How to find $d\phi(I)$ where $\phi(A)=AA^T$

Question

I'm trying to show that $so(3) = \{A\in M(3; \mathbb{R}): A = -A^T \} $ is the lie algebra of $SO(3)$. For this i am using the following fact:"The tangent space at the identity to a Lie subgroup of $GL(n,\mathbb{R})$, endowed with the matrix commutator, is isomorphic to its Lie algebra". I defined $\phi:GL(3, \mathbb{R}) \to Sym$ (Sym is the subset of symmetric matrices) $\phi(A)=AA^T$. I want to check that the kernel of $d\phi(I): gl(3,R) → Sym $ is the subspace of skew-symmetric matrices in $Gl(3,R)$. I'm having trouble finding $d \phi (I)$, how do I proceed?

Answer 1

The derivative at a matrix $A$ applied to a matrix $H$ is, by definition $$ d\phi_A(H)=\lim_{t\rightarrow 0}\frac{(A+tH)(A+tH)^T-AA^T}{t} = HA^T+AH^T $$ In particular, putting $A=I$ we get $$ \ker d\phi_I=\{H:H+H^T=0\} $$

Answer 2

$\phi(A_0+tA)=(A_0+tA)(A_0+tA)^T=A_0A_0^T+t(A_0A^T+AA^T_0)+t^2AA^T)$ implies that $d\phi(A_0)(A)=A_0A^T+AA_0^T$, we deduce that $\phi(I)(A)=A+A^T$ and $d\phi(I)$ is the set of antisymmetric matrices.