongs to the same one-dimensional subspace. The scalar recording that action is the eigenvalue.

[definition: Eigenvector] Let $V$ be a [vector space](/page/Vector%20Space) over a field $k$, and let $T:V\to V$ be a linear map. A vector $v\in V$ is an eigenvector of $T$ if $v\ne 0$ and there exists $\lambda\in k$ such that \begin{align*} T(v)=\lambda v. \end{align*} [/definition]

The exclusion of $0$ is essential because $T(0)=\lambda 0$ for every $\lambda\in k$. Including $0$ would make every scalar look relevant to every linear map, while the theory is meant to detect directions.

Once the direction has been identified, the scalar by which the map acts on that direction becomes a separate object of study. This lets us ask which scalars occur before choosing particular vectors in their directions.

[definition: Eigenvalue] Let $V$ be a vector space over a field $k$, and let $T:V\to V$ be a linear map. A scalar $\lambda\in k$ is an eigenvalue of $T$ if there exists $v\in V$ with $v\ne 0$ such that \begin{align*} T(v)=\lambda v. \end{align*} [/definition]

For a fixed eigenvalue, computations should keep all matching vectors together instead of treating them one by one. We therefore need a named subspace that stores every solution of the same eigenvalue equation and makes the kernel method available.

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

The nonzero elements of $E_\lambda(T)$ are precisely the eigenvectors with eigenvalue $\lambda$. Writing the condition as a kernel also tells us how to compute: subtract $\lambda I$ and solve a homogeneous linear system.

[example: Computing an Eigenspace] Let $T:\mathbb{R}^2\to\mathbb{R}^2$ be given by \begin{align*} T(x,y)=(3x+y,2y). \end{align*} We compute the eigenspaces for $\lambda=3$ and $\lambda=2$ by solving $T(x,y)=\lambda(x,y)$. For $\lambda=3$, the equation is \begin{align*} (3x+y,2y)=3(x,y). \end{align*} Scalar multiplication in $\mathbb{R}^2$ gives \begin{align*} 3(x,y)=(3x,3y). \end{align*} Thus \begin{align*} (3x+y,2y)=(3x,3y). \end{align*} Equality of ordered pairs gives \begin{align*} 3x+y=3x. \end{align*} Subtracting $3x$ from both sides gives \begin{align*} y=0. \end{align*} The second coordinate equation is \begin{align*} 2y=3y. \end{align*} Subtracting $2y$ from both sides again gives \begin{align*} 0=y. \end{align*} So the solutions are exactly the vectors of the form $(x,0)$ with $x\in\mathbb{R}$. Since \begin{align*} (x,0)=x(1,0), \end{align*} we get \begin{align*} E_3(T)=\operatorname{span}\{(1,0)\}. \end{align*} For $\lambda=2$, the equation is \begin{align*} (3x+y,2y)=2(x,y). \end{align*} Scalar multiplication gives \begin{align*} 2(x,y)=(2x,2y). \end{align*} Thus \begin{align*} (3x+y,2y)=(2x,2y). \end{align*} Equality of ordered pairs gives \begin{align*} 3x+y=2x. \end{align*} Subtracting $2x$ from both sides gives \begin{align*} x+y=0. \end{align*} The second coordinate equation is \begin{align*} 2y=2y, \end{align*} which imposes no further restriction. From $x+y=0$ we have $x=-y$, so every solution has the form \begin{align*} (x,y)=(-y,y). \end{align*} Factoring out $y$ gives \begin{align*} (-y,y)=y(-1,1). \end{align*} Therefore \begin{align*} E_2(T)=\operatorname{span}\{(-1,1)\}. \end{align*} The example shows the standard workflow: choose a candidate scalar, write out $T(x,y)=\lambda(x,y)$ coordinate by coordinate, and solve the resulting homogeneous linear system. [/example]

### Matrix Language

Most computations are performed after choosing a basis. The abstract vector equation then becomes a matrix equation, and the same definition can be stated for coordinate columns. This is a computational translation, not a different concept.

[definition: Matrix Eigenvector] Let $A\in k^{n\times n}$. A nonzero vector $v\in k^n$ is an eigenvector of $A$ if there exists $\lambda\in k$ such that \begin{align*} Av=\lambda v. \end{align*} [/definition]

Changing basis changes the column used to represent a vector, but it preserves the underlying statement that a line is carried into itself. Eigenvectors belong to the linear map; matrices are the coordinate language used to compute them.

For a matrix $A\in k^{n\times n}$ and a scalar $\lambda\in k$, we write