How to show that $G = \{X \in GL(n,\mathbb R): X^{t}AX=A\}$ is a Lie group?
$G = \{X \in GL(n,\mathbb R): X^{t}AX\}$ is the conjugation stabilizer of $A$, want to show that it is a Lie group.
I am having trouble showing that it is an embedded submanifold of $ GL(n,\mathbb R)$. I defined a map $f: GL(n,\mathbb R) \to GL(n,\mathbb R)$, $f(X) = X^{t}AX$, but I do not really think that the differential of this map will be surjective. How should deal with this problem?
Show that $G$ is closed by definition of your map. Then apply closed subgroup theorem i.e $G$ is a closed subgroup of $GL(n, \mathbb{C})$ which is a Lie group and so $G$ is a Lie group.