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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Best Answer
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.