[proofplan] The proof is a finite-dimensional linear algebra argument applied to the Hecke operators. We use the standard Hecke facts that the operators $T_n$ commute with one another and are self-adjoint for the Petersson inner product. A commuting family of [self-adjoint operators](/page/Self-Adjoint%20Operators) admits an orthogonal simultaneous eigenspace decomposition; then the level-$1$ Fourier coefficient formula for Hecke operators shows that every nonzero simultaneous eigenform has nonzero first Fourier coefficient, so each basis vector can be normalized. [/proofplan] [step:Use the self-adjoint commuting Hecke operators to build simultaneous eigenspaces] Write \begin{align*} V := S_k(SL_2(\mathbb{Z})). \end{align*} The Petersson inner product makes $V$ a finite-dimensional complex [inner product space](/page/Inner%20Product%20Space). We use the standard normalization of the level-$1$ Hecke operators, namely the operators \begin{align*} T_n: V &\to V \end{align*} whose Fourier coefficients satisfy \begin{align*} a_r(T_nf) = \sum_{d \mid \gcd(r,n)} d^{k-1} a_{rn/d^2}(f) \end{align*} for every $f \in V$ and every $r,n \in \mathbb{N}$. With this normalization, the Hecke operators are self-adjoint for the Petersson inner product and commute: \begin{align*} T_mT_n = T_nT_m \end{align*} for all $m,n \in \mathbb{N}$. We prove the [simultaneous diagonalisation](/theorems/2403) statement for the family $\mathcal{T} := \{T_n : n \in \mathbb{N}\}$. Let $W \subset V$ be a nonzero subspace invariant under every operator in $\mathcal{T}$. If every $T_n|_W$ is scalar, then $W$ is already a common eigenspace for the whole family. Otherwise choose $n_0 \in \mathbb{N}$ such that $T_{n_0}|_W$ is not scalar. Since $T_{n_0}$ is self-adjoint on $V$ and $W$ is invariant under $T_{n_0}$, the restriction $T_{n_0}|_W: W \to W$ is self-adjoint for the restricted Petersson inner product. The finite-dimensional spectral theorem gives an [orthogonal decomposition](/theorems/436) of $W$ into proper nonzero eigenspaces of $T_{n_0}|_W$. Each such eigenspace is invariant under every $T_n$. Indeed, if \begin{align*} E_\lambda := \{f \in W : T_{n_0}f = \lambda f\} \end{align*} for an eigenvalue $\lambda \in \mathbb{C}$ of $T_{n_0}|_W$, then for $f \in E_\lambda$ and $n \in \mathbb{N}$, \begin{align*} T_{n_0}(T_nf) = T_n(T_{n_0}f) = T_n(\lambda f) = \lambda T_nf. \end{align*} Thus $T_nf \in E_\lambda$. Starting from $W=V$, repeat this refinement only on summands on which some $T_n$ is not scalar. Every genuine refinement replaces a nonzero summand by an orthogonal direct sum of proper nonzero subspaces, so the maximal dimension of a non-final summand decreases after finitely many refinements at that dimension. Since dimensions are positive integers bounded by $\dim_{\mathbb{C}} V$, the process terminates by induction on dimension. We obtain an orthogonal decomposition \begin{align*} V = \bigoplus_{\alpha \in A} V_\alpha, \end{align*} where $A$ is a finite index set and, for every $\alpha \in A$ and every $n \in \mathbb{N}$, there is a scalar $\lambda_{\alpha,n} \in \mathbb{C}$ such that \begin{align*} T_nf = \lambda_{\alpha,n}f \end{align*} for all $f \in V_\alpha$. Choosing an orthogonal basis of each nonzero $V_\alpha$ gives an orthogonal basis of $V$ consisting of simultaneous Hecke eigenforms. [guided] Let \begin{align*} V := S_k(SL_2(\mathbb{Z})). \end{align*} This is a finite-dimensional complex [vector space](/page/Vector%20Space), and the Petersson inner product makes it a