IMC 2000 · Day 2 · P12

First we prove that for any polynomial $q$ and $m \times m$ matrices $A$ and $B$, the characteristic polinomials of $q(e^{AB})$ and $q(e^{BA})$ are the same. It is easy to check that for any matrix $X$, $q(e^X) = \sum\limits_{n=0}^{\infty} c_n X^n$ with some real numbers $c_n$ which depend on $q$. Let $$C = \sum_{n=1}^{\infty} c_n \cdot (BA)^{n-1} B = \sum_{n=1}^{\infty} c_n \cdot B (AB)^{n-1}.$$ Then $q(e^{AB}) = c_0 I + AC$ and $q(e^{BA}) = c_0 I + CA$. It is well-known that the characteristic polynomials of $AC$ and $CA$ are the same; denote this polynomial by $f(x)$. Then the characteristic polynomials of matrices $q(e^{AB})$ and $q(e^{BA})$ are both $f(x - c_0)$.

Now assume that the matrix $p(e^{AB})$ is nilpotent, i.e. $\bigl( p(e^{AB}) \bigr)^k = 0$ for some positive integer $k$. Chose $q = p^k$. The characteristic polynomial of the matrix $q(e^{AB}) = 0$ is $x^m$, so the same holds for the matrix $q(e^{BA})$. By the theorem of Cayley and Hamilton, this implies that $\bigl( q(e^{BA}) \bigr)^m = \bigl( p(e^{BA}) \bigr)^{km} = 0$. Thus the matrix $q(e^{BA})$ is nilpotent, too.