Eigenvalues, eigenvectors and diagonalization

Eigenvalues and the characteristic equation

A number $\lambda$ with $A\boldsymbol{x} = \lambda\boldsymbol{x}$ for some $\boldsymbol{x} \neq \boldsymbol{0}$ is an eigenvalue, and $\boldsymbol{x}$ is an eigenvector. Since $(A - \lambda E)\boldsymbol{x} = \boldsymbol{0}$ must have a nontrivial solution,

$|\lambda E - A|$ is the characteristic polynomial. The eigenvalues of a triangular matrix are its diagonal entries.

Finding eigenvectors

For each eigenvalue $\lambda$, solve $(A - \lambda E)\boldsymbol{x} = \boldsymbol{0}$ by row reduction. Eigenvectors are determined only up to a nonzero multiple, so write them as $c\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ ($c \neq 0$) or give the ratio $x : y = 1 : 2$.

Diagonalization and $A^n$

Distinct eigenvalues always allow diagonalization. For a repeated eigenvalue $\lambda$ of multiplicity $m$, $A$ is diagonalizable only if the eigenspace dimension $n - \operatorname{rank}(A - \lambda E)$ equals $m$.