[proofplan] The key point is that the first Fourier coefficient of $T_n g$ recovers the $n$-th Fourier coefficient of $g$. Therefore, if $g$ has the same Hecke eigenvalues as the normalized eigenform $f$, every Fourier coefficient of $g$ is forced to be $a_1(g)$ times the corresponding Fourier coefficient of $f$. Hence $g=a_1(g)f$ by equality of Fourier expansions, so the eigenspace is exactly the complex line spanned by $f$. [/proofplan] [step:Show that the normalized newform spans a line inside the simultaneous eigenspace] Since $f$ is assumed to satisfy \begin{align*} T_n f = a_n(f) f \end{align*} for every $n \in \mathbb{N}$, the definition of $E_f$ gives $f \in E_f$. Since $a_1(f)=1$, the form $f$ is not the zero cusp form. Therefore \begin{align*} \mathbb{C}f \subseteq E_f \end{align*} and $\dim_{\mathbb{C}} E_f \geq 1$. [/step] [step:Relate the Hecke eigenvalue condition to Fourier coefficients] Let $g \in E_f$. Write its Fourier expansion at the cusp $\infty$ as \begin{align*} g(z)=\sum_{m=1}^{\infty} a_m(g) q^m, \qquad q=e^{2\pi i z}. \end{align*} For each $n \in \mathbb{N}$, the defining Fourier-coefficient property of the Hecke operator $T_n$ on $S_2(\Gamma_0(N))$ gives \begin{align*} a_1(T_n g)=a_n(g). \end{align*} On the other hand, since $g \in E_f$, \begin{align*} T_n g = a_n(f) g. \end{align*} Taking the first Fourier coefficient on both sides gives \begin{align*} a_1(T_n g)=a_1(a_n(f)g)=a_n(f)a_1(g). \end{align*} Combining the two equalities, we obtain \begin{align*} a_n(g)=a_1(g)a_n(f) \end{align*} for every $n \in \mathbb{N}$. [guided] Let $g \in E_f$. We want to prove that $g$ is a scalar multiple of $f$, and the scalar should be forced by the first Fourier coefficient of $g$. Write \begin{align*} g(z)=\sum_{m=1}^{\infty} a_m(g)q^m, \qquad q=e^{2\pi i z}. \end{align*} Because $g$ is a cusp form, the expansion starts at $m=1$. The Hecke operator $T_n$ is normalized so that the first Fourier coefficient of $T_n g$ is the $n$-th Fourier coefficient of $g$: \begin{align*} a_1(T_n g)=a_n(g). \end{align*} This is the coefficient-extraction mechanism behind multiplicity one. Now use the defining property of $E_f$. Since $g \in E_f$, the same operator $T_n$ acts on $g$ through the scalar $a_n(f)$: \begin{align*} T_n g=a_n(f)g. \end{align*} Taking first Fourier coefficients gives \begin{align*} a_1(T_n g)=a_1(a_n(f)g)=a_n(f)a_1(g). \end{align*} Comparing the two formulas for $a_1(T_n g)$ yields \begin{align*} a_n(g)=a_1(g)a_n(f) \end{align*} for every $n \in \mathbb{N}$. [/guided] [/step] [step:Conclude that every vector in the eigenspace is a scalar multiple of $f$] Define the scalar \begin{align*} c:=a_1(g) \in \mathbb{C}. \end{align*} The previous step proves that \begin{align*} a_n(g)=c\,a_n(f) \end{align*} for every $n \in \mathbb{N}$. Hence the Fourier expansions of $g$ and $cf$ agree coefficient by coefficient: \begin{align*} g(z)=\sum_{n=1}^{\infty} a_n(g)q^n = \sum_{n=1}^{\infty} c\,a_n(f)q^n = c f(z). \end{align*} Thus $g \in \mathbb{C}f$. Since $g \in E_f$ was arbitrary, we have \begin{align*} E_f \subseteq \mathbb{C}f. \end{align*} Together with $\mathbb{C}f \subseteq E_f$, this gives \begin{align*} E_f=\mathbb{C}f. \end{align*} Therefore \begin{align*} \dim_{\mathbb{C}}E_f=1. \end{align*} [/step]