# Existence of a vector $v$ in $V$ such that the $T$-annihilator of $v$ is the minimal polynomial for $T$.

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?

• What is the definition of $T$-annihilator''?
– daw
Nov 23, 2018 at 13:04
• 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. Nov 23, 2018 at 13:51

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 $$g=p$$.