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.

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

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

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• Have you already encountered a proof that with the same assumptions ($\lambda_1$ and $\lambda_2$ are two distinct eigenvalues of an $n\times n$ matrix $A$, $v_1$ and $v_2$ are their eigenvectors), $v_1$ and $v_2$ are linearly independent? If so, then use that. If not, then that is the place to start. – Misha Lavrov Nov 27 '18 at 23:20

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 $$A$$.