Solution
= Solution
A <basis> of $E_n$ is $\{u\}$. For $n\geq2$,
$$
\boxed{E_0=\{v:\sum_jv_j=0\}=\operatorname{span}\{e_1-e_n,\ldots,e_{n-1}-e_n\}.}
$$
The displayed vectors are linearly independent by their first $n-1$ coordinates, and every vector with zero coordinate sum has exactly that expansion. At $n=1$ this list is empty and the <kernel> is $\{0\}$; it is not an additional <eigenspace> associated with a present <eigenvalue>. This is the <spectrum of an all-ones matrix>.