 Solution

The decomposition into $\mathrm{span}\{u\}$ and the zero-sum vector subspace gives characteristic polynomial $t^{n-1}(t-n)$. For $n\ge 2$, $n$ has algebraic multiplicity and geometric multiplicity one; zero has both multiplicities $n-1$. At $n=1$, only the eigenvalue 1 occurs, with both multiplicities one. Referring to a multiplicity zero for the absent root zero is merely a bookkeeping convention, not an assertion that it is an eigenvalue.