Prove that every nonzero vector in V is a maximal vector for T.

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Discussion Overview

The discussion revolves around proving that every nonzero vector in a finite-dimensional vector space \( V \) is a maximal vector for a linear operator \( T: V \rightarrow V \). Participants explore the implications of the minimal polynomial being irreducible and its relevance to the proof.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question how the assumption of the minimal polynomial being irreducible aids in proving that every nonzero vector is a maximal vector for \( T \).
  • One participant suggests that if the field \( F \) is algebraically closed, \( T \) must have \( n \) eigenvalues, and if at least two are distinct, the minimal polynomial would be reducible, implying all eigenvalues must be equal for the statement to hold.
  • Another participant notes that if \( F \) is the field of real numbers, the irreducibility of certain polynomials does not necessarily relate to primality, indicating a potential misunderstanding in the terminology used.
  • There is a suggestion to clarify the problem statement, as the title may lead to confusion regarding the specifics of the proof required.

Areas of Agreement / Disagreement

Participants express uncertainty about the implications of the minimal polynomial's properties and whether the proof holds for all fields. There is no consensus on how to approach the proof or the relevance of certain assumptions.

Contextual Notes

Participants highlight the need for clarity in the problem statement and the definitions used, particularly regarding irreducibility and primality in different fields.

toni07
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Let $T: V \rightarrow V$ be a linear operator on a fi nite-dimensional vector space $V$ over $F$. Assume that $_{\mu T}(x) \in F[x]$ is an irreducible polynomial.

I don't understand how assuming that the minimal polynomial is prime helps to prove the question. Please help.
 
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Re: Prove that every nonzero vector in $V$ is a maximal vector for $T$

crypt50 said:
Let $T: V \rightarrow V$ be a linear operator on a fi nite-dimensional vector space $V$ over $F$. Assume that $_{\mu T}(x) \in F[x]$ is an irreducible polynomial.

I don't understand how assuming that the minimal polynomial is prime helps to prove the question. Please help.

Prove which question?

Note that if the field is the field of the real numbers, then the polynomial $x^2+\pi$ is irreducible, which has little to do with primes.
 
Re: Prove that every nonzero vector in $V$ is a maximal vector for $T$

I like Serena said:
Prove which question?

Note that if the field is the field of the real numbers, then the polynomial $x^2+\pi$ is irreducible, which has little to do with primes.

The question in the title: Prove that every nonzero vector in $V$ is a maximal vector for $T$.
 
Re: Prove that every nonzero vector in $V$ is a maximal vector for $T$

crypt50 said:
The question in the title: Prove that every nonzero vector in $V$ is a maximal vector for $T$.

Ah. I missed that.
And I presume that with prime you mean irreducible.

Well. Let's see.
Suppose $V$ is n-dimensional.

Then, if $F$ is algebraically closed (such as the complex numbers), $T$ has n eigenvalues.
If at least 2 eigenvalues are distinct, then the minimal polynomial is reducible.
Therefore all eigenvalues have to be equal.
That means that each nonzero vector has to be a maximal vector for $T$.

If $F$ is the field of the real numbers, we can extend it to the complex numbers, and the same argument holds.

So we're left with all other fields that do not obey the same principles.
Are you supposed to prove it for any field?
 
Re: Prove that every nonzero vector in $V$ is a maximal vector for $T$

crypt50 said:
The question in the title: Prove that every nonzero vector in $V$ is a maximal vector for $T$.

The thread title should give a brief description of the problem, while the problem itself should be fully given within the body of the first post. As you can see, putting the question in the title leads to confusion. :D
 

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