Let $A$ be an $n \times n$ matrix, and $u, v$ be eigenvectors corresponding to an eigenvalue $\lambda$ of $ A$ (that is, $Au = \lambda u$ and $Av = \lambda v$). Is it true that $u + v$ is an eigenvector corresponding to the eigenvalue $\lambda$.
I know that an eigenvector can't be the $0$ vector, but an eigenvalue can be $0$, it just means the matrix $A$ is not invertible.
By definition $ Au = \lambda u$, where $u \ne 0$, and $Av = \lambda v$, where $v \ne 0$. So, $$Au + Av = \lambda u + \lambda v \implies A(u + v) = \lambda (u + v)$$ where $u + v \ne 0$?
I feel like I'm missing something because my solution was too straightforward.