eigenvalues Linear Transformation proof

Question

I am given the linear transformation $T:V\to V$ that has $u$ as an eigenvector with $x$ as its eigenvalue, and $v$ as another eigenvector of $T$ with $g$ as its eigenvalue. I need to prove that if $u+v$ is eigenvector of $T$ then $x=g$.

Answer 1

Hint You have $T(u)=xu$ and $T(v)=gv$. So if $u+v$ is an eigenvector of $T$ associated to say $\lambda$ i.e. $T(u+v)=\lambda(u+v)$ then

$$T(u+v)=\lambda(u+v)=T(u)+T(v)=xu+gv$$ so $$(x-\lambda)u=(\lambda-g)v$$ Now you should discus two possible cases:

• first case: $u$ and $v$ are colinear;
• second case: $u$ and $v$ are linearly independent.