Q:Let u and v be the eigen vectors of A corresponding to eigen values 1 and 3 respectively.Prove that u+v is not an eigen vector of A?
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.