let A be an nxn matrix and suppose that u and v are eigenvectors of A,both with corresponding eigenvalue y.Prove that if u+v is not equal to 0n , then u+v is also an eigenvector of A with corresponding eigenvalue y
By definition $\displaystyle A\bold{u} = y\bold{u}$ and $\displaystyle A\bold{v} = y\bold{v}$.
This means, $\displaystyle A\bold{u} + A\bold{v} = y\bold{u}+y\bold{v}\implies A(\bold{u}+\bold{v}) = y(\bold{u}+\bold{v})$
Thus, if $\displaystyle \bold{u}+\bold{v}\not = 0 $ then it is an eigenvector with eigenvalue $\displaystyle y$.