While studying, I recently came across the following interesting problem.
Let's say that the (level one) weight $k$ modular forms $M_k(\Gamma(1))$ have dimension $d$. We know by the ring structure of $M_*(\Gamma(1))$ that the map $M_k(\Gamma(1)) \rightarrow \mathbb{C}^d$, sending a form to its first $d$ Fourier coefficients, is an isomorphism. This suggests a natural basis for $M_k(\Gamma(1))$: the pre-images of the standard basis vectors $(e_1, \ldots, e_n)$ for $\mathbb{C}^d$. Let's call this basis $g_1, \ldots, g_d$.
The claim is that if $T_n$ is the $n$th Hecke operator, then
$$T_n = \sum_{j=1}^{d-1} a_n(g_j) T_j$$
acting on $S_k(\Gamma(1))$.
How can you prove this identity?
I had thought to approach this by using the formula for the Fourier coefficients of the Hecke operators. We know that it suffices to check that both sides agree on the basis vectors $g_k$, which we can check by comparing the first $d$ Fourier coefficients. In principle, we can write down a formula for $b_i(T_n g_k)$ and $\sum_{j=1}^{d-1} a_n(g_j) b_i(T_j(g_k))$, but I tried doing this and it seemed really messy.
EDIT: I originally accepted the answer below, but now I don't know that it's correct - see my comment.
The space of cusp forms $S_k(\Gamma(1))$ admits a basis of normalized ($a_1=1$) eigenforms, so it suffices to prove the identify for normalized eigenforms.
Let $f = \alpha_1 g_1 + \cdots \alpha_{d-1} g_{d-1}$ be a normalized eigenform in $S_k(\Gamma(1))$. Then, $$T_n(f) = a_n(f)f = [\alpha_1a_n(g_1) + \cdots + \alpha_{d-1}a_n(g_{d-1})]f,$$ where $a_n(\cdot)$ denotes the $n$th Fourier coefficient of the modular form.
On the other hand, since $a_j(f) = \alpha_j$ for $0 < j < d$, we have $$\sum_{j=1}^{d-1}a_n(g_j)T_j(f) = \sum_{j=1}^{d-1}a_n(g_j)a_j(f)f = \sum_{j=1}^{d-1}a_n(g_j)\alpha_jf.$$