Problem on the set of all orthogonal matrices If $\alpha : (-1,1) \to O(n, \mathbb{R})$ be any smooth map with $\alpha(0)=I$ then what can we say about $\alpha^{'}(0) ?$  Here $I$ is the identity matrix and $O(n,\mathbb{R})$ is the set of all $n \times n$ orthogonal matrices over $\mathbb{R}$.


*

*It is non-singular

*It is skew-symmetric

*It is singular

*It is symmetric.


If we take $\alpha(t) = \begin{pmatrix}
  \text{cos}t & \text{sin}t\\
 \text{ -sin}t & \text{cos}t
\end{pmatrix} 
$ then we see that $\alpha^{'}(0) $ is non-singular.
Again if we choose $\alpha(t) = I$ then $\alpha^{'}(0) $ is singular. Two different answer arises in two cases.
Where am I doing wrong ?
Any insight will be highly appreciated. Thank you.

You can say that $\alpha'(0)$ is ant-symmetric. You know that$$\bigl(\forall t\in(-1,1)\bigr):\alpha(t)^T.\alpha(t)=\operatorname{Id}_n.$$Therefore $\alpha'(0)^T.\alpha(0)+\alpha(0)^T.\alpha'(0)=0$. In other words, $\alpha'(0)^T+\alpha'(0)=0$, which means that $\alpha'(0)$ is anti-symmetric.
