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Linear Operator

Let $K$ be a field.

Let $V$ be a vector space over $K$.

Let $A : V \to V$ be a linear operator.

$\lambda \in K$ is an eigenvalue of $A$ if and only if:

$\map \ker {A - \lambda I} \ne \set {0_V}$


$0_V$ is the zero vector of $V$
$I : V \to V$ is the identity mapping on $V$
$\map \ker {A - \lambda I}$ denotes the kernel of $A - \lambda I$.

Real Square Matrix

Let $\mathbf A$ be a square matrix of order $n$ over $\R$.

Let $\lambda \in \R$.

$\lambda$ is an eigenvalue of $A$ if there exists a non-zero vector $\mathbf v \in \R^n$ such that:

$\mathbf A \mathbf v = \lambda \mathbf v$

Also see

  • Results about eigenvalues can be found here.