# Algebra (Warwick) Help!Watch

#1
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 and there exists a vector v in V such that (Firstly if somebody could confirm that it should be no equal to that would make my life better).
Prove that the vectors, 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 meaning 0 is an eigenvector of T.

Claim v,T(v) etc are L.I.
If not then there exists real numbers such that:

Which implies that

and so:
divides our minimal polynomial which also divides the minimal polynomial .

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

Finally, since 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
0
9 years ago
#2
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.)
9 years ago
#3
(Original post by zrancis)
Claim v,T(v) etc are L.I.
If not then there exists real numbers such that:

Which implies that

and so:
divides our minimal polynomial which also divides the minimal polynomial .
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-1)[v] is not equal to 0 so the minimal polynomial which brings a contradiction.
If P(T) = 0, then must divide P. So since T^n = 0, , so it must be a power of x, x^k, say. Then size , 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:

, where and each a_k is non-zero.

Then multiply by to get a non-trivial relation with a smaller number of non-zero terms. Contradiction.
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