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

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.