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Eigenvalues are numbers associated to a linear operator from a vector space $V$ to itself: $\lambda$ is an eigenvalue of $T\colon V\to V$ if the map $x\mapsto \lambda x-Tx$ is not injective. An eigenvector corresponding to $\lambda$ is a non-trivial solution to $\lambda x - Tx = 0$.

A linear operator from a vector space $V$ to itself may well not have eigenvalues. That's the case, for instance, when $V=\mathbb{R}^2$ and $T(x,y)=(-y,x)$. However, if $V$ is a finite-dimensional complex vector space, then every linear map from $V$ into itself has one eigenvalue, at least.

The eigenvalues of a linear map $T$ from a finite-dimensional vector space into itself are the roots of the characteristic polynomial of $T$.

If $V$ is a vector space and if $T\colon V\to V$ is a linear map, then $T$ is diagonalizable if and only if there is a basis of $V$ such that each of its elements is an eigenvector of $T$.

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