Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 4 / 6C / Solution

Using ${\chi }_M(t)=\det (tI-M)$, expansion or triangular reordering givesThe eigenspace for $1$ is spanned by $e_1$, so its algebraic multiplicity two requires a size-two Jordan block. A convenient Jordan chain isFor eigenvalue $2$, the eigenspace has basis $w_1=e_1+e_3$, $w_2=e_4$, hence both corresponding blocks have size one. Therefore a Jordan normal form and the minimal polynomial areThe ordered basis $(v_1,v_2,w_1,w_2)$ supplies the change-of-basis matrixIts determinant is one, and the chain equations give $MP=PJ$, proving $P^{-1}MP=J$. The square at $t=1$ in the minimal polynomial is essential: $M$ is not diagonalizable.