 Solution

ID: 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 \geq 1

. With

u = (1, \dots, 1)^{T}

, the all-ones matrix satisfies

A v = (\sum_{j} v_{j}) u

. Hence $n$ is an eigenvalue, and for $n \geq 2$ 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.

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