Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 1 / 7B (Vectors and Matrices)

ChooseThe Leibniz formula for determinants shows that its leading term is $(-z{)}^n$, as required. A scalar $\lambda$ is an eigenvalue exactly when $A-\lambda I$ has a nonzero kernel, equivalently when it is singular, soThe fundamental theorem of algebra gives a complex root of the degree-$n$ polynomial ${\chi }_A$, so every nonempty complex square matrix has at least one complex eigenvalue.

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.

Over $\mathbb{C}$, triangularize $B$ and denote its eigenvalues, with algebraic multiplicity, by ${\beta }_1,\dots ,{\beta }_n$. ThenThe numbers ${\omega }_l=e^{2\pi il/m}$ are all the root of unity solutions of $w^m=1$, soConsequentlyIf $\lambda$ is an eigenvalue of $B^m$, choose $z$ with $z^m=\lambda$. The displayed identity makes at least one factor ${\chi }_B({\omega }_{l}z)$ zero. Thus $\mu ={\omega }_{l}z$ is an eigenvalue of $B$ and