Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 3 / 7D / ii

Take the Jordan blockIts only eigenvalue is one. An eigenvector satisfies $(C-I)v=0$, so its second coordinate vanishes; the eigenspace is one-dimensional. A diagonalizable matrix of size two requires a basis of two independent eigenvectors, which this matrix lacks even over $\mathbb{C}$.