 Spectrum of an all-ones matrix

For $u=(1,\dots ,1{)}^T$, $J_n=u{u}^T$ maps $v$ to $({\sum }_i{v}_i)u$. Therefore $u$ is an eigenvector with eigenvalue $n$, while the zero-sum vector subspace is its kernel. For $n\ge 2$ a basis of the kernel is $e_1-e_n,\dots ,e_{n-1}-e_n$. These vectors and $u$ form an eigenbasis, proving diagonalization and characteristic polynomial $t^{n-1}(t-n)$. Both algebraic multiplicity and geometric multiplicity are one at $n$ and $n-1$ at zero. At $n=1$ the sole eigenvalue is 1; the zero-sum space has dimension zero and contributes no additional root.