Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Digital systems, linear algebra, introduction to formal proof resit / 6 / b / i / Solution

Assume $n\ge 1$. With $u=(1,\dots ,1{)}^T$, the all-ones matrix satisfies $Av=({\sum }_j{v}_j)u$. Hence $n$ is an eigenvalue, and for $n\ge 2$ the other eigenvalue is $0$. At $n=1$ the matrix is $[1]$, so 1 is its only eigenvalue; zero is not an eigenvalue with a nonzero eigenvector in that case.