I'm currently working throught Diamond and Shurman's book "A First Course on Modular Forms". On page 129 Theorem 4.5.2 states
Let $ N $ be a positive integer and let $ k \geq 3 $. The set $$ \left\{E_{k}^{\psi,\varphi,t}\ |\ (\psi,\varphi,t) \in A_{N,k} \right\} $$ is a basis of $ \mathcal{E}_{k}(\Gamma_{1}(N)) $,
where $ A_{N,k} $ is defined as the set of triples $ (\psi,\varphi,t) $ such that $ \psi $ and $ \varphi $ are primitive Dirichlet characters modulo $ u $ and $ v $ respectively such that $\varphi(-1) \psi(-1) = (-1)^k$, and $ t $ a positive integer such that $ tuv \mid N $. We are given that $ \textit{dim}(\mathcal{E}_{k}(\Gamma_{1}(N)) = |A_{N,k}|$, so all that remains is to prove linear independence. We have that $$ E_{k}^{\psi,\varphi,t} = \delta(\psi)L(1-k,\varphi) + 2\sum_{n=1}^{\infty}\left(\sum_{\substack{m \mid n \\ m > 0}} \psi(n/m)\varphi(m)m^{k-1}\right)q^{nt} $$ where $$ \delta(\psi) = \left\{\begin{array}{c} 1 \textit{ if } \psi \textit{ is trivial }\\ 0 \textit{ otherwise } \end{array} \right\} $$ and $ q = e^{2\pi iz} $. So I have started by observing that the first term is nice i.e $$ \delta(\psi)L(1-k,\varphi) + 2q^{t} .$$ I want to say that no further values of $ n $ cancel out the $ n = 1 $ value, and so just prove the linear independence of the $ n = 1 $ term. However I am unable to establish whether this is the case. The book skips the proof and I have been unable to find anything on the internet. If anybody is able to prove the linear independence or knows of a proof then I would be most grateful.
Following up on the hint in @cfairwea's comment, it's not too hard to check that if $(\varphi', \psi') \ne (\varphi, \psi)$, then for any $t$ and $t'$ the series $E_k^{\varphi, \psi, t}$ and $E_k^{\varphi', \psi', t'}$ have different eigenvalues for the Hecke operators away from $N$, and hence there can't be any nontrivial linear dependence relation between them. So it suffices to show that the $E_k^{\varphi, \psi, t}$ for a fixed pair $(\varphi, \psi)$ and different values of $t$ are linearly independent; and this is not so difficult (you've already made a good start on it).