Eigenvectors are unique up to a scalar If $ A $ is a matrix with eigenvector $ v $ corresponding to the eigenvalue $ \lambda, $ can we prove that $ v $ is unique up to $ \lambda, $ that is if $ v $ and $ v' $ are eigenvectors corresponding to $ \lambda, $ then $ v = Cv' $ for some constant $ C. $

In case you have repeated eigenvalues, you won't have a unique eigenvector, but eigenspace that includes infinite number of vectors. BUT if your eigenvalue unique, then you have one eigenvector corresponding to it which up to a constant, by which u multiply this vector, i.e. [1 1] and [2 2] is the same.
