TITLE: $f'$ has maximal rank
QUESTION [5 upvotes]: I'm trying to prove that the set of orthogonal matrices $O(n)$ is a manifold. (We identify the set of matrices of order $n$ with $\Bbb R^{n^2}$) In order to do so, I must show $f'$ has maximal rank in $O(n)$, having already defined $f:\Bbb R^{n^2}\longrightarrow\Bbb R^{n^2}$ by $f(A)=AA^t-I_n$. 
How can I do that? I proved it with $n=2,3$, but I don't know how to generalize the argument. A hint is enough, since this is part of a problem set.

REPLY [2 votes]: Baron VT's answer is more appropriate for your manifold focus, but here is a direct proof anyway:
It is is easy to check directly that $Df(A)(\Delta) = L(\Delta) = A \Delta^T +\Delta A^T$.
Using the rank nullity theorem we can focus on $\ker L$ rather than $\operatorname{rk} L$
Note that $A$ is invertible, and so the map $\phi(\Delta) = \Delta A$ is invertible (in fact, $\phi^{-1}(\Delta) = \Delta A^T$ ).
Let $L'(\Delta) = L(\Delta A) = \Delta^T+\Delta$. We note that $\ker L'$ is exactly the set of skew-symmetric matrices, which has dimension $\frac{1}{2}n(n-1)$, and hence $\dim \ker L = \frac{1}{2}n(n-1)$ also.