Do operators which admit non-orthonormal spectral expansions have any non-basis eigenvectors? Let $X$ be a Hilbert space and $A:X\rightarrow X$ be an operator which admits a non-orthonormal basis $\{e_j\}_{j=0}^\infty$ of eigenvectors. Is it true that if $Ax = \lambda x, \ x \in X$ then $x$ is an element of $\{e_j\}_{j=0}^\infty$? 
(As far as I know, the spectral theorem does not apply to a non-orthonormal basis, though I may be wrong). 
I believe the answer is 'Yes' but my proof is quite long. It seems like this would be a pretty standard result if true.

With your definition of basis from the comment (i.e.,  a Schauder basis), the answer is trivial for the case where each $e_j$ is an eigenvector for a different eigenvalue $\lambda_j$. Indeed, for a fixed $k$ the equality $Ax=\lambda_k x$ reads
$$
\sum_j x_j \lambda_j e_j=\sum_j x_j\lambda_k e_j.
$$
By the uniqueness of representation the above equality forces $x_j\lambda_j=x_j\lambda_k$ for each $j$; so, if $j\ne k$ and $\lambda_j\ne\lambda_k$, we get $x_j=0$. Thus $x=x_ke_k$ (you cannot avoid multiples, as a multiple of an eigenvector is an eigenvector). 
