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 we have and thus is 2-dimensional because obviously are linearly independent. this also implies that and thus and hence is solvable.
an indirect way to prove that have no common eigenvector is to suppose that they do, say and for some and then and so that implies which is false because (expand down the first column!).
i don't know. i didn't check the identity that's your job!
but if that's the right and if is the one you gave us, that is a diagonal matrix with on the main diagonal, then the second part of the problem would be false because then the transpose of would be a common eigenvector of (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 so everything is ok. you probably didn't consider the fact that in because the characteristic of is so you should check your calculation again!
in fact, the second that you gave us satisfies too but that is useless because it has a common eigenvector with as i said before.