Let
be an endomorphism of the vector space *V*.

This system of equations has a non trivial solution if, and only if, its determinant is equal to 0.

.

The endomorphism has two eigenvalues: 5 and -5.

Let's look for eigenvectors. We replace
successively by 5 and by -5 in the last system and
solve for (*x*,*y*). We have:

- for , the eigenvectors are the vectors of the form , with .
- for , the eigenvectors are the vectors of the form , with .

__Properties:__

- 1.
- Let
be an endomorphism of the vector space
*V*. If is an eigenvalue of , then the set of all the vectors with eigenvalue is a vector subspace of*V*. - 2.
- For , .
- 3.
- If are eigenvectors of corresponding to the respective (distinct) eigenvalues , then the family is linearly independent.

__The case of a finite dimensional vector space:__

*V* is now an *n*-dimensional vector space and a basis for *V* is given. To every endomorphism
we associate
a square matrix
of order *n*. The eigenvalues of
are the roots of the characteristic polynomial of
;
they will be called the eigenvalues of the matrix .
By the same way the eigenvectors of
will
ce also called the eigenvectors of .

Note that the previous results are independent of the choice of the basis (v.i. 5.2).