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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?

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1 Answer 1

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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 $$e^Ae^B=e^{A+B}$$

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  • $\begingroup$ actually it's looks amazing but we didn't learn that this map is continuous(doesn't part of the course). so I think that I need to prove it with more "algebraic" way. Do you have other suggestion ? what about the way I mentioned in the Question? tnx! $\endgroup$
    – GroundIns
    Jun 15, 2014 at 16:29
  • $\begingroup$ I didn't see an alternative answer. For the way you mentioned notice that the diagonalization of the skew-symmetric matrix is more complicated than you wrote. See this. $\endgroup$
    – user63181
    Jun 15, 2014 at 16:49
  • $\begingroup$ fixed it, now it is over C $\endgroup$
    – GroundIns
    Jun 15, 2014 at 16:52

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