Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 1

The minimal polynomial $m_T(t)$ is the unique monic polynomial of least positive polynomial degree satisfying $m_T(T)=0$. Such an annihilating polynomial exists without using the Cayley-Hamilton theorem: the vector space $\mathrm{End}(V)$ has dimension of a vector space equal to $n^2$, so $I,T,\dots ,T^{n^2}$ are linearly dependent. Normalize a nonzero annihilating polynomial to be monic and choose one of least degree. Two distinct monic choices of that degree would have an annihilating difference of lower degree, contradicting minimality. A nonzero constant cannot annihilate $T$ because $V\ne 0$.

Choose a basis $B$ and let $A=[T{]}_B$. The characteristic polynomial isIt is independent of the chosen basis. Indeed a change of basis replaces $A$ by a similar matrix $P^{-1}AP$, andby multiplicativity of the determinant. Thus this monic degree-$n$ polynomial is an invariant of the linear transformation itself.

Since $v$ is a cyclic vector and $V\ne 0$, $v\ne 0$. Finite dimension of a vector space forces a first dependence in the sequence $v,Tv,T^{2}v,\dots$. Let $k\ge 1$ be the smallest index with $T^{k}v$ in the linear span of its predecessors. Thenis linearly independent. Its linear span $W$ is invariant under $T$: all but the last basis candidate map to the next, and the last maps to $T^{k}v\in W$. Induction therefore puts every $T^{j}v$ in $W$. Cyclicity gives $W=V$, so $B$ is a basis and $k=n$.

WriteThe matrix of $T$ in this basis is the companion matrixFor $p(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$, the defining relation says $p(T)v=0$. Since $p(T)$ commutes with $T$, it also annihilates every $T^{j}v$ in $B$, hence $p(T)=0$. No nonzero polynomial of degree less than $n$ can annihilate $v$, because that would contradict linear independence of $B$. Thus $m_T=p$.

Finally the characteristic polynomial of a companion matrix is $p$, by direct determinant expansion. Explicitly, expansion along the first row of $tI-C$ gives the recurrencestarting from $D_1=t+a_0$. It yields $D_n=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$. ConsequentlyThis proves that a cyclic operator has equal minimal and characteristic polynomials by the cyclic basis and a direct determinant calculation, with no appeal to the Cayley-Hamilton theorem.

For $T_1$, choose the third standard basis vector $v=e_3$. Its first three orbit vectors areThey are linearly independent: their matrix of columns has determinant $-1$. They therefore span ${\mathbb{R}}^3$, proving $T_1$ has a cyclic vector, namely $e_3$.

For $T_2$, write $T_2=I+N$ where $N^2=0$. The binomial theorem gives $T_2^{j}v=v+jNv$ for every integer $j\ge 0$. The entire orbit of any $v$ lies in $\mathrm{span}\{v,Nv\}$, which has dimension of a vector space at most two. Therefore $T_2$ has no cyclic vector in the three-dimensional space. Equivalently, its minimal polynomial is $(t-1{)}^2$, while its characteristic polynomial is $(t-1{)}^3$.