finite-dimensional complex inner product space. We use the standard normalization of the level-$1$ Hecke operators: for each $n \in \mathbb{N}$, the operator \begin{align*} T_n: V &\to V \end{align*} is the Hecke operator whose Fourier coefficients satisfy \begin{align*} a_r(T_nf) = \sum_{d \mid \gcd(r,n)} d^{k-1} a_{rn/d^2}(f) \end{align*} for every $f \in V$ and every $r,n \in \mathbb{N}$. With this convention, the standard Hecke algebra facts used here are that each $T_n$ is self-adjoint for the Petersson inner product and that the Hecke operators commute: \begin{align*} T_mT_n = T_nT_m \end{align*} for all $m,n \in \mathbb{N}$. The point that needs proof is not just simultaneous diagonalisation for two operators, but simultaneous diagonalisation for the entire infinite family $\{T_n : n \in \mathbb{N}\}$. We handle this by refining invariant subspaces only when some operator in the family is not yet scalar on that subspace. Let $W \subset V$ be a nonzero subspace invariant under every $T_n$. If every restricted operator $T_n|_W$ is scalar, then every nonzero vector in $W$ is already an eigenvector for every Hecke operator, so no further work is needed on $W$. If not, choose $n_0 \in \mathbb{N}$ such that $T_{n_0}|_W$ is not scalar. Because $T_{n_0}$ is self-adjoint on $V$ and maps $W$ into $W$, its restriction \begin{align*} T_{n_0}|_W: W &\to W \end{align*} is self-adjoint with respect to the restricted Petersson inner product: for $f,g \in W$, \begin{align*} (T_{n_0}f,g) = (f,T_{n_0}g), \end{align*} and $T_{n_0}g \in W$. The finite-dimensional spectral theorem therefore decomposes $W$ as an orthogonal direct sum of eigenspaces of $T_{n_0}|_W$. Since $T_{n_0}|_W$ is not scalar, this decomposition is a genuine refinement into proper nonzero subspaces. We must check that this refinement is compatible with every other Hecke operator. Let \begin{align*} E_\lambda := \{f \in W : T_{n_0}f = \lambda f\} \end{align*} be one eigenspace of $T_{n_0}|_W$. For $f \in E_\lambda$ and $n \in \mathbb{N}$, commutativity gives \begin{align*} T_{n_0}(T_nf) = T_n(T_{n_0}f) = T_n(\lambda f) = \lambda T_nf. \end{align*} Hence $T_nf \in E_\lambda$. Thus every eigenspace produced by the spectral theorem remains invariant under the entire Hecke family. Now start with $W=V$ and repeat this procedure on any summand on which at least one $T_n$ is not scalar. The termination argument is finite-dimensional: a genuine refinement replaces a summand by proper nonzero subspaces of smaller dimension, and dimensions are positive integers at most $\dim_{\mathbb{C}}V$. Induction on the dimension of the summands shows that after finitely many refinements no summand admits further refinement. Therefore each final summand $V_\alpha$ has the property that every $T_n|_{V_\alpha}$ is scalar. We have obtained an orthogonal decomposition \begin{align*} V = \bigoplus_{\alpha \in A} V_\alpha, \end{align*} where $A$ is a finite index set and, for every $\alpha \in A$ and every $n \in \mathbb{N}$, there exists $\lambda_{\alpha,n} \in \mathbb{C}$ such that \begin{align*} T_nf = \lambda_{\alpha,n}f \end{align*} for all $f \in V_\alpha$. Finally, choose an orthogonal basis inside each nonzero $V_\alpha$. Since every vector in $V_\alpha$ is a simultaneous eigenvector for all $T_n$, the union of these bases is an orthogonal basis of $V$ consisting of simultaneous Hecke eigenforms. [/guided] [/step] [step:Show that a nonzero simultaneous eigenform has nonzero first Fourier coefficient] Let $f \in V$ be a nonzero