Problem. Let $f:(-\epsilon, \epsilon)\to \mathbb{R}^{n^{2}}$ differentiable in $t=0$, with $f(0)=I_{n\times n}$ and $f'(0) = A$. If $f(t)$ is, for all $t$, an orthogonal, then $A$ is antisymmetric. If $\det f(t) = 1$ for all $t$ then the trace of $A$ is zero.
I already showed the first part. I have a small question: $\det f(t)$ couldn't be any constant?
In a previous question I prove that:
Claim. If $f: (-\epsilon,\epsilon) \to \mathbb{R}^{n^{2}}$ is a differentiable path, with $f(0)=I$ and $g: (-\epsilon, \epsilon) \to \mathbb{R}$ is defined by $g(t) = \det f(t)$, then $g'(0) = \mathrm{tr}(A)$ where $A = f'(0)$.
So, would not it suffice take $g(t) = 1$? I know that $f(t)$ is orthogonal, then $\det f(t) = \pm 1$, but for the second part, don't seems to me that the question requires $f(t)$ orthogonal.
Thanks for the advance!
Yes, the second part of the problem follows immediately from the claim. In the second part, you are not meant to assume that $f(t)$ is orthogonal: the assumptions of the two parts are meant to be separate. (Note that if $f(t)$ is orthogonal for all $t$ that actually immediately implies $\det f(t)=1$ for all $t$ by continuity, since the determinant of an orthogonal matrix can only be $1$ or $-1$ and $\det f(0)=1$.)