Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 3 / 10G / Solution

(1) True. Over $\mathbb{C}$, $A$ has an eigenvalue $\lambda$ with nonzero eigenspace $E$. Commutativity makes $E$ invariant under $B$: $A(Bv)=B(Av)=\lambda Bv$. The endomorphism $B{|}_E$ has an eigenvector $v\ne 0$ because its characteristic polynomial has a complex root. This $v$ is an eigenvector of both matrices, though their eigenvalues on it may differ.

(2) False. Take $A=I$ and $B=2I$. They commute, but their only eigenvalues are respectively $1$ and $2$.

(3) True. Associativity gives $(BA{)}^{n+1}=B(AB{)}^{n}A=0$, so $T=BA$ is nilpotent. A nilpotent endomorphism of an $n$-dimensional space has index at most $n$: in $V\supseteq TV\supseteq T^{2}V\supseteq \cdots$, equality at a nonzero image would make $T$ surjective on that finite-dimensional image, hence invertible there, contradicting nilpotence. Thus dimensions strictly decrease until zero, in at most $n$ steps. Therefore $(BA{)}^n=0$.

(4) False. Take $T=0$ on a two-dimensional vector space and $\lambda =0$. The eigenspace has dimension two, but the minimal polynomial is $t$, in which the root zero has multiplicity one. Eigenspace dimension counts independent eigenvectors; the exponent in the minimal polynomial measures the longest corresponding Jordan chain, not that dimension.

(5) True. Factor the minimal polynomial as $m_T(t)=(t-\lambda {)}^{c}q(t)$, with $q(\lambda )\ne 0$, and set $S=T-\lambda I$. If $v\in W_{c+1}$, then $w=S^{c}v$ satisfies $Sw=0$, so $q(T)w=q(\lambda )w$. But $m_T(T)v=0$ also gives $q(T)w=0$. Thus $w=0$, so $v\in W_c$. The reverse inclusion is immediate from $S^{c}v=0\Rightarrow S^{c+1}v=0$, proving