Prove that every non-zero vector in V is a maximal vector

[Ninty64]

Homework Statement

Let V be a finite dimensional vector space and T is an operator on V. Assume $μ_{T}(x)$ is an irreducible polynomial. Prove that every non-zero vector in V is a maximal vector.

Homework Equations

$μ_{T}(x)$ is the minimal polynomial on V with respect to T.

The Attempt at a Solution

I am not sure how to go about solving this problem. I know that if T is cyclic then $\exists \vec{v} \in V$ such that $<T,\vec{v}>=V$, then for any $\vec{w}\in V$,
$\exists f(x) \in F[x]$ such that $\vec{w} = f(T)(\vec{v})$
and it follows that $μ_{T,\vec{w}}(x) = \frac{μ_{T}(x)}{gcd(μ_{T}(x),f(x))}$
However, since $μ_{T}(x)$ is an irreducible polynomial, then $gcd(μ_{T}(x),f(x)) = μ_{T}(x)$ or $gcd(μ_{T}(x),f(x)) = 1$
If $gcd(μ_{T}(x),f(x)) = μ_{T}(x)$, then $μ_{T,\vec{w}}(x) = 1$, which implies that $\vec{w} = I(\vec{w}) = \vec{0}$
If $gcd(μ_{T}(x),f(x)) = 1$, then $μ_{T,\vec{w}}(x) = μ_{T}(x)$, which means that $\vec{w}$ is a maximal vector. Thus every non-zero vector in V is a maximal vector.

However, I'm not sure how to prove that T is cyclic.

I also attempted to find the minimal polynomials with respect to the basis vectors for V.
Since $μ_{T}(x) = lcm(μ_{T,\vec{v_1}}(x),μ_{T\vec{v_2}}(x),...,μ_{T,\vec{v_n}}(x))$
$\Rightarrow μ_{T}(x)=μ_{T,\vec{v_1}}(x)=μ_{T,\vec{v_2}}(x)=...=μ_{T,\vec{v_n}}(x)$ since the minimal polynomial is irreducible. But then I didn't know where to go from there either.

 

[mighty2000]

I'm not sure how to use the fact that μ_{T}(x) is an irreducible polynomial to prove that every non-zero vector in V is a maximal vector. I would appreciate any guidance or suggestions on how to approach this problem.

Dear fellow scientist,

Thank you for your post. The key to solving this problem lies in understanding the properties of irreducible polynomials and their relationship to linear operators. First, let's define what it means for a vector to be maximal in a vector space.

A vector \vec{v} is maximal in a vector space V if and only if there exists no non-zero vector \vec{w} in V such that \vec{v} + \vec{w} \in V. In other words, \vec{v} is not a proper subset of any other vector in V.

Now, let's consider the minimal polynomial μ_T(x) of an operator T on V. As you correctly stated, this polynomial is irreducible. This means that it cannot be factored into a product of two non-constant polynomials. In other words, μ_T(x) cannot be written as μ_T(x) = f(x)g(x), where f(x) and g(x) are non-constant polynomials.

Now, let's suppose that there exists a non-zero vector \vec{v} in V that is not maximal. This means that there exists a non-zero vector \vec{w} in V such that \vec{v} + \vec{w} \in V. We can then define a polynomial f(x) as follows:

f(x) = μ_T(x) - μ_T(\vec{v}+ \vec{w})

Note that f(x) is a polynomial of degree less than or equal to the degree of μ_T(x). Also note that f(x) is not the zero polynomial, since μ_T(x) is irreducible and μ_T(\vec{v}+ \vec{w}) is a non-zero constant. Therefore, f(x) is a non-zero polynomial of degree less than or equal to the degree of μ_T(x).

Now, we can use the fact that μ_T(x) is the minimal polynomial on V with respect to T to show that f(x) is not a multiple of μ_T(x). If f(x) were a multiple of μ_T(x), then we could write f(x) = μ_T(x)g(x)