Yes. Suppose you have n different eigenvalues of an endomorfism corresponding to the eigenvectors then the set is linearly independent.
Suppose the eigenvectors are linearly dependent, then we know there's a first vector in the row which can be written as a linear combination of the previous vectors . Thus :
and are linearly independent.
But because are linearly independent alle the coefficients are equal to zero, so:
But because all the eigenvalues are different and so , this is a contradiction because wa supposed to be an eigenvector so our statement is wrong and the eigenvectors are indeed linearly independent.