Let $A$ and $B$ be real $n \times n$ matrices.

Show that $(A B)^{T}=B^{T} A^{T}$.

For any square matrix, the matrix exponential is defined by the series

$$e^{A}=I+\sum_{k=1}^{\infty} \frac{A^{k}}{k !}$$

Show that $\left(e^{A}\right)^{T}=e^{A^{T}}$. [You are not required to consider issues of convergence.]

Calculate, in terms of $A$ and $A^{T}$, the matrices $Q_{0}, Q_{1}$ and $Q_{2}$ in the series for the matrix product

$$e^{t A} e^{t A^{T}}=\sum_{k=0}^{\infty} Q_{k} t^{k}, \quad \text { where } t \in \mathbb{R}$$

Hence obtain a relation between $A$ and $A^{T}$ which necessarily holds if $e^{t A}$ is an orthogonal matrix.