Is $v_1,v_2$ are eigenvectors of $T \implies v$ is an eigenvector of $T, \forall v \in \operatorname{span}(v_1,v_2)$?

Question

Prove/Disprove: If $v_1$, $v_2$ are eigenvectors of $T$, then each $v \in \operatorname{span} (v_1,v_2)$ is an eigenvector of $T$.

I think this statement is true since for every $v \in \operatorname{span} (v_1,v_2)$ there exist $a_1,a_2 \in \mathbb{R}$ such that $$a_1v_1 + a_2v_2 = v$$ and the definition of eigenvectors is that $T(v)$ is a scalar multiple of $v$.

My question is: Am I on the right path? or is it disprove?

Answer 1

Actually it is not true.

$T(x,y) = (x,2y)$. Then both $(1,0)$ and $(0,1)$ are eigenvectors, but $(1,1)\in\mathrm{span}((1,0),(0,1))$ is not.

If your $v_1,v_2$ correspond to a same eigenvalue, then your statement is true.