I am not quite clear what you mean..A^n=A^k*A^(n-k)=0??does it not suffice?
Hello!
The course is Linear Algebra, the subject is vector-spaces, basis, dimensions.
The question is:
A (from the field) F^nxn is a nilpotent matrix.
(Nilpotent matrix - Wikipedia, the free encyclopedia)
A square matrix is nilpotent from the order k if are all different from zero, but
Prove: A^n=0
Now, there's a clue :
let k be the smallest number for which . There is a 'v' for which isn't equal to zero.
prove that is a linear independent.
Okay, now I didn't want to look for any other way to solve this (meaning without using the clue), because I also wanted to use what I've learned and practiced.
So, what I did was:
Let , =/0 (is not zero)
therefore, let there be v=/0, for which =/0, and for any other number that is lower than k. (since A is a nilpotent matrix)
I have to prove that is linear independent. So, let {α1, α2, . . . , αk} be parameters. Let's see when (for which alphas):
α1v+α2Av+. . . + α(k)A^(k-1)v=0
(in order to prove that it's linear independent I need to show that the ONLY case that it's equal to zero is when α1=α2=α3=...=αk=0)
So, I multiply it by the matrix A:α1Av+α2vA^2+ . . . + α(k-1)A^(k-1)v+α(k)A^k=0
^according to what we assumed, A^k=0, so it's equal to:
α1Av+α2A^2*v+ . . . + α(k-1)A^(k-1)v=0
we'll multiply again by A:
α1A^2*v+α2A^3*v+ . . . + α(k-2)A^(k-1)v=0
.
.
.
and so on. In total, we multiply it by A (k-1) times, and in the end we get:
α1A^(k-1)*v=0
In such case, α1 can be equal to zero, while the other alphas can be everything. which means this is not necessarily linear independent! Argh...Can anyone please help me with this ?
Thank you very much !!!
I'm sorry, I don't understand what you mean.
Where should I use :
?
I have no 'n', since I only use alpha(i), v, k, and A.
I need a mathematical proof for this, and more than that - I want one that is based on the clue (because that's connected to the subject I'm practicing).
Thank you
Thank you !!!
I got it!
I was one step away from this...
I was supposed to do this: assume that A^n is not 0, and therefore there is k>n for which A^k=0. (so A^k-1 isn't 0).
Then, I need to prove what I mentioned here (that these vectors are independent), and then - since I found K independent vectors in the space F^(nxn), then there must be:
#B([<] or [=] ) dim(F^nxn).
therefore, K([<] or [=])n, which is false, since we assumed in the beginning that there should be a k>n.
I like this one !!!!
there are different ways to do this. one way is to note that the minimal polynomial m(A) has degree ≤ n, since the characteristic polynomial p(x) = det(xI - A) has degree ≤ n,
and p(A) = 0, so m(x) divides p(x). but q(x) = x^k is a polynomial for which q(A) = 0, so m(x) divides q(x) as well. hence m(x) = x^t, for some t ≤ k.
but k is the smallest power of A equal to 0, so t = k ≤ n.
i'm not entirely sure i agree with the above proofs, since they only show that k ≤ n^2 = dim(F^(nxn)).