$\exp(A)$ is orthogonal iff $A$ is skew symmetric

Question

Let $A \in M_n(\mathbb R)$ and $\exp(A)=\sum_{i=0}^{\infty}\frac{A^i}{i!}$. How to prove $\exp(A)$ is an orthogonal matrix iff $A^t=-A$?

I can get that $\exp(A)$ is an orthogonal matrix iff $\exp(A^t)=\exp(-A)$, but how to get $A^t=-A$ from it?

What's more, is $\exp(A)$ injective or not?

Answer 1

The implication that $\exp(A)$ orthogonal would imply $A^\top=-A$ is simply wrong. Take a matrix for which $\exp(A)=I$, for instance $A=2\pi\pmatrix{0&-1\\1&0}$. Now that one is antisymmetric, but antisymmetry is not conserved under arbitrary conjugation, while being the identity is. For instance conjugating by a diagonal matrix with entries $(1,a)$ for some $a\neq0,1,-1$ gives a counterexample $$A'=2\pi\pmatrix{0&-a^{-1}\\a&0}:$$ it still has $\exp(A')=I$, but it is not antisymmetric.

Answer 2

Yet, the result is true over $\mathbb{Q}$.

Proposition. Assume that $A\in M_n(\mathbb{Q})$ and $e^A$ is orthogonal; then $A+A^T=0$.

Proof. Here $e^{A^T}=e^{-A}$.

Case 1. The spectrum of $A^T$ (or that of $A$) is $2i\pi$ congruence-free (for every $u,v\in spectrum(A),u-v\notin 2i\pi \mathbb{Z}^*$); according to a result of Hille, $A^TA=AA^T$, that is $A$ is normal. Moreover $e^{A+A^T}=I$ and $A+A^T$ is real symmetric, diagonalizable and has only real eigenvalues $(\lambda_i)$. Consequently, $e^{\lambda_i}=1,\lambda_i=0,A+A^T=0$ and we are done.

Case 2. Otherwise, let $K$ be the field-decomposition of the characteristic polynomial of $A$ -it is a finite algebraic extension of $\mathbb{Q}$-. There is a non-zero integer $p$ s.t. $2i\pi p\in K$; consequently $\pi^2\in K$, a contradiction.