[proofplan] The proof compares the coefficient of $q$ on both sides of the eigenvalue equation $T_n f=\lambda_n f$. The standard Fourier coefficient formula for the Hecke operator $T_n$ shows that the coefficient of $q$ in $T_n f$ is exactly $a_n(f)$, because the only divisor of $\gcd(1,n)$ is $1$. The normalization $a_1(f)=1$ then identifies the coefficient of $q$ in $\lambda_n f$ with $\lambda_n$. [/proofplan] [step:Define the coefficient extraction used in the comparison] For each cusp form \begin{align*} g(z)=\sum_{m=1}^{\infty} a_m(g)q^m, \end{align*} define the first Fourier coefficient functional \begin{align*} c_1:S_k(\operatorname{SL}_2(\mathbb{Z})) &\to \mathbb{C} \\ g &\mapsto a_1(g). \end{align*} This functional is complex-linear because Fourier coefficients are extracted termwise from the Fourier expansion. [/step] [step:Compute the coefficient of $q$ in $T_n f$ using the Hecke coefficient formula] Fix $n \geq 1$. By the standard Fourier coefficient formula for Hecke operators on cusp forms of weight $k$ (citing a result not yet in the wiki: Fourier coefficient formula for Hecke operators), the coefficient of $q^m$ in $T_n f$ is \begin{align*} a_m(T_n f) = \sum_{d \mid \gcd(m,n)} d^{k-1} a_{mn/d^2}(f). \end{align*} Applying this formula with $m=1$, we have $\gcd(1,n)=1$, so the only positive divisor $d$ of $\gcd(1,n)$ is $d=1$. Therefore \begin{align*} a_1(T_n f) &= \sum_{d \mid \gcd(1,n)} d^{k-1} a_{n/d^2}(f) \\ &= 1^{k-1}a_n(f) \\ &= a_n(f). \end{align*} [guided] Fix $n \geq 1$. We want to recover the eigenvalue $\lambda_n$ from the equation $T_n f=\lambda_n f$, and the normalization of $f$ tells us that the coefficient of $q$ is the right coefficient to inspect. We use the standard Fourier coefficient formula for Hecke operators on cusp forms of weight $k$ (citing a result not yet in the wiki: Fourier coefficient formula for Hecke operators). It states that, for every $m \geq 1$, the coefficient of $q^m$ in $T_n f$ is \begin{align*} a_m(T_n f) = \sum_{d \mid \gcd(m,n)} d^{k-1} a_{mn/d^2}(f). \end{align*} Now set $m=1$. Since $\gcd(1,n)=1$, the divisor condition $d \mid \gcd(1,n)$ permits exactly one value, namely $d=1$. Thus the sum collapses to a single term: \begin{align*} a_1(T_n f) &= \sum_{d \mid \gcd(1,n)} d^{k-1} a_{n/d^2}(f) \\ &= 1^{k-1}a_n(f) \\ &= a_n(f). \end{align*} So the coefficient of $q$ in $T_n f$ is the $n$th Fourier coefficient of $f$. [/guided] [/step] [step:Compute the coefficient of $q$ in the eigenvalue equation] Since $f$ is a Hecke eigenform, the hypothesis gives \begin{align*} T_n f = \lambda_n f. \end{align*} Apply the linear functional $c_1$ to both sides. The previous step gives $c_1(T_n f)=a_n(f)$, while linearity of $c_1$ gives \begin{align*} c_1(\lambda_n f)=\lambda_n c_1(f)=\lambda_n a_1(f). \end{align*} Because $f$ is normalized, $a_1(f)=1$. Hence \begin{align*} a_n(f)=\lambda_n a_1(f)=\lambda_n. \end{align*} Since $n \geq 1$ was arbitrary, the equality $a_n(f)=\lambda_n$ holds for every $n \geq 1$. [/step]