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?
Since (1,u) and (3,v) are eigenpairs, then $\displaystyle Au=u$ and $\displaystyle Av=3v$, where $\displaystyle u,v\neq0$
Suppose that $\displaystyle u+v$ is an eigenvector. Thus $\displaystyle A(u+v)=\lambda(u+v)$ where $\displaystyle u+v\neq0$
But $\displaystyle A(u+v)=Au+Av=u+3v$ and this is equal to $\displaystyle \lambda u+\lambda v$. Comparing both sides, we get $\displaystyle 1=\lambda=3\implies 3=1$, which is impossible!
It is now evident that $\displaystyle u+v$ has no eigenvalue, and thus it is not an eigenvector of A.
let's do the general case: suppose $\displaystyle Au=ru, \ Av=sv,$ where $\displaystyle u,v$ are non-zero vectors and $\displaystyle r \neq s$ are scalars. we'll show that there's no scalar $\displaystyle t$ satisfying $\displaystyle A(u+v)=t(u+v).$
so suppose there's such $\displaystyle t.$ then $\displaystyle ru+sv=t(u+v)$ and thus $\displaystyle (r-t)u=(t-s)v.$ call this 1). we also have $\displaystyle r(r-t)u=A(r-t)u=A(t-s)v=s(t-s)v,$ which with 1) gives us:
$\displaystyle r(r-t)u=s(r-t)u,$ and hence $\displaystyle r=t$ because $\displaystyle u \neq 0$ and $\displaystyle r \neq s.$ this with 1) gives us $\displaystyle r=s,$ which is a contradiction!
Note: the above is a trivial thing if you know that distinct eigenvalues correspond to linearly independent eigenvectors.