Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 6C (Vectors and Matrices)

$A(A^2-4A+5I)=2I$, so $A$ has inverse.

$A^{-1}=\frac{1}{2}(A^2-4A+5I)$.

The relation $A^3=4A^2-5A+2I$ reduces $A^k$ inductively to a linear combination of $I,A,A^2$.

$Bx=0⟺ASx=0⟺x\in S^{-1}(\ker A)$; invertibility of $S$ preserves dimension.

Rank-nullity theorem gives $\mathrm{rank}T+\dim \ker T=n$. Equal nullities therefore give equal ranks.

$B^k=S^{-1}{A}^{k}S$, so the preceding argument applied to $A^k,B^k$ gives equal ranks.

Cayley-Hamilton theorem says ${\chi }_B(B)=0$. Since $B$ is nilpotent, all eigenvalues and hence every non-leading coefficient of ${\chi }_B(t)$ vanish, so ${\chi }_B(t)=t^n$ and $B^n=0$.