Finding the Lie Algebra of a Lie Group

Question

I am having a hard time finding the set of all $X \in M(n, \mathbb{R})$ such that $e^{tX^T}Be^{tX} = B$ for all $t \in \mathbb{R}$. where $b$ is any matrix in $M(n, \mathbb{R})$.

Answer 1

Hint: Differentiate both sides of $$e^{tX^T}Be^{tX} = B\tag{1}$$ with respect to $t$ and evaluate the derivatives at $t=0$, you get $$X^TB+BX=0.\tag{2}$$ So, all solutions to $(1)$ must be solutions to $(2)$. Conversely, by using the power series expansion of the matrix exponential function, show that whenever $(2)$ is satisfied, $(1)$ is satisfied too. Now equation $(2)$ can be rewritten as: $$\left[(B^T\otimes I)K+(I\otimes B)\right]\operatorname{vec}(X)=0,\tag{3}$$ where $\otimes$ denotes Kronecker product, $\operatorname{vec}(X)$ is the vectorisation of $X$ and $K$ is the commutation matrix. So, in principle, we can find all solutions to $(1)$ by solving the system of homogeneous linear equations $(3)$.