Not sure about my proof that orthogonal matrices are a manifold in ${\rm Mat}_{n \times n}(\mathbb{R})$

Question

Not sure about my proof that orthogonal matrices are a manifold in ${\rm Mat}_{n \times n}(\mathbb{R})$

I know that the manifold is the zero set of the function $f(A) = AA^T - I_n$. The thing I don't know how to prove is that $D_f = 2A \neq 0$ , at least I'm pretty sure that it's true...

Any help would be welcome!

Answer 1

You have a map $$f:{\rm Mat}(n,\Bbb R)\to S(n,\Bbb R),\quad A\mapsto AA^T-I$$ where $S(n,\Bbb R)$ are the symmetric matrices. Now, the derivative $Df_A:{\rm Mat}(n,\Bbb R)\to S(n,\Bbb R)$ is given for a point $A\in f^{-1}(0)$ and a vector $V\in{\rm Mat}(n,\Bbb R)$, by $$Df_A(V)=\frac{d}{dt}\Big|_{t=0}f(A+tV)=\frac{d}{dt}\Big|_{t=0}((A+tV)(A^T+tV^T)-I)=VA^T+AV^T.$$ This is surjective since if $Y\in S(n,\Bbb R)$ and $A\in f^{-1}(0)$ then $$Df_A(\frac{1}{2}YA)=\frac{1}{2}YAA^T+\frac{1}{2}AA^TY^T=Y.$$ Hence, $0$ is a regular value of $f$, so $O(n)=f^{-1}(0)$ is a smooth manifold.