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

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- Thread starter
- #1

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

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- #2

- Mar 5, 2012

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Prove which question?

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

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.

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

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- Mar 5, 2012

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

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

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- #5

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