Assume $n\geq1$. With $u=(1,\ldots,1)^T$, the <all-ones matrix> satisfies $Av=(\sum_jv_j)u$. Hence \b[$n$ is an <eigenvalue>, and for $n\geq2$ 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.

Solution

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

 Solution (source code)