Results 1 to 10 of 10

Math Help - Linear Independence, Nilpotent Matrices

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    168

    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 A,A^2,...,A^(k-1) are all different from zero, but A^k=0

    Prove: A^n=0

    Now, there's a clue :
    let k be the smallest number for which A^k=0. There is a 'v' for which A^{k-1}*v isn't equal to zero.
    prove that {v, Av, A^2v, . . . . , A^{k-1}*v} 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 A^k=0, A^k=/0 (is not zero)
    therefore, let there be v=/0, for which A^{k-1}*v=/0, and for any other number that is lower than k. (since A is a nilpotent matrix)


    I have to prove that {v, Av, A^2v, . . . , A^{k-1}v} 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 !!!
    Last edited by adam63; August 14th 2009 at 09:31 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    ynj
    ynj is offline
    Senior Member
    Joined
    Jul 2009
    Posts
    254
    I am not quite clear what you mean..A^n=A^k*A^(n-k)=0??does it not suffice?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2009
    Posts
    168
    I'm sorry, I don't understand what you mean.

    Where should I use :
    A^n=A^k*A^{n-k}=0

    ?

    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    ynj
    ynj is offline
    Senior Member
    Joined
    Jul 2009
    Posts
    254
    Quote Originally Posted by adam63 View Post
    I'm sorry, I don't understand what you mean.

    Where should I use :
    A^n=A^k*A^{n-k}=0

    ?

    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
    oh,you just let a1=0 and continue do the same thing for a2,a3...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2009
    Posts
    168
    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 !!!!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2011
    Posts
    11

    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    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)).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Nov 2011
    Posts
    11

    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    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?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Nov 2011
    Posts
    11

    Re: Linear Independence, Nilpotent Matrices

    Any other starting tips.
    Last edited by CalBear12; November 30th 2011 at 03:48 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. linear operator nilpotent
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 20th 2011, 08:21 PM
  2. JNF of nilpotent matrices
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 1st 2010, 03:49 AM
  3. Matrices- Nilpotent
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 22nd 2009, 06:42 PM
  4. checking matrices for linear independence
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 29th 2009, 08:02 PM
  5. nilpotent matrices
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 30th 2008, 03:16 AM

Search Tags


/mathhelpforum @mathhelpforum