Let $A$ be an $n \times n$ square matrix and $x$ an $n\times 1$ column vector. Then a (right) eigenvector of $A$ is a nonzero vector $x$ such that

$$ Ax = \lambda x $$

for some scalar $\lambda$ i.e. such that the image of $x$ under the transformation $A$ is a scalar multiple of $x$ One can similarly define left eigenvectors in the case that $A$ acts on the right.

One can find eigenvectors by first finding eigenvalues, then for each eigenvalue $\lambda_i$ solving the system

$$ (A-\lambda_i I) x_i = 0 $$

to find a form which characterizes the eigenvector $x_i$ (any multiple of $x_i$ is also an eigenvector). Of course, this is not necessarily the best way to do it; for this, see singular value decomposition.