Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 27 / 1 / b / Solution

For a prime number $p$, the defining finite sum for the Hecke operator simplifies toThe first summand has no $q^1$ term. In the second, summing the $p$th roots of unity selects precisely the terms whose index is divisible by $p$, and its $q^1$ coefficient is $a_p$. Comparing with $T_{p}f=\lambda (p)f$, using $a_1=1$, gives $\boxed{\lambda (p)=a_p}$. This also follows from the general coefficient computation in part (a).