Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 2 / c / ii / Solution

Set $N=M_i-I$. Direct multiplication giveswith $N\ne 0$. If $M_i$ were diagonalisable, then subtracting the identity in the same basis would diagonalise $N$. A diagonal matrix whose cube is zero has every diagonal entry zero and is itself zero. Similarity would then imply $N=0$, a contradiction. HenceThis proves failure directly; merely observing that it is not self-adjoint would not suffice, because non-self-adjoint matrices can still be diagonalisable.