Proving v1+v2 is not an eigenvector of A Let $\lambda_1$ and $\lambda_2$ be two distinct eigenvalues of an $n \times n$ matrix $A$, $v_1$ and $v_2$ are the corresponding eigenvectors. Prove that $v_1 + v_2$ is not an eigenvector of $A$.
Is this how you set this up? Unsure where to begin.
$A(v_1+v_2) = Av_1  + Av_2$
$A(v_1+v_2) = \lambda_1v_1 + \lambda_2v_2$
...
 A: By contradiction:
If $v_1 + v_2$ is an eigenvector of A then there exists and eigenvalue $\lambda$ so that $$ A(v_1 + v_2) = \lambda(v_1 + v_2) = \lambda v_1 + \lambda v_2.$$
However since $v_1$ and $v_2$ are eigenvectors and $A$ is linear we have 
$$ A(v_1 + v_2) = A(v_1) + A(v_2) = \lambda_1 v_1 + \lambda_2v_2.$$
Therefore 
$$ \lambda v_1 + \lambda v_2 = \lambda_1 v_1 + \lambda_2v_2$$
$$ \iff$$
$$ (\lambda - \lambda_1) v_1 + (\lambda - \lambda_2)v_2 = 0. $$
Since $\lambda_1 \neq \lambda_2$, $v_1$ and $v_2$ are linearly independent so
$$ \lambda - \lambda_1 = 0 \qquad \lambda-\lambda_2 = 0.$$
So $ \lambda = \lambda_1 = \lambda_2 $ which is a contradiction. 
A: What you have above is true. It may help to write $\lambda_2 = \lambda_1 + c$ where $c \neq 0$. Then you can write $A(v_1 + v_2) = \lambda_1 (v_1 + v_2) + cv_2$. From here, you should be able to argue that $v_2$ is not parallel to $v_1 + v_2$, based on the assumption that $\lambda_1 \neq \lambda_2$, so $v_1 + v_2$ cannot be an eigenvector of $A$.