simultaneous Hecke eigenform. Write its Fourier expansion at infinity as \begin{align*} f(z) = \sum_{r=1}^{\infty} a_r(f)q^r, \qquad q := e^{2\pi iz}. \end{align*} For each $n \in \mathbb{N}$, let $\lambda_n(f) \in \mathbb{C}$ be the scalar satisfying \begin{align*} T_nf = \lambda_n(f)f. \end{align*} Using the coefficient formula for the normalized level-$1$ Hecke operator stated above with $r=1$, the only divisor of $\gcd(1,n)$ is $1$, so \begin{align*} a_1(T_nf) = a_n(f). \end{align*} On the other hand, using the eigenvalue equation and linearity of the coefficient functional $a_1$, we get \begin{align*} a_1(T_nf) = a_1(\lambda_n(f)f) = \lambda_n(f)a_1(f). \end{align*} Therefore \begin{align*} a_n(f) = \lambda_n(f)a_1(f) \end{align*} for every $n \in \mathbb{N}$. If $a_1(f)=0$, then $a_n(f)=0$ for every $n \in \mathbb{N}$, so every Fourier coefficient of $f$ vanishes. Hence $f=0$, contradicting the choice of $f$. Thus \begin{align*} a_1(f) \neq 0. \end{align*} [guided] Let $f \in V$ be a nonzero simultaneous Hecke eigenform. Since $f$ is a cusp form for $SL_2(\mathbb{Z})$, it has a Fourier expansion at the cusp infinity of the form \begin{align*} f(z) = \sum_{r=1}^{\infty} a_r(f)q^r, \qquad q := e^{2\pi iz}. \end{align*} The coefficient $a_r(f)$ denotes the coefficient of $q^r$ in this expansion. Because $f$ is a simultaneous eigenform, for each $n \in \mathbb{N}$ there is a scalar $\lambda_n(f) \in \mathbb{C}$ such that \begin{align*} T_nf = \lambda_n(f)f. \end{align*} We now compare the first Fourier coefficient of both sides after applying $T_n$. The normalized level-$1$ Hecke coefficient formula stated in the previous step says \begin{align*} a_r(T_nf) = \sum_{d \mid \gcd(r,n)} d^{k-1} a_{rn/d^2}(f). \end{align*} Taking $r=1$, the set of positive divisors of $\gcd(1,n)$ is $\{1\}$, so the formula becomes \begin{align*} a_1(T_nf) = a_n(f). \end{align*} The eigenvalue equation gives another expression for the same coefficient. Since $a_1$ is a complex-linear coefficient functional, \begin{align*} a_1(T_nf) = a_1(\lambda_n(f)f) = \lambda_n(f)a_1(f). \end{align*} Combining these two identities gives \begin{align*} a_n(f) = \lambda_n(f)a_1(f) \end{align*} for every $n \in \mathbb{N}$. This identity shows that all Fourier coefficients of a simultaneous eigenform are determined by its first Fourier coefficient and its Hecke eigenvalues. If $a_1(f)=0$, then \begin{align*} a_n(f)=0 \end{align*} for every $n \in \mathbb{N}$. Therefore the entire Fourier expansion of $f$ is zero, so $f=0$. This contradicts the assumption that $f$ is nonzero. Hence \begin{align*} a_1(f) \neq 0. \end{align*} [/guided] [/step] [step:Normalize the orthogonal simultaneous eigenbasis] Let \begin{align*} \mathcal{B} = \{f_1,\dots,f_d\} \end{align*} be the orthogonal simultaneous Hecke eigenbasis constructed above, where $d=\dim_{\mathbb{C}}V$. If $d=0$, the assertion is vacuous. If $d>0$, then each $f_j$ is nonzero, so the previous step gives \begin{align*} a_1(f_j) \neq 0 \end{align*} for every $j \in \{1,\dots,d\}$. Define \begin{align*} g_j := a_1(f_j)^{-1}f_j. \end{align*} Then $a_1(g_j)=1$. Since scalar multiplication by a nonzero complex number preserves eigenspaces, each $g_j$ remains a simultaneous eigenform for every $T_n$. Since scalar multiplication also preserves orthogonality between distinct vectors, the family \begin{align*} \{g_1,\dots,g_d\} \end{align*} is an orthogonal basis of $V$ consisting of normalized simultaneous Hecke eigenforms. This proves the theorem. [/step]