1. ## eigenvector(lie algebra)

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

$x=\begin{bmatrix} 0~1~0~...~0\\ 0~0~1~...~0\\ .~~~.~~~.~.\\ .~~~.~~~.~.\\ 0~0~0~...~1\\ 1~0~0~...~0\\ \end{bmatrix}$ $y=\begin{bmatrix} 0~~0~~...~~0~~~~~0\\ 0~~1~~...~~0~~~~~0\\ .~~.~~...~~.~~~~~.\\ .~~.~~...~~.~~~~~.\\ .~~.~~...~~.~~~~~.\\ 0~~0~~...~~p-2~0\\ 0~~0~~...~~0~p-1\\ \end{bmatrix}$

( we have $[x,y]=xy-yx=x$).

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

2. ## Re: eigenvector(lie algebra)

Originally Posted by vernal
Let p be prime and let F be a field of characteristic p. consider the following $p\times p$ matrices:

$x=\begin{bmatrix} 0~1~0~...~0\\ 0~0~1~...~0\\ .~~~.~~~.~.\\ .~~~.~~~.~.\\ 0~0~0~...~1\\ 1~0~0~...~0\\ \end{bmatrix}$ $y=\begin{bmatrix} 0~~0~~...~~0~~~~~0\\ 0~~1~~...~~0~~~~~0\\ .~~.~~...~~.~~~~~.\\ .~~.~~...~~.~~~~~.\\ .~~.~~...~~.~~~~~.\\ 0~~0~~...~~p-2~0\\ 0~~0~~...~~0~p-1\\ \end{bmatrix}$

( we have $[x,y]=xy-yx=x$).

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!).

3. ## Re: eigenvector(lie algebra)

i think
in question x wrong.
Should

$\begin{bmatrix} 0~~1~~0~~...~~0\\ 0~~0~~1~~...~~0\\ .~~.~~.~~...~~.\\ .~~.~~.~~...~~.\\ .~~.~~.~~...~~.\\ 0~~0~~...~~0~~1\\ {\color{Red} 0}~~0~~...~~0~~0\\ \end{bmatrix}$

is it true?

4. ## Re: eigenvector(lie algebra)

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

$\begin{bmatrix} 0~~1~~0~~...~~0\\ 0~~0~~1~~...~~0\\ .~~.~~.~~...~~.\\ .~~.~~.~~...~~.\\ .~~.~~.~~...~~.\\ 0~~0~~...~~0~~1\\ {\color{Red} 0}~~0~~...~~0~~0\\ \end{bmatrix}$

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?).

5. ## Re: eigenvector(lie algebra)

i check it and $[x,y]\neq x$,, so probably x=$\begin{bmatrix} 0~~1~~0~~...~~0\\ 0~~0~~1~~...~~0\\ .~~.~~.~~...~~.\\ .~~.~~.~~...~~.\\ .~~.~~.~~...~~.\\ 0~~0~~...~~0~~1\\ {\color{Red} 0}~~0~~...~~0~~0\\ \end{bmatrix}$,and $y=\begin{bmatrix} 0~~0~~...~~0~~~~~0\\ 0~~1~~...~~0~~~~~0\\ .~~.~~...~~.~~~~~.\\ .~~.~~...~~.~~~~~.\\ .~~.~~...~~.~~~~~.\\ 0~~0~~...~~p-2~0\\ 0~~0~~...~~0~p-1\\ \end{bmatrix}$. then $[x,y]= x$. i think Such conditions are...

with this condition ,

Originally Posted by NonCommAlg
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.))

6. ## Re: eigenvector(lie algebra)

Originally Posted by vernal
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.

7. ## Re: eigenvector(lie algebra)

Originally Posted by NonCommAlg
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. $1-p=1$. and Everything is ok. thank you for your help

8. ## Re: eigenvector(lie algebra)

Originally Posted by vernal
oh, sorry. yes . Here I was wrong
!اگر ایرانی نبودی، می گذاشتم توی اشتباهت بمونی

9. ## Re: eigenvector(lie algebra)

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

خیلییییییی ممنونم ازت