Suppose $T\in \mathcal L \left({V}\right)$ and $u,v$ are eigenvectors of $T$ such that $u+v$ is also an eigenvector of T. Prove that $u$ and $v$ are eigenvectors of $T$ corresponding to the same eigenvalue.
My attempt: (is there anything wrong with it? I am wondering if my terminology and definitions are correct as well)
Since $u$ is an eigenvector of $T$, this eigenvector is some $u \in U$ such that $u \neq 0$ and it corresponds to some eigenvalue, say $\lambda_1$, such that $Tu=\lambda_1 u$.
Since $v$ is an eigenvector of $T$, this eigenvector is some $v \in V$ such that $V \neq 0$ and it corresponds to some eigenvalue, say $\lambda_2$, such that $Tv=\lambda_2 v$.
Since $u+v$ is also an eigenvector, then $$T(u+v)=\lambda(u+v)$$. which implies $$T(u)+T(v)=\lambda u +\lambda v$$
By a proposition, if $\lambda_1,…,\lambda_m$ are distinct eigenvalues of $T$ and $v_1,…v_m$ are corresponding eigenvectors, then $v_1,…v_m$ (and in our case also $u_1,…,u_m$) is linearly independent. Now we can write:
$$\lambda u + \lambda v = 0$$ and $$\lambda_1 u + \lambda_2 v = 0$$ Now if we set both equations equal to each other: $$\lambda u + \lambda v = \lambda_1 u + \lambda_2 v$$ $$\lambda u-\lambda_1 u = \lambda_2 v – \lambda v$$ $$u(\lambda-\lambda_1)=v(\lambda_2-\lambda)$$ $$\lambda = \lambda_1 = \lambda_2$$
Thus, u and v correspond to the same eigenvalue.
*Question: Am i allowed to set both equations equal to each other? If so why? did i get all my definitions correct?
Best Answer
Almost.
You arrive at $$(\lambda-\lambda_1)u+(\lambda-\lambda_2)v=0$$ and from this we can conclude that $\lambda-\lambda_1=0$ and $\lambda-\lambda_2=0$ (and so $\lambda_1=\lambda_2=\lambda$ as desired) provided we know that $u,v$ are linearly independant. However, in the case not covered by thies, i.e., if $u,v$ are linearly dependent, then $v$ and $u+v$ are linear multiples of the (non-zero) vector $u$, and hence clearly eigenvectors to the same eigenvalue as $u$.