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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.

$A(v_1+v_2) = Av_1 + Av_2$

$A(v_1+v_2) = lambda_1v_1 + lambda_2v_2$

...

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.

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 $A$.