= Prime-power old block at a bad prime
{title2=$V=\operatorname{span}\{p^{jk/2}f(p^jz):0\le j\le r\}$}
If $p$ divides the primitive level $M$ of a trivial-character <newform> and $N=Mp^r$, put $e_j=p^{jk/2}f(p^jz)$. Then $U_pe_0=a_pe_0$ and $U_pe_j=p^{k/2}e_{j-1}$ for $j\ge1$. The normalized <Fricke involution> acts by $e_j\mapsto\varepsilon e_{r-j}$, where $W_Mf=\varepsilon f$. 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.
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