Algebra (Warwick) Help! Watch

zrancis
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Hi, I'm trying to prove the following question:

Let T be a linear function that maps V to itself, where Dim(V)=n and suppose T^n=0 and there exists a vector v in V such that T^{n-1}(v)=0_V (Firstly if somebody could confirm that it should be no equal to that would make my life better).
Prove that the vectors, v,T(v),T^2(v),...,T^{n-1}(v) are linearly independent and their nullity is 1.

Okay, so I'm trying to adapt a proof I already have without fully understanding it. Here goes:

The rank of T must be less than n since T(T^{n-1}(v))=0 meaning 0 is an eigenvector of T.

Claim v,T(v) etc are L.I.
If not then there exists real numbers a_i such that:
a_0v+a_1T(v)+...a_{n-1}T^{n-1}(v)=0
Which implies that
(a_0v+a_1T(v)+...a_{n-1}T^{n-1}(v))v=0
and so:
a_0+a_1x+...+a_{n-1}x^{n-1} divides our minimal polynomial \mu_{T(v)} which also divides the minimal polynomial \mu_{T}.

But since T^n(v)=0, T^(n-1)[v] is not equal to 0 so the minimal polynomial \mu_{T}=x^n which brings a contradiction.

Finally, since T(v),...,T^{n-1}(v) are L.I. this implies the that dim(T) is greater than or equal to n-1 and hence the rank of T is n-1 which also shows the nullity of T is 1.

Now I follow this right upto the contradiction, why does that imply the minimal polynomial is equal to x^n?

Sorry about some of the notation, if you need me to clarify I will.

Thanks
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Kolya
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I have posted this question before. Unfortunately, the search function is f***ed so I can't retrieve that thread for you. (Although someone suggested a different method, so it wouldn't be of much use to you anyway.)
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DFranklin
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(Original post by zrancis)
Claim v,T(v) etc are L.I.
If not then there exists real numbers a_i such that:
a_0v+a_1T(v)+...a_{n-1}T^{n-1}(v)=0
Which implies that
(a_0v+a_1T(v)+...a_{n-1}T^{n-1}(v))v=0
and so:
a_0+a_1x+...+a_{n-1}x^{n-1} divides our minimal polynomial \mu_{T(v)} which also divides the minimal polynomial \mu_{T}.
I'm not seeing why this has to be true. Consider the case S = I (I know that won't satisfy the original conditions). Then we have S^2+S-2 = 0. But S has minimal polynomial (x-1), and it's not true that x^2+x-2 divides (x-1).

But since T^n(v)=0, T^(n-1)[v] is not equal to 0 so the minimal polynomial \mu_{T}=x^n which brings a contradiction.
If P(T) = 0, then \mu_T must divide P. So since T^n = 0, \mu_T | x^n, so it must be a power of x, x^k, say. Then size T^{n-1} \neq 0, k must be at least n, so done.

Personally, what I'd do is:

If the vectors are lin. dep., then pick a relation with a minimal number of non-zero terms:

a_0 T^{n_0}(v) + a_1T^{n_1}(v) + ... + a_k T^{n_k}(v), where 0 \leq n_0 \leq n_1 \leq ... \leq n_k < n and each a_k is non-zero.

Then multiply by T^{n-n_k} to get a non-trivial relation with a smaller number of non-zero terms. Contradiction.
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