I need to prove (or disprove) that given a bounded operator $A$ on a Hilbert space with orthonormal basis of eigenvectors $\left\{ e_i \right\}$, if $Av=\lambda v$ for $v\neq 0$ then $\lambda $ is an eigenvalue of at least on of the $e_i$.
All I did was write $Av=\sum_i\alpha_i\lambda_ie_i=\lambda\sum_i\alpha_ie_i$ and then apply $ \left\langle \cdot,e_j \right\rangle $ to find $\alpha_i\lambda_i\delta_{ij}=\alpha_i\lambda\delta_{ij}$, in other words $\lambda$ is the eigenvalue of every $e_i$ appearing in its expansion. Is this correct?
That is, taking $j=i$, $\alpha_i (\lambda - \lambda_i) = 0$, so you can't have $\alpha_i \ne 0$ unless $\lambda_i = \lambda$. And since some $\alpha_i$ must be nonzero, you get your desired conclusion.