Hey so I have this question:



What I have so far is..

Since we know N^r = 0, then by the Cayley-Hamilton Theorem (I'm not sure if this is a correct assumption), then we know x^r is a factor of the characteristic polynomial of N. We know the minimal polynomial divides the characteristic polynomial so the minimal polynomial is x.

I'm kind of stuck here and not sure how to move forward. I know the answer should be x^n using a proof with eigenvalues but I do not think I can use it as it is not explicitly stated we are working over the complex numbers.
Can anyone help?

Thanks

Assume r is the degree of N (ie : the minimal integer k such that N^k = 0).
You know that the mimimal polynomial of N is X^r, and that the only eigenvalue is zero.
You are working on a field, that is not explicitly algebraically closed (for example the set of real numbers).

But any field has an algebraic closure (set of complex numbers for real numbers), and your very same matrix N is still nilpotent in any algebraic closure : N^r is still zero !
You can conclude that zero is the only eigenvalue in the algebraic closure, and, since the characteristic polynomial is splitable, and of degree n, it must be X^n in that algebraic closure.

Question : is there any difference between the characteristic polynomial in your initial field and in its closure ?
det ( X.Id - N ) doesn't change when you consider the coefficients as elements of a given field, or as elements of its closure!

Is the characteristic polynomial able to split into linear terms regardless of the field chosen?

It splits into linear terms in the closure. And since any of its root is an eigenvalue, and since only zero is an eigenvalue, then it must be X^n.

Once you know that in the closure, the characteristic polynomial is X^n, it's safe to say it is the same in the initial field.

And since it is X^n in the initial field, then yes, it splits into linear terms also here. But you couldn't say that in the first place.

Makes sense, I guess I can use eigenvalues then
thanks for clearing that up