I am trying the following exercise.
Let $V=S_2(\Gamma _1 (16))$ and we are given the following basis of $V$, their $q$ expansion at the $\infty $ cusp. \begin{align} f_1 =& q − 2q^3 − 2q^4 + 2q^6 + 2q^7 + 4q^8 − q^9 + O(q^{10}), \\ f_2 =& q^2 − q^3 − 2q^4 + q^5 + 2q^7 + 2q^8 − q^9 + O(q^{10}). \end{align} Show that $S_2(\Gamma _1 (8))=\{0\}$ and $S_2(\Gamma _1(16))_{new}=V$.
I don't know how to prove that $S_2(\Gamma _1 (8)=\{0\}$. The dimension formula involves genus and I don't know how to compute the genus of $X_1(8)$. But once this is done, it follows immediately that $V= S_2(\Gamma _1 (16))_{new}$.
Next exercise is to
Compute the matrix of the operator $T_2$.
Now, I could use the formula $$ a_m(T_n)=\sum _{d|(m,n)} a_{(m,n)/d^2}(\langle d \rangle f)$$ since I have the $q$-expansions and look at first few coefficients. But how is this enough since I only know the $q$-expansions till $O(q^{10})$?