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 $\mathbf{e}=(1,1,1{)}^T$ is an eigenvector with eigenvalue six. Since $M$ is a real symmetric matrix, its other two eigenvalues, say ${\lambda }_2,{\lambda }_3$, are real. The trace givesAlso, for a real symmetric matrix, $\mathrm{tr}(M^2)$ is the sum of the squares of all its entries. HereThus ${\lambda }_2^2+{\lambda }_3^2=42-36=6$. Together with their zero sum, this implies