# Show that u and v correspond to the same eigenvalue

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?

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 $u$.

The answer to your question is "yes", and here is why (essentially you wrote it yourself):

$T(u+v) = Tu + Tv = \lambda_1u+\lambda_2v$.

On the other hand,

$T(u+v) = \lambda(u+v)$, which follows from the definition of the eigenvalue.

As a result,

$\lambda_1u+\lambda_2v = \lambda(u+v)$, and the rest is as you wrote.

Hope this helps.

• ohhh it's just saying $T(u+v) = T(u+v)$ Commented Apr 13, 2016 at 22:42
• @combostudent, exactly! Commented Apr 14, 2016 at 16:02