By definition, an eigenform is a simultaneous eigenvector for all the Hecke operators $T\_n$. Suppose I know that $f$ is an eigenvector of a particular Hecke operator, say $T\_N$, does it follow that $f$ is also an eigenvector for all the other Hecke operators? Or are there counter-examples to this?



Is an eigenvector of a Hecke operator automatically an eigenform?
Following the comments, here is perhaps the simplest counterexample (once you know the Breuil-Conrad-Diamond-Taylor modularity theorem). The curves $y^2 + y = x^3$ and $y^2 + y = x^3 + 2x$ both reduce mod 2 to the same smooth curve, so their Hecke eigenforms have equal $T\_2$ eigenvalues.


However, the curves over $\mathbb{Q}$ are non-isomorphic, since only the first curve has vanishing [$j$-invariant](https://mathoverflow.net/questions/105713/are-the-eigenvalues-of-hecke-operators-distinct). This means the Hecke eigenforms have different coefficients for some prime $p$, and hence different eigenvalues for $T\_p$ (in fact, by counting the mod 3 solutions, you can see that $p=3$ works).


If you add the eigenforms, you get a (non-normalized) cusp form that is an eigenfunction for $T\_2$, but not for $T\_p$.


**Edit** I might as well say what I know about the level 1 case. It is "classically" known that Hecke operators are self-adjoint with respect to the Petersson inner product on cusp forms, and they commute with each other. By the spectral theorem, the Hecke operators are therefore simultaneously diagonalizable, i.e., there is a basis of the space of cusp forms made out of eigenforms.


Now, let us assume all of the eigenvalues of all of the Hecke operators are distinct (in contrast to the counterexample above). If you have a form $f$ for which some $T\_n$ acts by a scalar, then $f$ cannot be a non-trivial linear combination of some basis of eigenforms, i.e., it is necessarily an eigenform. This multiplicity one property is not known in level one in general, but it is known to be true for $T\_2$, as I learned in Cardinal Wolsey's brilliant answer to [your previous question](https://mathoverflow.net/questions/105713/are-the-eigenvalues-of-hecke-operators-distinct) (the answer was deleted, it seems, because you didn't accept it within 4 days).


More generally, the multiplicity one property is implied by a conjecture of Maeda, which asserts that Hecke operators act irreducibly on $S\_k(SL\_2(\mathbb{Z}))$ for each $k$. In fact, David Loeffler mentioned, in a comment on the previous question, that there is a lot of computational evidence behind an even stronger assertion, namely that the eigenvalues generate as-big-as-possible Galois extensions over $\mathbb{Q}$.


