Linear Independence, Nilpotent Matrices

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 (Itwasntme)?

Thank you very much :)!!!

Re: Linear Independence, Nilpotent Matrices

Bumping this. I have basically the same problem. Can anyone clarify or further elaborate on how to start this problem? I'm not really following adam's explanation or work.

Re: Linear Independence, Nilpotent Matrices

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)).

Re: Linear Independence, Nilpotent Matrices

I appreciate the response. I'm still not really following, any chance of an explanation in english where I can try and understand what is trying to be proven as a way to get started.

Re: Linear Independence, Nilpotent Matrices

there are different ways to approach this, depending on what you have to work with. i don't know what you do or do not know. where is the first place you get lost?

Re: Linear Independence, Nilpotent Matrices