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Math Help - eigenvector(lie algebra)

  1. #1
    Member vernal's Avatar
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    eigenvector(lie algebra)

    Let p be prime and let F be a field of characteristic p. consider the following matrices:



    ( we have ).

    deduce that x and y span a 2-dimensional solvable subalgebra L of gl(p,F). show that x and y have no common eigenvector...

    thanks
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  2. #2
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    Re: eigenvector(lie algebra)

    Quote Originally Posted by vernal View Post
    Let p be prime and let F be a field of characteristic p. consider the following matrices:



    ( we have ).

    deduce that x and y span a 2-dimensional solvable subalgebra L of gl(p,F). show that x and y have no common eigenvector...

    thanks
    since [x,y] = x \in Fx + Fy, we have L=Fx+Fy and thus L is 2-dimensional because obviously x,y are linearly independent. this also implies that L'=[Fx+Fy, Fx+ Fy] = Fx and thus [L',L']=\{0\} and hence L is solvable.
    an indirect way to prove that x,y have no common eigenvector is to suppose that they do, say xv=\alpha v and yv = \beta v for some 0 \neq v \in F^p and \alpha, \beta \in F. then \alpha v = xv = [x,y]v = \alpha \beta v - \alpha \beta v = 0 and so \alpha = 0. that implies \det x = 0, which is false because \det x = 1 (expand x down the first column!).
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  3. #3
    Member vernal's Avatar
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    Re: eigenvector(lie algebra)

    i think
    in question x wrong.
    Should



    is it true?
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    Re: eigenvector(lie algebra)

    Quote Originally Posted by vernal View Post
    i think
    in question x wrong.
    Should



    is it true?
    i don't know. i didn't check the identity [x,y] = x. that's your job!
    but if that's the right x and if y is the one you gave us, that is a diagonal matrix with 0, 1, ... , p-1 on the main diagonal, then the second part of the problem would be false because then the transpose of [1,0, \cdots ,0] would be a common eigenvector of x,y (why?).
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  5. #5
    Member vernal's Avatar
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    Re: eigenvector(lie algebra)

    i check it and ,, so probably x=,and . then . i think Such conditions are...
    Please see the answer to this requirement.


    with this condition ,

    Quote Originally Posted by NonCommAlg View Post
    an indirect way to prove that x,y have no common eigenvector is to suppose that they do, say xv=\alpha v and yv = \beta v for some 0 \neq v \in F^p and \alpha, \beta \in F. then \alpha v = xv = [x,y]v = \alpha \beta v - \alpha \beta v = 0 and so \alpha = 0. that implies \det x = 0, which is false because \det x = 1 (expand x down the first column!).
    is false. is it true? So what we do?

    ((also this question want to tell ,in lie's theorem we need our field to have charactristice zero...and The question is, with conditions, lie's theorem fail.))
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    Re: eigenvector(lie algebra)

    Quote Originally Posted by vernal View Post
    i check it and [x,y] \neq x.
    you made a mistake in your calculation! i just checked and i got [x,y] = x. so everything is ok. you probably didn't consider the fact that p - 1 = -1 in F because the characteristic of F is p. so you should check your calculation again!
    in fact, the second x that you gave us satisfies [x,y]=x too but that x is useless because it has a common eigenvector with y, as i said before.
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    Member vernal's Avatar
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    Re: eigenvector(lie algebra)

    Quote Originally Posted by NonCommAlg View Post
    you probably didn't consider the fact that p - 1 = -1 in F because the characteristic of F is p. so
    i
    oh, sorry. yes . Here I was wrong. . and Everything is ok. thank you for your help
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    Re: eigenvector(lie algebra)

    Quote Originally Posted by vernal View Post
    oh, sorry. yes . Here I was wrong
    !اگر ایرانی نبودی، می گذاشتم توی اشتباهت بمونی
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  9. #9
    Member vernal's Avatar
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    Re: eigenvector(lie algebra)

    Quote Originally Posted by NonCommAlg View Post
    !اگر ایرانی نبودی، می گذاشتم توی اشتباهت بمونی

    خیلییییییی ممنونم ازت
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