 Solution

Take a nonzero eigenform of positive weight, and put $h(\tau )=f(p\tau )$ and $C=\chi (p)p^{k-1}$. The original form belongs to level $Np$ by inclusion of groups. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(Np)$, the conjugate $\left(\begin{array}{cc}a& pb\\ c/p& d\end{array}\right)$ lies in ${\Gamma }_0(N)$, soIts Dirichlet character is therefore the specified reduced Dirichlet character; modular cusp holomorphy is permitted in the question. This is the Dirichlet character version of an oldform by argument dilation.

A nonzero positive-weight modular form cannot be constant, because the matrix $\left(\begin{array}{cc}1& 0\\ N& 1\end{array}\right)$ would force a nonzero constant to equal $(N\tau +1{)}^k$ times itself. Let $n_0>0$ be its first nonzero positive Fourier index. In a relation $Af+Bh=0$, the $q^{n_0}$ coefficient forces $A=0$, then $B=0$. Hence their span is two-dimensional, even when the original constant term is nonzero.

At the new level $p$ is a bad prime, so its operator is $U_p$. The good-prime eigenrelation at the old level and part (c) giveThus the matrix in the ordered basis $(f,h)$ iswith characteristic polynomial $X^2-\lambda X+C$. For its distinct roots,This is prime stabilization of an oldform.

The nonzero positive-weight qualification is necessary for the two-dimensional assertion. The zero form gives no such span, and if weight zero is allowed, $f=1$ at trivial Dirichlet character has distinct good-prime roots $1$ and $p^{-1}$, while $f(\tau )=f(p\tau )$ spans only one dimension. The asserted result uses the usual nonzero positive-weight eigenform setting.