Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 1 / 7B / b / Solution

Eigenvectors belonging to distinct eigenvalues are linearly independent. Under the hypothesis they therefore form a basis $v_1,\dots ,v_n$. For every $j$,because ${\chi }_A({\lambda }_j)=0$. A linear map that vanishes on a basis is zero, henceThis proves the Cayley-Hamilton theorem in the distinct-eigenvalue case.