Let $\phi: GL(3,\Bbb R)\to S$ defined by $\phi(A)=AA^t$ where $S$ is the subset of symmetric matrices. I have two questions:
Q1: prove that $\ker d\phi(A)=\{H\in \mathfrak{gl}(3,\Bbb R)|A^{-1}H\in \ker d\phi(I)\}$
I know that $d\phi(A)(H)=AH^t+HA^t$. Thus $\ker d\phi(A)=\{H\in \mathfrak{gl}(3,\Bbb R)|AH^t=-HA^t\}.$
Q2: Prove that $\mathfrak{so}(3)=\ker d\phi(I)=T_ISO(3).$
$\phi$ is a quadratic map and $d\phi_A(H)=HA^T+AH^T$. We deduce that $ker(d\phi_A)=\{H:HA^T+AH^T=0\}$ $ker(d\phi_I)=\{H:H+H^T=0\}$.
$HA^T+AH^T=0$, is equivalent to $A^{-1}(HA^T+AH^T)(A^{-1})^T=A^{-1}H+(A^{-1}H)^T=0$ and this is equivalent to saying that $A^{-1}H$ is an element of $ker(d\phi_I)$.