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Definition: $T$-annihilator of a vector $\alpha$ (denoted as $p_\alpha$) is the unique monic polynomial which generates the ideal such that $g(T)\alpha = 0$ for all $g$ in this ideal.

I'm trying to prove the below statement without invoking the Cyclic Decomposition Theorem.

Let $T$ be a linear operator on a finite-dimensional vector space $V$. Then there exists a vector $v$ in $V$ such that the $T$-annihilator of $v$ is the minimal polynomial for $T$.

Attempt: Assume that there is no such $v$. Then every vector has a $T$-annihilator of degree less than that of the minimal polynomial. Define a monic polynomial $h$ which is the sum of $T$-annihilators of given basis elements. Then $h(T)v=0$ for all $v\in V$. But this contradicts the definition of minimal polynomial since the degree of $h\lt$ the degree of the minimal polynomial.

Can someone verify my argument?

cqfd
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  • What is the definition of ``$T$-annihilator''? – daw Nov 23 '18 at 13:04
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    No that proof doesn't work. There is no reason that $h(T)v = 0$. If $h$ is the sum $g_1+g_2+\cdots+g_n$ then $h(T)v$ equals $g_1(T)v+g_2(T)v+\cdots + g_n(T)v$ with possibly only ONE term missing. – Matthew C Nov 23 '18 at 13:51

1 Answers1

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Let's first show the result when the minimal polynomial has the form $p^n$, with $p$ irreducible. We know that $p(T)^n=0$ , $p(T)^{n-1}\neq0$ so there exist a vector $\alpha \in V$ such that $p(T)^{n-1}\alpha\neq0$, $p(T)^n\alpha=0$. Thus the T-annihilator $g$ of $\alpha$ divides $p^n$ and since $p(T)^{r}\alpha\neq0$ for $r\leq n-1$, $g=p^n$.

Now consider the general case and let $p=p_{1}^{r_1}...p_{k}^{r_k}$ be the minimal polynomial for $T$ where the $p_i$ are distinct irreductible monic polynomials. Then applying the primary decomposition for $T$ we obtain $V=W_1 \oplus\cdots\oplus W_k$ , and denoting by $T_i$ the restriction of $T$ to $W_i$ the minimal polynomial for $T_i$ is $p_{i}^{r_i}$. Now we can use the result above : there exist $\alpha_i \in W_i$ such that the T-annihilator $g_i$ of $\alpha_i$ is $p_{i}^{r_i}$

Let $\alpha = \sum_{i=1}^k\alpha_i $. We know that the T-annihilator $g$ of $\alpha$ divides $p$. Let $f$ be any polynomial such that $f(T)\alpha=0$. Then $\sum_{i=1}^k f(T)\alpha_i =0 $ which implies $f(T)\alpha_i =0$ for each $i$ ($\alpha_i \in W_i$ and the $W_i$ are invariant under $T$ so $f(T)\alpha_i \in W_i$, and finally the $W_i$ are independant). Thus $p_{i}^{r_i}$ divides $f$ for each $i$ so $p$ divides $f$. Now this shows that $p$ divides $g$ which gives us $g=p$.

Mathieu Eury
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