Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 18 / b / Solution

The matrix is symmetric under interchange of the first and third coordinates. For an antisymmetric vector $(1,0,-1)$, direct multiplication gives $A(1,0,-1{)}^{\mathsf{T}}=-6(1,0,-1{)}^{\mathsf{T}}$. For a symmetric vector $(p,q,p)$, the linear transformation reduces toThe reduced characteristic polynomial is $\lambda (\lambda -12)$. At $\lambda =0$, its equations give $q=2p$; at $\lambda =12$, they give $q=-p$. Normalizing the three eigenvectors givesTheir norms are one. Their unnormalized pairwise dot products are $1-1=0$, $1-1=0$ and $1-2+1=0$, proving mutual orthogonality. The eigenvalues sum to six and have product zero, agreeing with the matrix trace and determinant from part (a).