Results 1 to 5 of 5

Math Help - Linear Algebra help

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    2

    Linear Algebra help

    Hello.

    M,N \in M_n(\mathbb{R}) and P \in \mathbb{R}[X] (P is a real polynomial)

    We suppose that P(exp(MN)) is nilpotent (where exp(MN)=matrix exponential of MN).

    Have we got P(exp(NM)) nilpotent too ?

    Thanks for any help. I don't see how to start.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by NicaaN View Post
    Hello.

    M,N \in M_n(\mathbb{R}) and P \in \mathbb{R}[X] (P is a real polynomial)

    We suppose that P(exp(MN)) is nilpotent (where exp(MN)=matrix exponential of MN).

    Have we got P(exp(NM)) nilpotent too ?

    Thanks for any help. I don't see how to start.
    this problem has been bugging me for hours! i don't know the answer for the general case, but i can prove that the answer is "yes" if

    at least one of M or N is invertible. i'll assume that N is invertible. (a similar proof works for M) first a Lemma which is always true:


    Lemma: NP(e^{MN})=P(e^{NM})N.


    Proof: clearly (MN)^j=M(NM)^{j-1}N, \ j \geq 1. thus for any positive integer k we have:

    Ne^{kMN}=N+N\sum_{j=1}^{\infty}\frac{k^j(MN)^j}{j!  }=N+\sum_{j=1}^{\infty}\frac{k^j(NM)^j}{j!}N=e^{kN  M}N. so if P(t)=\sum_{k=0}^nc_kt^k, then:

    NP(e^{MN})=\sum_{k=0}^n c_kNe^{kMN}=\sum_{k=0}^nc_ke^{kNM}N=P(e^{NM})N. \ \ \ \square


    now if N is invertible, then by the Lemma: N P(e^{MN})N^{-1}=P(e^{NM}). so: N(P(e^{MN}))^{\ell}N^{-1}=(P(e^{NM}))^{\ell}, for any positive integer \ell.

    it's obvious now that if P(e^{MN}) is nilpotent, then P(e^{NM}) is nilpotent too. Q.E.D.
    Last edited by NonCommAlg; July 20th 2008 at 04:30 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2008
    Posts
    2
    Quote Originally Posted by NonCommAlg View Post
    this problem has been bugging me for hours! i don't know the answer for the general case, but i can prove that the answer is "yes" if

    at least one of M or N is invertible.
    Thanks a lot for your help NonCommAlg.
    I'm not sure but if M and N are not invertible, can't we use the fact that the set of invertible matrices GL_n (\mathbb{R}) is dense (we can find a sequence of invertible matrices which has M as a limit point) and the contuinity of the product matrices and the exponential ?
    So the result remains true.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by NicaaN View Post
    Thanks a lot for your help NonCommAlg.
    I'm not sure but if M and N are not invertible, can't we use the fact that the set of invertible matrices GL_n (\mathbb{R}) is dense (we can find a sequence of invertible matrices which has M as a limit point) and the contuinity of the product matrices and the exponential ?
    So the result remains true.
    it's not that simple: first we need a sequence of invertible matrices \{N_k\} that has N as the limit and also P(e^{MN_k}) is nilpotent for

    almost all k. even if we can find such a sequence, then we'll have another difficulty: if (P(e^{MN_k}))^{r_k}=0, then as i showed in my

    previous post, we'll get (P(e^{N_kM}))^{r_k}=0. but then we'll need the sequence \{r_k\} to be bounded in order to prove that P(e^{NM})

    is nilpotent (by taking limit as k \rightarrow \infty).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    I'm not an algebraist, but this argument seems to me to work (If not, NonCommAlg will soon put me right!)

    If P(exp(MN)) is nilpotent then (P(exp(MN)))^k = 0 for some k. Therefore the polynomial (P(X))^k is a multiple of the minimal polynomial of exp(MN). So the roots of P(X) will include all the eigenvalues of exp(MN). But exp(MN) and exp(NM) have the same eigenvalues. So for r large enough, (P(X))^r will contain all these roots with at least the multiplicity that they have in the minimal polynomial of exp(NM), and hence (P(exp(NM)))^r = 0.

    [I hope that's all legible. The forum's server is still having troubles, and the LaTeX facility isn't working.]
    Last edited by Opalg; July 23rd 2008 at 01:44 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 1st 2011, 11:00 PM
  2. Basic Linear Algebra - Linear Transformations Help
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: December 7th 2010, 04:59 PM
  3. Replies: 2
    Last Post: December 6th 2010, 04:03 PM
  4. Replies: 7
    Last Post: August 30th 2009, 11:03 AM
  5. Replies: 3
    Last Post: June 2nd 2007, 11:08 AM

Search Tags


/mathhelpforum @mathhelpforum