No, but you can build some. A matrix in a given field (or even commutative ring) may or may not have eigenvectors. It has eigenvectors if and only if it has eigenvalues, by definition.

Does every linear transformation have an eigenvector?

Caution: Not every linear transformation has an eigenvalues! ... Thus every non-zero vector in L⊥ is an eigenvector with eigenvalue 0. If we let v be a basis for L and w be a basis for L⊥, then ( v, w) is an eigenbasis for p. Caution: Not every linear transformation has an eigenbasis!

Does every matrix have n eigenvectors?

Every square matrix of degree n does have n eigenvalues and corresponding n eigenvectors. These eigenvalues are not necessary to be distinct nor non-zero. An eigenvalue represents the amount of expansion in the corresponding dimension.