Let $A_{nXn}(\mathbb{R})$ Skew-symmetric matrix $A=-A^t$ prove that $e^A(e^A)^t=I$
while:
$e^A=\sum_{i=0}^{\infty} \frac{A^n}{n!}$
I tried this:
$A=-A^t \Rightarrow A$ is Diagonalizable with orthogonal basis over
$\mathbb{C}$ $\Rightarrow A=PDP^*$and $P^*=P^{-1}$
and D is:
\begin{pmatrix}
\lambda_1 &0 & \cdots & 0 \\
0 & \lambda_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_n
\end{pmatrix}
we know that :$\lambda_i=0$ or $\lambda_i=ib$ for $b \in R$ because A is Skew-symmetric matrix.
so: $e^A=Pe^DP^t$
so if I will prove that $\lambda_i=0$ we will get $e^A=PP^t=I$. Is it possible?
Best Answer
We can see easily that the linear map
$$\mathcal M_n(\Bbb R)\rightarrow \mathcal M_n(\Bbb R),\quad A\mapsto A^t$$ is continuous hence we have
$$\left(e^A\right)^t=\left(\lim_{n\to\infty}\sum_{k=0}^n\frac{A^k}{k!}\right)^t=\lim_{n\to\infty}\left(\sum_{k=0}^n\frac{A^k}{k!}\right)^t=e^{A^t}=e^{-A}$$ and the result follows from the fact: if $AB=BA$ then $$e^Ae^B=e^{A+B}$$