Hello.
M,N and (P is a real polynomial)
We suppose that is nilpotent (where =matrix exponential of MN).
Have we got 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 is invertible. (a similar proof works for M) first a Lemma which is always true:
Lemma:
Proof: clearly thus for any positive integer we have:
so if then:
now if N is invertible, then by the Lemma: so: for any positive integer
it's obvious now that if is nilpotent, then is nilpotent too. Q.E.D.
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 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 that has N as the limit and also is nilpotent for
almost all k. even if we can find such a sequence, then we'll have another difficulty: if then as i showed in my
previous post, we'll get but then we'll need the sequence to be bounded in order to prove that
is nilpotent (by taking limit as ).
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.]