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.