An exponential map of a matrix computation

Question

Suppose the $n\times n$ matrices $A$ and $M$ satisfy $AM+MA^{T}=0.$ Show by direct computation that the product $\mathrm{exp}(At)~M~\mathrm{exp}(A^{T}t)=M$ for all $t\in \mathbb{R}.$

Note: By definition of exponential map to have $$LHS=(I+tA+\frac{t^2A^2}{2!}+\frac{t^3A^3}{3!}+\cdots)M(I+tA^{T}+\frac{t^2(A^{T})^2}{2!}+\frac{t^3(A^{T})^3}{3!}+\cdots)=(M+tAM+\frac{t^2A^2 M}{2}+\frac{t^3A^3M}{3!}+\cdots)(I+tA^{T}+\frac{t^2(A^{T})^2}{2!}+\frac{t^3(A^{T})^3}{3!}+\cdots)=M+B.$$ In which B consists of some chunks of matrices. How to efficently show by computation that $B =0$?

Answer 1

Hint. Differentiate with respect to $t$.

Answer 2

To proceed with direct calculation, first start with the relation $AM = -MA^T$ and show by induction that $A^n M = (-1)^n M(A^T)^n$ for all $n \ge 1$. With these identities at hand, we compute

\begin{align}\exp(At)M\exp(A^T t) &= \exp(At) \cdot \sum_{m = 0}^\infty \frac{M(A^T)^m t^m}{m!} \\ &= \exp(At) \cdot \sum_{m = 0}^\infty \frac{(-1)^m A^m M}{m!}t^m \\ &= \exp(At)\exp(-At)M\\ &= \exp(0t)M\\ &= M \end{align}