Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 1 / 8B / a / i / Solution

The symmetry between the last two coordinates gives the eigenvector $(0,1,-1{)}^T$ with eigenvalue $2$. On vectors $(s,t,t{)}^T$, the remaining eigenvalue equation iswhose characteristic polynomial is ${\lambda }^2-5\lambda +4=(\lambda -1)(\lambda -4)$. Corresponding eigenvectors are $(1,-1,-1{)}^T$ and $(2,1,1{)}^T$. Thus all eigenvalues and eigenspaces areNonzero scalar multiples give all eigenvectors in each eigenspace. The three displayed vectors are mutually orthogonal and have squared norms $3,2,6$, respectively.