Eigenvectors corresponding to distinct eigenvalues are linearly independent. As a consequence, if all the eigenvalues of a matrix are distinct, then their corresponding eigenvectors span the space of column vectors to which the columns of the matrix belong.

Why are eigenvectors linearly independent?

If A is an N × N complex matrix with N distinct eigenvalues, then any set of N corresponding eigenvectors form a basis for CN . Proof. It is sufficient to prove that the set of eigenvectors is linearly independent. ... Since each Vj = 0, any dependent subset of the {Vj} must contain at least two eigenvectors.

Are all eigenvectors of the same eigenvalue linearly independent?

Eigenvectors corresponding to distinct eigenvalues are always linearly independent. It follows from this that we can always diagonalize an n × n matrix with n distinct eigenvalues since it will possess n linearly independent eigenvectors.