Proposition: Prove that orthogonal matrices $\mathrm O_n(\mathbb{R})$ is a regular submanifold of $\operatorname{GL}_n(\mathbb{R})$.
Proof: Denote $f:\operatorname{GL}_n(\mathbb{R})\to \operatorname{GL}_n(\mathbb{R}),A\mapsto AA^t-I_n$
We show that $\forall A \in \operatorname{GL}_n(\mathbb{R}) $the image of the tangent map $f_{*{A}}$ is all the $n\times n$ symmetric matrices.
Take a curve $\sigma:(-\varepsilon,\varepsilon)\rightarrow \operatorname{GL}_n(\mathbb{R}) $ satisfying $\sigma(0)=A$ then $\frac{d}{dt}|_{t=0}f(\sigma(t))=A^t\dot{\sigma}(0)+(\dot{\sigma}(0))^tA$,which is a symmetric matrix.
Conversely, given any symmetric matrix $B=C+C^t$, let $\sigma (t)=A+(CA^{-1})^t$, it is easy to check that $\sigma (0)=A$ and that $\frac{d}{dt}|_{t=0}f(\sigma(t))=B$. Hence we get $$\operatorname{rank}f=\operatorname{rank}f_{*{A}}$$
By Regular Value Theorem ,we know that $O_n(\mathbb{R})$ is a regular submanifold of $\operatorname{GL}_n(\mathbb{R})$ with dimension of $\frac{n^2-n}{2}$.
Question:I have no problem with the proof. My question here is that is there any other way to calculate the rank of $f$ without using $\operatorname{rank}f=\operatorname{rank}f_{*{A}}$ ? For example, use Kronecker product to explicitly show what $f\in \operatorname{map}(\mathbb{R}^{n^2},\mathbb{R}^{n^2})$ looks like.
Upgrade: I think I should restate my question. For example, define $T:GL_n(\mathbb{R})\rightarrow GL_n(\mathbb{R}),X\mapsto AX$, then under a basis $T$ has the form $A\otimes I_n$
My question is: does $T':X\mapsto XX^t$ have a similar form?