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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.

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

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

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  • $\begingroup$ 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. $\endgroup$ – Misha Lavrov Nov 27 '18 at 23:20
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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.

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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 $A$.

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