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.

We have:

and

Thus:

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.