Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 8F / Solution

For an $n\times n$ matrix $A$, the characteristic polynomial isThe Cayley-Hamilton theorem states that ${\chi }_A(A)=0$.

Over $\mathbb{C}$, choose a basis in which $A$ is upper triangular, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$. For the standard invariant flag $V_j=⟨e_1,\dots ,e_j⟩$,The factors $A-{\lambda }_{j}I$ commute, so applying their product in descending order sends $V_n$ successively into $V_{n-1},\dots ,V_0=0$. Hencewhich proves the theorem.

Direct expansion gives the commutator product rule:

Put $C=[B,A]$. Since $C$ commutes with $A$, repeated use of the product rule givesBy linearity, for every polynomial $\phi$,

Let $D(X)=[B,X]$ and suppose $f(A)=0$. For $k=1$,Assume inductively thatBoth $u$ and $C$ are polynomials in, or commute with, $A$, so $u{C}^m=C^{m}u=0$. Apply the derivation $D$ to $u{C}^m=0$ and multiply on the left by $C^m$:Since $D(u)=f^{(k+1)}(A)C$, this saysThe induction is complete. Taking $f={\chi }_A$ and $k=n$ givesThus $[B,A]$ is nilpotent, which is the Jacobson lemma for a commuting commutator.

Solved by gpt-5.6-sol high.