Theorem 1, Section 7.1 of Hoffman’s Linear Algebra

Question

Let $\alpha$ be any non-zero vector in $V$ and let $p_\alpha$ be the $T$-annihilator of $\alpha$.

(i) The degree of $p_\alpha$ is equal to the dimension of the cyclic subspace

(ii) If the degree of $p_\alpha$ is $k$, then the vectors $\alpha,T(\alpha), T^2(\alpha),…,T^{k-1}(\alpha)$ form a basis for $Z(\alpha ;T)$.

(iii) If $U$ is the linear operator on $Z(\alpha; T)$ induced by $T$, then the minimal polynomial for $U$ is $p_\alpha$.

My attempt: (i) & (ii) Suppose $p_{\alpha}$ is a monic generator of the ideal $S_T(\alpha ;\{0\})$, and $\deg (p_{\alpha})=k$. Let $B=\{\alpha ,T(\alpha),…,T^{k-1}(\alpha)\}$. We show $B$ is basis of $Z(\alpha; T$).

Suppose $c_0\cdot \alpha +c_1\cdot T(\alpha)+…+c_{k-1}\cdot T^{k-1}(\alpha)=0$, for some $c_i\in F$.

Then $$f=\sum_{i=0}^{k-1}c_i\cdot x^i\in S_T(\alpha; \{0\}).$$

But $\deg (f)=k-1\lt k=\deg (p_\alpha)$, so $c_i=0$, $\forall 0\leq i\leq k-1$. Thus $B$ is independent.

We claim $T^j(\alpha)\in \text{span}(B)$, $\forall j\geq k$.

Base case: $j=k$. Let $$p_\alpha =x^k+\sum_{i=0}^{k-1}a_i\cdot x^i.$$ Then $$[p_{\alpha}(T)](\alpha)= T^k(\alpha)+\sum_{i=0}^{k-1}a_i\cdot T^i(\alpha)=0.$$ So $T^k(\alpha)= \sum_{i=0}^{k-1}-a_i\cdot T^i(\alpha)$. Thus $T^k(\alpha) \in \text{span}(B)$.

Inductive step: Suppose $T^j(\alpha)\in \text{span}(B)$, for some $j\geq k$. Then $T^j(\alpha)=\sum_{i=0}^{k-1}b_i\cdot T^i(\alpha)$. So $$T^{j+1}=T(T^j)=T\left(\sum_{i=0}^{k-1}b_i\cdot T^i(\alpha)\right)= \sum_{i=0}^{k-1}b_i\cdot T^{i+1}(\alpha).$$ Since $T^k(\alpha)\in \text{span}(B)$, we have $T^{j+1}= \sum_{i=0}^{k-1}b_i\cdot T^{i+1}(\alpha)\in \text{span}(B)$. By principle of mathematical induction, $T^j(\alpha)\in \text{span}(B)$, $\forall j\geq k$.

Let $g(T)(\alpha)\in Z(\alpha;T)$, where $g=\sum_{i=0}^nd_i\cdot x^i\in F[x]$. Then $$g(T)(\alpha)=\sum_{i=0}^nd_i\cdot T^i(\alpha)\in \text{span}(\{\alpha,T(\alpha),…,T^n(\alpha)\}).$$ By the above claim, $\text{span}(\{\alpha,T(\alpha),…,T^n(\alpha)\})\subseteq \text{span}(B)$. So $g(T)(\alpha)\in \text{span}(B)$ and $Z(\alpha ;T)\subseteq \text{span}(B)$. Since $\alpha,T(\alpha),...,T^{k-1}(\alpha)\in Z(\alpha;T)$ and $Z(\alpha;T)$ is subspace of $V$, we have $\text{span}(B)\subseteq Z(\alpha;T)$. Thus $\text{span}(B)=Z(\alpha;T)$. Hence $B$ is basis of $Z(\alpha;T)$ and $\dim Z(\alpha ;T)=|B|=k$.

