I'm reading Lee's Introduction to Smooth Manifolds and trying to show that $$T_A {\rm O}(n)=\{B\in {\rm M}(n,\mathbb{R}):BA^T=-AB^T\}.$$
Here is my attempt about this question:
Take any smooth curve $\gamma(t)$ in ${\rm O}(n)$ satisfying $\gamma(0)=A$, then $$\gamma(t)\cdot\gamma(t)^T=I_n.$$ By taking the derivative of the above formula with respect to $t$, we can get $$\gamma'(t)\cdot\gamma(t)^T+\gamma(t)\cdot(\gamma(t)^T)'=0,$$ then $$\gamma'(0)\cdot\gamma(0)^T=-\gamma(0)\cdot(\gamma(t)^T)'|_{t=0}=-\gamma(0)\cdot\gamma'(0)^T.$$ Since $\gamma(0)=A$, we obtain that $\gamma'(0)$ satisfies $$\gamma'(0)\cdot A^T=-A\cdot\gamma'(0)^T,$$ and $\gamma'(0)\in T_A {\rm O}(n)$ is arbitrary, therefore $$T_A {\rm O}(n)\subset\{B\in {\rm M}(n,\mathbb{R}):BA^T=-AB^T\}.$$
Now it suffices to prove that $$\dim\{B\in {\rm M}(n,\mathbb{R}):BA^T=-(BA^T)^T\}=\frac{n(n-1)}{2}$$ where $A\in {\rm O}(n)$, because $\dim T_A {\rm O}(n)=\dim{\rm O}(n)=\frac{n(n-1)}{2}$, but I'm stuck here.
I know that the dimension of the vector space composed of all skew-symmetric matrices is $\frac{n(n-1)}{2}$, and the above set is very similar to the vector space composed of all skew-symmetric matrices in form, but I don't know what the connection between them is and how to connect them. Any help would be great appreciated!
While there is a comprehensive answer referenced in the comment by Anne Bauval, there is also a simple direct answer to your question: Let $A \in O(n)$ and define $$S_A :=\{B\in {\rm M}(n,\mathbb{R}):BA^T=-AB^T\}.$$ Then $B \in S_A$ iff $BA^T$ is skew-symmetric. Since $A^T$ is invertible, the map $B \mapsto BA^T$ preserves dimension.