Prime-power old block at a bad prime ($V=\mathrm{span}\{p^{jk/2}f(p^{j}z):0\le j\le r\}$)

If $p$ divides the primitive level $M$ of a trivial-character newform and $N=M{p}^r$, put $e_j=p^{jk/2}f(p^{j}z)$. Then $U_p{e}_0=a_p{e}_0$ and $U_p{e}_j=p^{k/2}{e}_{j-1}$ for $j\ge 1$. The normalized Fricke involution acts by $e_j\mapsto \epsilon e_{r-j}$, where $W_{M}f=\epsilon f$. These two operators act irreducibly: a nonzero invariant subspace contains an eigenvector of $U_p$; its eigenvalue is $a_p$ or zero. Either it contains $e_0$ immediately or applying the reversal and a power of $U_p$ produces $e_0$, using $a_p^2\ne p^k$. Reversal then produces $e_r$, and its successive images produce every basis vector.