(iii) It’s easy to check, $[U]_B=\begin{bmatrix} & & & -a_0\\ 1& & &-a_1\\ &\ddots & & \vdots \\ & & 1& -a_{k-1}\\ \end{bmatrix}$. Characteristic polynomial of $U$ is $f:F\to F$ such that $f(x)=\det (xI_k-[U]_B)=x^k+\sum_{i=0}^{k-1}a_i\cdot x^i=p_{\alpha}(x)$, $\forall x\in F$. So $f=p_{\alpha}$. By Cayley-Hamilton theorem, $f(U)=p_{\alpha}(U)=0$. Let $g\in F[x]$ such that $g(U)=0$. Then $g(U)(\alpha)=g(T)(\alpha)=0$. So $g\in S_T(\alpha ;\{0\})$. Thus $\deg (p_{\alpha})\leq \deg (g)$. Hence $p_{\alpha}$ is minimal polynomial of $U$. Is my proof correct?

---

Hoffman’s proof: (i) & (ii) Let $g\in F[x]$. By theorem 4 section 4.4, $\exists q,r\in F[x]$ such that $g=p_{\alpha}q+r$, and either $r=0$ or $\deg (r)\lt \deg (p_\alpha)=k$. The polynomial $p_\alpha q$ is in the $T$-annihilator of $\alpha$, and so $g(T)(\alpha)=r(T)(\alpha)$. Since $r = 0$ or $\deg (r)\lt k$, the vector $r(T)(\alpha)$ is a linear combination of the vectors $\alpha, T(\alpha),…, T^{k-1}(\alpha)$, and since $g(T)(\alpha)$ is a typical vector in $Z(\alpha; T)$, this shows that these $k$ vectors span $Z(\alpha; T)$.

(iii) If $g\in F[x]$, then $p_{\alpha}(U)g(T)\alpha =p_{\alpha}(T)g(T)\alpha =g(T)p_{\alpha}(T)\alpha=g(T)0=0$. Thus the operator $p_\alpha (U)$ sends every vector in $Z(\alpha ; T)$ into $0$ and is the zero operator on $Z(\alpha ;T)$.

Answer 1

I find both your proof and Hoffman's needlessly complicated. Also the terminology is somewhat confusing me (I've difficulty reconciling "let $p_\alpha$ be the $T$-annihilator of $\alpha$" with "$p_\alpha q$ is in the $T$-annihilator of $\alpha$". Let me call $p_\alpha$ the $\alpha$-minimal polynomial of $T$, and any polynomial in the ideal $S_T(\alpha;\{0\})$ it generates a $\alpha$-annihilating polynomial of $T$.

This all becomes easy if you start like this. The infinite sequence of vectors $\alpha,T(\alpha),T^2(\alpha),\ldots$ cannot be linearly independent (in finite dimension) so there is some $k$ such that $\def\B{\mathcal{B}}\B=[\alpha,T(\alpha),T^2(\alpha),\ldots,T^{k-1}(\alpha)]$ is linearly independent, but adding $T^k(\alpha)$ makes it linearly dependent (we have $k>0$ since $\alpha\neq0$). Now the span $Z$ of $\B$ is $T$-stable (as $T$ maps all vectors of $\B$ into $Z$) and $\B$ is a basis of $V$ (nothing left to prove here). Clearly all powers $T^i(\alpha)$ remain inside the $T$-stable subspace $Z$, so $Z$ is the cyclic subspace $Z(\alpha;T)$.

The coordinates of$~T^k(\alpha)$ with respect to the basis$~\B$ directly determine the $\alpha$-minimal polynomial $p_\alpha$ of $T$, which is of degree$~k$. What remains to be shown is part of (iii): that $p_\alpha$ is an annihilating polynomial of the restriction$~U$ of$~T$ to$~Z$ (knowing that it is $\alpha$-minimal, it will then also be minimal on$~U$). But $\ker(p_\alpha[T])$ contains the vector $\alpha$ and is $T$-stable (since $p_\alpha[T]$ commutes with $T$, see the last line of Hoffman's proof), so it contains the cyclic subspace $Z$ generated by$~\alpha$; this proves the point.

Just to add a cross reference, see also this question.