 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/6/b/solution

Every row sums to six, so

e = (1, 1, 1)^{T}

is an eigenvector with eigenvalue six. Since

M

is a real symmetric matrix, its other two eigenvalues, say

λ_{2}, λ_{3}

, are real. The trace gives

λ_{2} + λ_{3} = tr M - 6 = 0.

Also, for a real symmetric matrix,

tr (M^{2})

is the sum of the squares of all its entries. Here

tr (M^{2}) = 1^{2} + 3^{2} + 2^{2} + 2 (2^{2} + 3^{2} + 1^{2}) = 42.

Thus

λ_{2}^{2} + λ_{3}^{2} = 42 - 36 = 6

. Together with their zero sum, this implies

spec M = {6, 3, - 3} .

 New to topics? Read the docs here!