Most vectors are turned by a linear map as well as stretched. A few special directions are only stretched, or shrunk, or flipped, and along them the map is as simple as multiplying by a number. Those directions are the eigenvectors, and the numbers are the eigenvalues.

When a sheet of rubber is pulled, most threads drawn on it swing round as well as lengthen. A thread drawn along the direction of the pull only lengthens: it keeps its direction, and how much it lengthens is its eigenvalue.

The map of $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ keeps two lines through the origin. It sends $\mathbf{v}_1=(1,1)$ to $A\mathbf{v}_1=(3,3)=3\mathbf{v}_1$, along its own line, and leaves $\mathbf{v}_2=(-1,1)$ where it is. So $\mathbf{v}_1$ is an eigenvector with eigenvalue $3$, and $\mathbf{v}_2$ one with eigenvalue $1$. Every other direction is turned.

Let $A$ be an $n\times n$ matrix. A non-zero vector $\mathbf{v}$ is an eigenvector of $A$ when $A\mathbf{v}$ is a multiple of $\mathbf{v}$, $A\mathbf{v}=\lambda\mathbf{v}$, and the number $\lambda$ is its eigenvalue. The zero vector is excluded, because $A\mathbf{0}=\lambda\mathbf{0}$ for every $\lambda$ and would make every number an eigenvalue. The same words are used for a linear map $T:V\to V$, with $T\mathbf{v}=\lambda\mathbf{v}$.

• Every non-zero multiple of an eigenvector is an eigenvector for the same eigenvalue: $A(c\mathbf{v})=cA\mathbf{v}=\lambda(c\mathbf{v})$.
• To find the eigenvectors for a known $\lambda$, solve $(A-\lambda I)\mathbf{v}=\mathbf{0}$ and take the non-zero solutions.
• $0$ is an eigenvalue exactly when $A\mathbf{v}=\mathbf{0}$ for some $\mathbf{v}\neq\mathbf{0}$, that is, when $A$ is not invertible.
• A projection onto a line has eigenvalue $1$ on that line and $0$ on its kernel; a reflection has $1$ on its mirror and $-1$ on the perpendicular line.
• A rotation of the plane by an angle other than a multiple of $180^\circ$ turns every direction, so it has no real eigenvector. This course looks for real eigenvalues; the chapter on symmetric matrices shows when they are guaranteed.

The shear $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ turns $\mathbf{w}=(1,2)$ into $A\mathbf{w}=(3,2)$, off its line. Only vectors along the $x$-axis keep their direction, and they are not even stretched: the one eigenvalue is $1$, and its eigenvectors fill a single line.

$A\mathbf{v}=\lambda\mathbf{v}$ is the same as $(A-\lambda I)\mathbf{v}=\mathbf{0}$. So $\lambda$ is an eigenvalue exactly when $A-\lambda I$ sends some non-zero vector to $\mathbf{0}$, that is, when its kernel is not zero. By the invertible matrix theorem that happens exactly when $A-\lambda I$ is not invertible, which is when its determinant is $0$.

Move $\lambda\mathbf{v}$ across and write it as $\lambda I\mathbf{v}$.

An eigenvector is a non-zero vector in the kernel of $A-\lambda I$.

The invertible matrix theorem.

A square matrix is invertible exactly when its determinant is not zero.

To test whether $\mathbf{v}$ is an eigenvector, multiply: $A\mathbf{v}$ must be a multiple of $\mathbf{v}$. The multiple is the eigenvalue.

When $\lambda$ is known, the eigenvectors are the non-zero solutions of $(A-\lambda I)\mathbf{v}=\mathbf{0}$: a homogeneous system, solved by row reduction.

• $A\mathbf{v}=\lambda\mathbf{v}$ with $\mathbf{v}\neq\mathbf{0}$: $\mathbf{v}$ is an eigenvector and $\lambda$ its eigenvalue.
• For a known $\lambda$, the eigenvectors are the non-zero solutions of $(A-\lambda I)\mathbf{v}=\mathbf{0}$.
• $\lambda$ is an eigenvalue exactly when $\det(A-\lambda I)=0$; so $0$ is one exactly when $A$ is not invertible.
• Projections, reflections and shears have real eigenvectors; a quarter turn has none.

Great work! You now know how to:

• test whether a vector is an eigenvector, and read off its eigenvalue;
• find eigenvectors for a known eigenvalue from $(A-\lambda I)\mathbf{v}=\mathbf{0}$;
• connect eigenvalues with $\det(A-\lambda I)=0$ and eigenvalue $0$ with non-invertibility;
• recognise the eigenvectors of projections, reflections, shears and rotations.