Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 13

The characteristic polynomial isThus the eigenvalues and corresponding eigenvectors over $\mathbb{C}$ areAll nonzero scalar multiples of these vectors are also eigenvectors. Distinct eigenvalues make them linearly independent. Put them in the columns ofThe equations $A{v}_{\pm }={\lambda }_{\pm }{v}_{\pm }$ are exactly $AB=BC$, giving the diagonalizationThere is no diagonalization over $\mathbb{R}$, since the matrix has no real eigenvalues.

Expanding the determinant gives $\det (\lambda I-M)=(\lambda -1)[(\lambda -1{)}^2-25]$. Solving each eigenvalue equation yieldsTheir pairwise inner products are $16-25+9=0$, $12-12=0$ and $12-12=0$. Their squared Euclidean norms are $50,50,25$, respectively, so division by $\sqrt{50},\sqrt{50},5$ gives an orthonormal basis. This also follows from the spectral theorem for real symmetric matrices: symmetry makes eigenvectors for distinct eigenvalues orthogonal.