 Solution

An eigenform is understood to be nonzero; the zero form would put no constraint on the alleged eigenvalues. The coefficient of $q$ in $T_w(n)f$ is $c(n)$, since $\gcd (n,1)=1$. Comparing with the coefficient in $\lambda (n)f$ givesThe constant coefficient comparison gives $\lambda (n)c(0)={\sigma }_{w-1}(n)c(0)$, so

For completeness derive the operator multiplication law underlying the last relation. On formal Fourier series defineThese auxiliary maps need not separately preserve modularity, but coefficient identities are valid and $U_p{V}_p=I$. Part (a) givesMultiplication on the left by $U_p+A{V}_p$ yields the $T_w(p^{e+1})$ sum and, from $U_p{V}_p^j=V_p^{j-1}$ for $j\ge 1$, the remaining sum $A{T}_w(p^{e-1})$. HenceInduction on the smaller exponent, applying this recurrence and cancelling its two overlapping sums, now givesTo make the induction explicit, write $T_r=T_w(p^r)$ and assume $2\le r\le s$. Substituting the formulas already established for $r-1$ and $r-2$ into $T_r{T}_s=(T_1{T}_{r-1}-A{T}_{r-2})T_s$ givesThe last term is exactly the missing $j=r$ term. The cases $r=0,1$ are the identity operator and the prime recurrence above.

For distinct primes, their $U$ and $V$ maps commute, and the divisor formula in (a) factors independently at each prime. It follows that $T_w(m)T_w(n)=T_w(mn)$ for coprime $m,n$. Multiplying the prime-power identities therefore gives the general Hecke multiplication relationsApply this identity to the same nonzero eigenform and cancel the vector $f$, obtainingThis reasoning does not silently divide by $c(1)$, which is important when considering noncuspidal weight-zero forms.