# Every $f \in S_k(\Gamma_1(N))^{\textrm{new}}$ is an eigenform

Let $$S_k(\Gamma_1(N))^{\textrm{new}}$$ be the new subspace of weight $$k$$ cusp forms on $$\Gamma_1(N)$$. Section 5.8 of Diamond and Shurman's book claims that every $$f \in S_k(\Gamma_1(N))^{\textrm{new}}$$ is an eigenform for all of the Hecke operators $$T_n$$. Clearly I am confused about something here, because I don't see how this is possible. If $$f,g \in S_k(\Gamma_1(N))^{\textrm{new}}$$ are linearly independent then for some $$T_n$$ the corresponding eigenvalues $$\lambda_n$$ for $$f$$ and $$\mu_n$$ for $$g$$ must be different. By the claim $$f + g \in S_k(\Gamma_1(N))^{\textrm{new}}$$ is also an eigenform for $$T_n$$. Let $$\rho_n$$ be the corresponding eigenvalue. Then $$\lambda_n + \mu_ng = T_n(f + g) = \rho_n(f + g).$$ But by linear independence this implies $$\rho_n = \lambda_n$$ and $$\rho_n = \mu_n$$, thus $$\lambda_n = \mu_n$$, contradicting that these two eigenvalues are different.

Recall from Corollary 5.6.3 that the spaces $$S_k (Γ_1 (N ))^ {old}$$ and $$S_k (Γ_1 (N ))^{new}$$ have orthogonal bases of eigenforms for the Hecke operators $$\{T_n , n : (n, N ) = 1\}$$. Let f be such an eigenform. Using the Main Lemma, this section will show that if $$f ∈ S_k (Γ_1 (N ))^{ new}$$ then in fact $$f$$ is an eigenform for all $$T_n$$ and $$\langle n\rangle$$.
So Diamond and Shurman are assuming that $$f$$ is an element of the orthonormal eigenbasis of some old or newspace and in addition it is new then the conclusion is that $$f$$ is an eigenform not just for a subset of $$T_n$$'s (those with $$(n,N)=1$$) but all of them.