1. ## Linear Algebra help

Hello.

M,N $\displaystyle \in M_n(\mathbb{R})$ and $\displaystyle P \in \mathbb{R}[X]$ (P is a real polynomial)

We suppose that $\displaystyle P(exp(MN))$ is nilpotent (where $\displaystyle exp(MN)$=matrix exponential of MN).

Have we got $\displaystyle P(exp(NM))$ nilpotent too ?

Thanks for any help. I don't see how to start.

2. Originally Posted by NicaaN
Hello.

M,N $\displaystyle \in M_n(\mathbb{R})$ and $\displaystyle P \in \mathbb{R}[X]$ (P is a real polynomial)

We suppose that $\displaystyle P(exp(MN))$ is nilpotent (where $\displaystyle exp(MN)$=matrix exponential of MN).

Have we got $\displaystyle P(exp(NM))$ 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 $\displaystyle N$ is invertible. (a similar proof works for M) first a Lemma which is always true:

Lemma: $\displaystyle NP(e^{MN})=P(e^{NM})N.$

Proof: clearly $\displaystyle (MN)^j=M(NM)^{j-1}N, \ j \geq 1.$ thus for any positive integer $\displaystyle k$ we have:

$\displaystyle Ne^{kMN}=N+N\sum_{j=1}^{\infty}\frac{k^j(MN)^j}{j! }=N+\sum_{j=1}^{\infty}\frac{k^j(NM)^j}{j!}N=e^{kN M}N.$ so if $\displaystyle P(t)=\sum_{k=0}^nc_kt^k,$ then:

$\displaystyle NP(e^{MN})=\sum_{k=0}^n c_kNe^{kMN}=\sum_{k=0}^nc_ke^{kNM}N=P(e^{NM})N. \ \ \ \square$

now if N is invertible, then by the Lemma: $\displaystyle N P(e^{MN})N^{-1}=P(e^{NM}).$ so: $\displaystyle N(P(e^{MN}))^{\ell}N^{-1}=(P(e^{NM}))^{\ell},$ for any positive integer $\displaystyle \ell.$

it's obvious now that if $\displaystyle P(e^{MN})$ is nilpotent, then $\displaystyle P(e^{NM})$ is nilpotent too. Q.E.D.

3. Originally Posted by NonCommAlg
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.
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 $\displaystyle GL_n$$\displaystyle (\mathbb{R}) is dense (we can find a sequence of invertible matrices which has M as a limit point) and the contuinity of the product matrices and the exponential ? So the result remains true. 4. Originally Posted by NicaaN Thanks a lot for your help NonCommAlg. I'm not sure but if M and N are not invertible, can't we use the fact that the set of invertible matrices \displaystyle GL_n$$\displaystyle (\mathbb{R})$ 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 $\displaystyle \{N_k\}$ that has N as the limit and also $\displaystyle P(e^{MN_k})$ is nilpotent for

almost all k. even if we can find such a sequence, then we'll have another difficulty: if $\displaystyle (P(e^{MN_k}))^{r_k}=0,$ then as i showed in my

previous post, we'll get $\displaystyle (P(e^{N_kM}))^{r_k}=0.$ but then we'll need the sequence $\displaystyle \{r_k\}$ to be bounded in order to prove that $\displaystyle P(e^{NM})$

is nilpotent (by taking limit as $\displaystyle k \rightarrow \infty$).

5. 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.]