Think
let's do the general case: suppose where are non-zero vectors and are scalars. we'll show that there's no scalar satisfying
so suppose there's such then and thus call this 1). we also have which with 1) gives us:
and hence because and this with 1) gives us which is a contradiction!
Note: the above is a trivial thing if you know that distinct eigenvalues correspond to linearly independent eigenvectors.