A seven-dimensional counterexample to derivative realizability

Let $C_k$ be the permutation matrix of a directed $k$-cycle, and let $A=\operatorname{diag}(\tfrac12,C_2,C_4)$. This is an entrywise nonnegative real $7\times7$ matrix with characteristic polynomial $p(z)=(z-\tfrac12)(z^2-1)(z^4-1)$. The normalized derivative $q=p'/7$ is not the characteristic polynomial of any entrywise nonnegative real $6\times6$ matrix. Newton's identities give the power sums $s_k$ of the roots of $q$ as exact rationals, and $s_7=-8593/823543<0$. If a nonnegative $B$ had characteristic polynomial $q$, then $\operatorname{tr}(B^7)=s_7$. This is impossible, because every power of a nonnegative matrix has nonnegative trace, and the trace identity needs no diagonalizability. Hence the critical points of a realizable list need not be realizable at the smaller order. Appending zero eigenvalues leaves $s_7$ unchanged, so zero padding does not realize them either. The example has positive trace and is reducible.