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
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 $\displaystyle [x,y] = x \in Fx + Fy,$ we have $\displaystyle L=Fx+Fy$ and thus $\displaystyle L$ is 2-dimensional because obviously $\displaystyle x,y$ are linearly independent. this also implies that $\displaystyle L'=[Fx+Fy, Fx+ Fy] = Fx$ and thus $\displaystyle [L',L']=\{0\}$ and hence $\displaystyle L$ is solvable.
an indirect way to prove that $\displaystyle x,y$ have no common eigenvector is to suppose that they do, say $\displaystyle xv=\alpha v$ and $\displaystyle yv = \beta v$ for some $\displaystyle 0 \neq v \in F^p$ and $\displaystyle \alpha, \beta \in F.$ then $\displaystyle \alpha v = xv = [x,y]v = \alpha \beta v - \alpha \beta v = 0$ and so $\displaystyle \alpha = 0.$ that implies $\displaystyle \det x = 0,$ which is false because $\displaystyle \det x = 1$ (expand $\displaystyle x$ down the first column!).
i don't know. i didn't check the identity $\displaystyle [x,y] = x.$ that's your job!
but if that's the right $\displaystyle x$ and if $\displaystyle y$ is the one you gave us, that is a diagonal matrix with $\displaystyle 0, 1, ... , p-1$ on the main diagonal, then the second part of the problem would be false because then the transpose of $\displaystyle [1,0, \cdots ,0]$ would be a common eigenvector of $\displaystyle x,y$ (why?).
i check it and ,, so probably x=,and . then . i think Such conditions are...
Please see the answer to this requirement.
with this condition ,
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.))
you made a mistake in your calculation! i just checked and i got $\displaystyle [x,y] = x.$ so everything is ok. you probably didn't consider the fact that $\displaystyle p - 1 = -1$ in $\displaystyle F$ because the characteristic of $\displaystyle F$ is $\displaystyle p.$ so you should check your calculation again!
in fact, the second $\displaystyle x$ that you gave us satisfies $\displaystyle [x,y]=x$ too but that $\displaystyle x$ is useless because it has a common eigenvector with $\displaystyle y,$ as i said before.