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

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In summary, the conversation discusses the concept of irreducible polynomials and their relationship with proving the question of whether every nonzero vector in a finite-dimensional vector space is a maximal vector for a linear operator. The assumption of the minimal polynomial being prime helps to prove this question by showing that all eigenvalues must be equal, leading to the conclusion that each nonzero vector is a maximal vector for the operator. This proof holds for fields such as the complex numbers, but may not hold for all other fields.
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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.
 
  • #3
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 $T$.
 
  • #4
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?
 
  • #5
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
 

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

What does it mean for a vector to be "maximal" for a transformation?

For a vector to be maximal for a transformation, it means that it cannot be extended or enlarged any further without violating the properties of the transformation. In other words, the vector is already as large as it can be while still satisfying the transformation.

How do you prove that every nonzero vector in V is a maximal vector for T?

This can be proven by showing that for any nonzero vector in V, there is no other vector in V that can be multiplied by a scalar and still result in a larger vector that satisfies the transformation. Alternatively, it can also be proven by showing that the dimension of the vector space V is equal to the dimension of the image of the transformation T, meaning that there is no room for any larger vectors in the image of T.

Can a zero vector be a maximal vector for T?

No, a zero vector cannot be a maximal vector for T because it is defined as a vector that has magnitude and direction of 0. This means that it cannot be extended any further without violating the properties of the transformation.

Does this statement hold true for all types of transformations?

Yes, this statement holds true for all types of transformations. As long as the transformation satisfies the properties of a linear transformation, the statement will hold true.

Why is it important to prove that every nonzero vector in V is a maximal vector for T?

Proving this statement is important because it helps us understand the behavior of the transformation T on the vector space V. It also allows us to make conclusions about the structure and properties of the vector space and the transformation. Additionally, it can be useful in solving problems and making predictions in various fields such as physics, engineering, and computer science.

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